Understanding Mechanics for JEE (Main Advanced) Vol 1

Prévia do material em texto

Dr. P.K. Sharma
UNDERSTANDING
PHYSICS
MECHANICS
(Part-A)
ForllT-JEE&
all other Engineering Entrance Examinations
PRAKASH
PUBLICATIONS 
muzaffarnagar
■H
Dedicated to my father 
Shri Bipin Bihari Sharma and his family 
who are the part and parcel of 
Indian Teachers Community.
(Part-A)
2010
Published By:
$
J
For IIT-JEE
& all other Engineering Entrance Examinations
Dr. Pradeep Kumar Sharma
(M.Sc., Ph.D.)
UNDERSTANDING PHYSICS
MECHANICS
PRAKASH PUBLICATIONS
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□ First Edition : 2008
□ Second Edition : 2009
□ Reprint :2010
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PREFACE TO THE SECOND EDITION.
It goes without saying that a teacher is worth of millions of books. The teachers are the evolutes of 
the mental energy of potential students. A teacher without students is lame and a student without teachers 
is blind. The teachers and students are interrelated, interconnected and interdependent in the world of 
knowledge. The association of teachers and students chum the ocean of knowledge to explore the 
riddles of nature which forms the basis of our lives. It is not always expected to enjoy the personal 
presence of a teacher. In that time, an ideal book which personifies thousands of the brains of students 
and enviable experience and expertise of many teachers takes the credit of impersonal guidance to the 
students. Let me present my effort in this regard as following :
Expressing my deepest gratitude to my revered teachers and the directors of the organizations for 
giving me the chance to cultivate my knowledge, extending my love and sincere thanks to all my 
students, unhesitatingly I can proclaim that by the raft of unswerving devotion to teaching and philosophy, 
the unalloyed motivation and support of my teachers, students, colleagues and well wishers, I am able 
to present this text named as Understanding Physics. It has four parts; Part-1: Particle Mechanics; Part- 
2: Rigid Body Mechanics, Fluid Mechanics, Gravitation, Properties of Matter and Simple Harmonic 
Motion; Part-3: Waves, Optics, Heat and Thermodynamics; Part-4: Electromagnetism and Modem 
Physics. Each part has nearly ten chapters. Thus, the total General Physics is divided into forty chapters 
approximately.
Understanding Physics MECHANICS (Part-A), has ten chapters covering the fundamentals of 
particle mechanics. Chapter-1 deals with the Vectors. In Chapter-2, using the vector ideas, I explained 
the kinematics of particles moving in arbitrary curves. Using the ideas of general kinematics described 
in Chapter-2, four new chapters are developed such as Straight Line Motion, Relative Motion, Circular 
Motion and Projectile Motion. After completing particle kinematics in Chapter-6,1 introduced the concept 
of Force in Chapter-7 and explained the applications of Newton’s laws for a particle in various physical 
systems. In Chapter-8, I described the properties of frictional force and extended the applications of 
Newton’s laws for a particle. In Chapter-9, roles of different forces in circular motion of particles are 
explained. Finally, Chapter-10 describes the concepts of work, energy, power, work-energy theorem 
and law of conservation of energy, in particle mechanics.
In this second edition, I have done following modifications to make the book more accessible and 
understandable to the students.
a. The complicated examples are removed and modified.
b. The language is made more lucid.
c. Some tabulated theories and points are reduced and simplified.
d. An adequate number of introductory problems are added to make the student comfortable and 
confident before solving final set of problems.
e. Many conceptual examples are introduced for deeper theoretical understanding.
C hapter’of this book starts with a lucid introduction justifying the significance of the chapter 
in practical hfe and its relation with the previous chapters. The contents of each chapter are divided into
(Vi)
April, 2009 . Pradeep Kumar Sharma
well defined sections. Each section builds a new concept from the scratch. The student friendly discussion 
and labelled diagrams in the section make the thing easy to grasp. After the explanation in each section, 
the main points are highlighted for your quick reference. The mathematical formulae related with physical 
concept are derived from the basics. 1 emphasized on the physical interpretations of each derived result 
(formula) rather than just getting the answer. Sometimes micro-interpretations are provided to understand 
the subtlety of macroscopic phenomena. At the end of an example, I ask few questions as the student 
task for deep understanding of the example. Furthermore, some finer points of the examples are tabulated 
by critically analyzing them which may help the students in broadening their understanding. After studying 
all sections of a chapter, you are expected to learn different tricks and techniques of problem solving, 
with a suitable level of analytical ability.
Then you are advised to refine your understanding by doing the exercises. To make the book 
balanced with theories and problems, I have tried to put varieties of thought provoking questions in 
different forms such as discussion questions, MCQ, fill in the blanks, match the columns, true or false, 
assertion and reason, comprehensions and numerical problems. While doing the exercises, you may find 
some difficult ideas conveyed repeatedly through different types of questions. This may help you to 
remember them longer and understand them deeper. Some difficult sections may seem to be bit elaborative 
for the sake of understanding. Thus, an average student will enjoy a'methodical self study without an 
extra labor in browsing different sources.
At last, I strongly'suggest the students to read the book without skipping intermediate sections 
because all sections are interrelated and continuous. You should not try to design your own method of 
solution without mastering the concepts and methods discussed in the book. Always try to understand 
the underlying concepts rather than just finding an answer which is the prime basis of this book. You 
should not directly attack the problems without referring to the relevant sections. You may lose the 
confidence if you will get wrong answer. Always try to get an answer qualitatively in terms of the given 
parameters; then put the given values to get the numerical answer. In this process, not only you can save 
your time without repeated numerical calculations but also can check the relevant steps more easily. 
Even though some students are deprived of getting proper guidance, sincere reading of this book may 
guide them step by step to improve their analytical skills to the highest level. The prime motto of this 
book is to make the General Physics more interesting and accessible to those who think that physics is 
tough and rigorous. If my effort will stimulate your interest in General Physics, I think it as the greatest 
reward for me.
I extend my sincere thanks to Mr. Prakash Chand Bathla, Proprietor Prakash Publications for 
undertaking this publishing work. Furthermore, I am grateful to my colleague Mr. S.K. Singh to review 
the matter scrupulously.product: Since A • B = IA || B | cos0 =| B \\ A | cos© = B ■ A, a scalar product is 
commutative. :■
A«B
B
V v/
A«B=-BxA 
Fig.1.68
Z6>
Z x /
V/
following the above two orders of vector addition we
—> —> —► —~
have (A + B) + C = A + (B + C). This is known as associative law.
Distributive law
The following identities can be proved.
A(B+ C)=AB+A-C
-> -> ->
Ax(B + C) = AxB + AxC
-> -> -> ->
a(A + B) = aA+aB
 Retry the above example using the equation Ax BC = 0
3------ ‘ B
A*
Fig. 1.7$
Since Ax B is perpendicular to both A and B , if A , B and C are coplanar. Ax B must 
be perpendicular to C. Hence, AxBC = Q.
Basically Ax B C represents the volume of a parallelepiped formed by the vectors A , B 
and C . If the volume is zero, A , B and C must lie in the same plane.
§1.12 Derivative (or Differentiation) of a Vector (Optional)
Need of derivative of vectors: In the next chapter we will learn that, when a particle moves, 
generally its position vector ~r . velocity 7. acceleration a change with time. When a body rotates, in 
space generally its angular quantities like angular displacement, angular velocity © , angular acceleration
C = pA + qB 
where p and q are scalars.
—> —> —>
A, B and C are coplanar if any vector —> —> —> —>
C say, is given as C = pA + qB 
where p and q are scalars
Fig.1.70
•“> „ „ A A A • A A A
Ex. 13. If A - i + 2j - 3A. B = 3i - j + k and C = i + j-nk ,find “n "for the vectors to be coplanar.
Using C = pA+qB, substituting A, B and C, we have p(i+2J-3k)
+q(3i -j + k) = i + j-nk. Comparing the coefficients of i,j and 
k, we have, p + 3q = 1, 2 p - q = 1 and -3p + q = -n. Solving first
4 1
two equations, we have P = ~ and 9 = y- Substituting/? and q in third 
11equation, we have n = —.
Finding
dA 
dt
Vectors ! 31
—> —>
momentum L, torque t etc of the moving bodies may change with time. When a charge particle 
moves, electric and magnetic field intensities at any point, i.e., E and H respectively change. To find 
the nature of motion of a particle or any object we must account for the time rate of change of the all 
vectors associated with it. That is what we call derivative or differentiation of vectors.
Definition of derivative of a vector : If a vector A changes with time, A at any instant t can 
be given as Al = A(t). After a time interval Ar, it can be given as A2 = A(t + Ar). Then, the change of 
the vector during the time Ar is A/l = A(t + Ar) - A(t). Hence the change of vector A with respect to 
time can be given as
2(r + Ar)~ J(r)i- rlim —= hm
Az—>0 Ar Az—>o
d~A -»
We can write the above term as — called time derivative of A.
dt
—>
d A.You can call — “time rate of change of the vector A
dA dAv dA. -♦
where —-»—— and are the time derivatives of the magnitudes of the components of A 
dt dt dt
along x, y and z-axes respectively.
The derivative of a vector A with respect to time is given as 
dA ..A(t + At)-A(t) — = hm------------------
dt a/->o Ar
d~A dA—: We can find — using many coordinate systems. Here we mention the use of 
dt dt
—>
two coordinate system for evaluation of ~
—>
(i) In cartesian coordinate system A can be given as
—> A A *
A = Axi + Ayj + A-k
Since i,j and k are constant (fixed) unit vectors, their derivative is zero. Then, we have
dA, ? ^Ay *. dA. ' —Li +—-] + —-k, 
dt dt dt
dA, dA,
dt ' dt
Sol.
we have
Putting t = i, we have
Ans.
Student Task^
0
Since hi
» 
I
dt
dA
Jt
-----iZ.--------[dX]
A
increases
Fig. 1.73
= /’ — (KCOSK)_/ + 2(l)£ 
/=!
= i + nj + 2k
32 GR B- Understanding Physics MECHANICS (Part-A)
Ex. 14. The vector A varies with time as A = ti -sinittj + t2k. Find the derivative of the vector at 
t = \.
Differentiating both sides of A =ti - sinKt j + t2k
(ii) Any vector A at any instant t can be given as
A =| A | a ’ •••(*)
where a = unit vector of A at that instant. In general, / 
changes its magnitude and direction in space with respect to
—► 
time. Let us bring the tails of the vector A at different instants
(snap- shots), to a point 0, as shown in the Fig. 1.72. We can
—> A •
see that A , and a rotate in space. During a small time interval
A/, the change in vector is associated with both change in its length and 
orientation. While changing the angle of orientation of the vector through 
an angle A0, the magnitude (length) of the vector increases by A | X |;
—> —► —
we call AMI “component of A A along (parallel to) the vector a
(AJ)||=AM|o,we have
SA = (Av4)|| +(A A)^ 
Dividing both side by St. we have
—>
> In the above example, find the value of A*x — at t = \ 
dt
dA d~ . -- = -(„-S,„K,7 + U)
= i - kcosk/ j + 2tk
Ans. V2ti2 + I
^A(t)
AA = AAh + AA±
Fig. 1.72
Vectors 33
+
- (ii) 0
a = rate of change of magnitude of Awhere
dA 
dt
dA 
dt
AX
Ar
when A/ -> 0, we have
dA 
dt
d\A\ 
dt
AX
11
■ + 
>11
/
AX
changes along the line of action of A . Now we need to understand the second term which means the 
—>
rate of change in direction of A .
The direction of A is characterised by the unit vector a. Hence 
the change of direction of A is followed by a rotation (change of angle 
of orientation) of a. As the unit vector rotates through a small angle 80, 
its tip moves in a circle of radius unity. Then the change in a, that is, 
Aa during a small time interval has a magnitude equal to the “length of 
the small arc AB” which is given as
XB=|Aa| = 2|a(t)|siny=:80
l5*l 50zThen, —1 = —(=©)
8t dt
Since the small arc AB is tangential, the time derivative of the 
rotating unit vector a is
da _dQ~ 
dt dt
where Q is a unit vector perpendicular to A (or a) taken in 
the increasing direction of 0. It is evident from the figure 1.77 (i) 
dothat — is perpendicular to a and leads a by 90°. Similarly, we can 
dt
dB dQ. , dQ
p‘ dt dt dt
and leads 0 by 90°
G.R.B. Understanding Physics MECHANICS (Part-A)34
A
...(iii)
A
Since
we have
Hence
SinceSol.
we have
Using
we have
Since
we have
u V V
where — = co = the rate of change in angle of orientation of r .
Then, we have Ans.
dA 
.dtj±
d\r\r_ 
dt
dr
---- = t
dt
da
dt
(
^4
dr dr .d(f)
— = t— + r—— 
dt dt dt
dr_ 
dt
Let us now substitute A =| A | a from eq. (i) in 
eq. (ii), to obtain
dlAj 
dt
+ 
'll
X 
dA 
. dt J||
dA 
Ex. 15. A particle moves such that its position vector varies with time as r =
dt dt
tr. Find the velocity of
the particle which is defined as the time rate of change of position vector, if ^ = a). 
dt
—-
A 
leads a by^ radian
Fig. 1.77 (i)
d-^a, 
dt
dG
dt
dr
dt
dt ’
35
numerically.
Ans. r0o)ew/(r + r0e“'9)
dW = F-dr
The total work done W can be givenby integrating the elementary work dW. Then, 
W = jdW = jF-dr*
Vectors
Student Task^
> In the above example, if r -
§1.13 Integration of a vector (Optional)
Definition of time integral: When the time derivative of a vector A is given as B, we can 
write
B = —
dt—> —> , —>
If B = B(f) is known, the change in A during a time dt is given as
d A = Bdt
—>
Then, integral the total change of A during time t is
a2= fdAJo
Substituting dA = Bdt,
we have bA = ^Bdt
—> —> —>
If initial A is given as A$ , at any instant A can be given as
A = Aq + J B dt
where ^Bdt is called time integral of the vector B.
Let us discuss the method of evaluating the time integral.
Evaluating the integral : First of all choosing cartesian coordinate system, let us write
B = Bxi + Byj + B:k
Then integrating both'sides with time, we have
j Bdt = i jBxdt + j fBydt + k js.dt
Since i,j and k are constant unit vectors, we need to integrate the scalars Bx,By and B. with 
respect to time when Bx,By and B. are given as the function of time.
Space integral of a vector : Let us-take the example of work done. We know that, work done 
by a force F acting on a particle when it undergoes a displacement dr is
r^'r-fi^d 7 assuming
G.R.B. Understanding Physics MECHANICS (Part-A)
Substituting
This gives
we have f A -dI
By using the concept of dot product, we have
^A-dl = j(Axdx + Aydy + A.dz)
jBdA= fBxdAx + fBydAy + jB.dAz
V/e will explain how to evaluate j B • dA by using other coordinate system in later chapters.
| A d I = ^Axdx + IAydy + ^A.dz
which is the sum of three simple integrals where Ax,Ay and A. are the components of A along 
x, y and z-axes respectively.
Surface integral of a vector : Apart from the space integral we have surface integral of any 
vector.-For instance, flux of liquid passing through an area can be stated as = J v dA, where 7 = 
velocity of liquid. Similarly, electric current passing through any surface can be given as i = I J dA, 
where J = current density. Electric flux $e/ and magnetic flux passing through an area A can be 
defined in a similar way; fyel = J D • dA and Find the flux of liquid passing through the faces 1,2 and 3 of
® o’the cube if
Fig. 1.78
(0.1.0)
(O.1.1)Z-v = (i + 2j + 3k) m/s ■
= -4J
xdx+ £ydy + ^zdz
 = J
(1.1.0)
1.1)
/ |/(i.o.or)X
(0.0.1)v - dA = fvxdAx + ^vydAy + fv.dA.
Ans. 1 m3/s, 2 m3/s and 3 m3/s z
G.R.B. Understanding Physics MECHANICS (Part-A)38
Assignments
7.
12.
13.
14.
15.
16.
20.
8.
9.
10.
11.
18.
19.
1.
2.
3.
4.
5.
6.
Discussion Type Questions
Can the magnitude of sum of two vectors be greater than the magnitude of their difference?
Can we add or subtract 2 N force with a velocity of 3 m/s?
Can we multiply 2 N force and velocity of 3 m/s?
What is the minimum number of coplanar vectors required to give a zero resultant?
How many minimum non-coplanar vectors can give a zero resultant?
In polygon law of vectors, the closing side gives us the resultant vector. When 
one side of a polygon containing eight sides as shown in the figure crosses any 
of its side, then, how do you choose its closing side?
Component of any vector A along a line is given as | A | = J cos 6, where 0 is the angle be­
tween A and the line. Is it always correct? Explain.
If C = A + B, does it mean that C Z A and C>B?
If two vectors A and B give a resultant C, show that |/|-|B|(b) 0°
If A + B = B, A*B is:
(a)0
(c) pseudo vector
If A*B = 0 and Z?x? = 0, the angle between A and C is: 
n 
4
(b) 0
(d) unpredictable
n
(d>I
Multiple Choice Questions
---► AM*
If A = 0.5 j + 0.8 j + ck is dimensionless, the value of c is:
(a) 1 (b) VOJI (c)
If /i+B = Cand A = B + C, the angle between and B is:
71
(a) 0 (b) - (c) n
Which of the following vectors, when added, give a zero resultant ?
(a) 2 N, 3 N, 6 N (b) 1 N, 2 N, 3 N
1 
(c)2m/s,3N,4N (d) 2 N, 3 N, ~N
If A + B + C = 0, which of the following can be correct?
(a) A + B>C (b) A-BT4
L x ^1. Comment on these formulae.
AB
G.R.B. Understanding Physics MECHANICS (Part-A)
0 = tan 11
(a) 0 (b) n (c) -
If | A x B | = A • B, the angle between A and B is:
(a) 0 (b) j (c)4 4
If A-B = AB,{A + B)-(1-B) is:
(a) 0 (b) A1-B1 (q) A2\B‘
If A x B = 0 and ^-B= -AB, the angle between J? and $ is:
(c) 7t/2
The angle between A and B is given as 0 = cos
41
(b) 73 (d) 75(a) 2
12.
(c) negative (d) infinite
13.
14.
15.
(a) sin (b) sin
If ^-B = 0 and A-C = O, then A can be parallel to:16.
(c) 7xC (d) (fi + C)(b) C
17.
(a) a + b = 27 (b)
Q0(c) b
18.
19.
20.
21.
(a) B
If PR = RQ, then:
A
B
(b) Finite angle
(d) Moment of inertia
The sum of two unit vectors is a unit vector. Then the magnitude of their difference is:
1 
(c)
(d)
Vectors
11.
_> —> _ 
a - b = 2 c
a - b = c
A---- can be :
dt
(a) 1 (b) 0
Which of the following is/are vector/s?
(a) Pressure
(c) Cunent density
The vectors of equal magnitude have a resultant equal to either of them. The angle between them 
is:
(a) 60° (b) 120° (c) 90° (d) 0°
The resultant of the forces 7 and B is perpendicular to A. Then the angle between them is:
A
B B (d) a or signifies a vector.
(d) Pseudo vector has no direction.
42
22.
23.
(d) none of these(c) unpredictable
(T+fl)x(T-S) =24.
(d) none of these
25.
(c) I(b) -1(a) 0
An pass through a polygon of n sides,1.
2.
R : A, B and C are coplanar.
3.
4.
A : Two vectors A and B are such that, aA+bB = C.
Al+A2 + ...+ A„=0.
R : This is polygon law of vectors.
A : When “n” number of vectors Ah A2,
(a) 0 (b) A2-B2 (c) A2 + B2+2AB
If the vectors 2i + 3j + ck and -3i+6k are orthogonal, the value of c is:
Assertion-Reason Type Questions
(a) Both assertion A and reason R are true and R is the correct explanation of A.
(b) Both assertion and reason are true but R is not the correct explanation of A.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.-
(e) Both assertion and reason are false.
(d)l
A : If three vectors A, B and C represent the sides of a triangle, A + B+C = 0.
R:
A : If Z A, = 0 and \A, | = | 4/ |, the angle between two consecutive vectors representing a
f 2^regular polygon of n-sides is k I -- .
\ n J
R : Each vector subtends an angle — at the centre of the polygon.
n
GR.B. Understanding Physics MECHANICS (Part-A)
Two forces P and $ are acting at a point. If P is reversed, the new resultant becomes 
perpendicular to the initial resultant. Then:
(a) P = Q (b) P = 2Q (c) 2P = Q WP--QU
Referring to the Q. no-22, if P is doubled, the new resultant makes an angle cf with the initial 
resultant. If P = Q, —> —>
If A, B and C represent the sides of a triangle :
‘Of
T
(c) ||r/,x —>
and multiplication) of two vectors A and B acting at a point simultaneously.
-» /“>----- j------------ 7 , . . -i Bsin0I A ± B\ = >JA2 +2r ±2XScos0 and —
(d) 10
(ii) The integration of a vector A with respect to time t can be given as 
J Adt = i $Axdl + j jAydl + k fA:dt,
(b) (i-3/ + Jt)N-m
(d) (2/+3;-jt)N m
dA 
where —— = dt
where e = angle between A and B
Using the above concepts, answer the following questions:
Area of the quadrilateral is :
V194 
(al­
lot 9
A(t + &t)-A(t)
At
^4
d11
c/e-= —0 and 
dt
^4 
k dt Jy
dA— = hm 
dt
dt
Passage-2 (Differentiation and Integration of Vectors) (Optional)
Let us summarize the calculus of vectors as follows :
45Vectors
1.
e (b)
(C)
2.
If A = (2 + t2)a rotates with an angular velocity co , the angle between A and —, at t = 1 is:3.
dt
(b) tan(a) 90°
(d) tan
4.
(b) a*d = add
(d) alio
When a vector A varies with time as A = eli-tj + sin tk, the angle between the time derivative5.
(b) zero(a) cos
dA 
dt
1_
/3
co
3co
T
dA 
dt
(a) — a + 
dt
d\A\
dt
where Ax,Ay and A. are the function of time.
Referring to the above points, answer the following questions
a + A—0 
dt
(d) cos
If a unit vector a rotates in x-y plane, then : 
da . .(a) J
(c) —and a are of same nature
dt
of A and the vector A at t = 0, is:
2
73
• -i 1
(c) sm
(c) cot- co
The magnitude of a vector A changes at a constant rate a and the vector rotates with an 
angular speed co. At any instant t, the time derivative of the vector has magnitude :
(a) 7a2 + co2 (b) a + Ao
(c) 7?l2a2 +co2 (d) 7 a2 +co2/l2
dt dt
(d) +
dt dt
G.R.B. Understanding Physics MECHANICS (Part-A)
8.
9.
1.
y
c„
X
vector.
4.
5.
6.
2.
3.
5.
6.
7.
Pseudo-vectors are imaginary vector.
Zero-vector has no direction.
Current is a vector.
The subtraction of two vectors of equal magnitude is zero.
—> —> —> —► ,
If we take the derivative of a vector A , we get another vector 2?. Then A ■ B is zero.
Polar vector changes when we rotate the axes.
Scalar quantity is that which has same value for different observers (with different orientations 
of axes).
Vectors may vary with reference frames.
Sum of two negative vectors is a 
If A + B = A - B, then B is a vector.
If | A + B 1 = | A - B |, then A and B are vectors.
If A and B represent the diagonals of a parallelogram, the magnitude of area of the parallelogram 
is 
If A,B and C represent the position vectors of the vertices of a triangle, | Ax B + B xC + C x A\ 
is equal to
(O.b.O)y^O.O.C)
ko.o)*x
46
True Or False
1.
2.
3.
4.
Fill In The Blanks
The magnitude of sum of three vectors A,B and c representing the sides of a cube of length
A is equal to
. B
,! J ■ 
A.y
Introductory Exercise
Vector Addition and Subtraction
1. A block of dimensions ax frxc is placed on the ground. If the comer 
0 is taken as the origin of the coordinate system, find the vectors 
Ar?SenUng dia&onals and body diagonals passing through the 
vi I LI 11*
Vectors 47
2.
3.
4.
(iii) average acceleration oav
5.
6.
7.
8.
(-2/+y-3A) m/s.9.
10.
11.
v = -^3 i + j, find its
A particle moves in a circle with an angular velocity to = 2A relative 
to the centre of the circle. If the velocity of the particle at any instant is 
centripetal acceleration ar(= iox v).
—> a A —> a *
Two particles A and B having velocities VA = (/ - j) m/s and VB = (j - k) m/s are moving in 
space. Find the velocity of:
(i) A relative to B (= VAB =VA-VB)
(ii) B relative to A (= VBA = VB-VA)
Three forces F\=(i-J + A)N, F2 =(~‘+ ) - *)N and /^ = (i + j + A)N act on a particle. Find 
the resultant (net) force acting on the particle.
The components of a particles velocity along x, y and z axis are 2 m/s, 3 m/s, 4 m/s respectively. 
Find the velocity of the particle and its direction cosines.
Vector Multiplication
A force F = (/+_/ +A)N acts on a particle which moves from positions P(2, 1, -3) m to 
Q (1, -1, 2) m. Find the work done by the force F during the displacement 5 from P to Q, 
using the formula W = F ~s = F- (r2- n).
--- > A A ---->
A force F = (/-j)N acts on a particle which moves with a velocity v =
Find the power delivered by the force. Use the formula P = F-~v .
Water flows through the tube with a velocity v = 1 m/s. Find the 
rate of flow (flux ) of water flowing through the given area A =
2 m2. Use the formula $ = A-7.
A = 2mz a n
■> v = 1 m/s
Prove that (i) | A+ S| |/f|-|S(.
Two vectors a = i - j + lie and b =2i - j -k are given. Find (i) a+ b (ii) 2 a- b .
A particle changes its velocity from Pj = 2i-J + 3k m/s to V2 -i + j-k m/s during a time 
A/ = 10 s. Find the:
(i) average velocity (for constant acceleration assume pav - )
(ii) change in velocity a V(= V2 -pf)
= r2-Ki
At '
48
12.
13.
14.
x
15.
16.
19.
20.
21.
22.
23.
24.
25.
26.
dx 
dt
17.
18.
GR.B. Understanding Physics MECHANICS (Part-A)
----► A 
If angular acceleration of a particle in circular motion about its centre of revolution is a = 2k 
rad/s2 when its radius vector is r = (2i + 3 j)m , find its tangential acceleration at(= a * r).
An insect of mass m = I gm moves with a velocity v = i - 2 j + k m/s. Find its angular momentum
L about the origin 0 when the insect passes through the point P(-1, 1,2)m. Use = r v . 
An apple of mass m = 200 gm falls from a tree such that its position 
is at a distance x = 1 m and y = 2 m from a person O. Find the y|
torque t due to gravity F = mg acting on the apple about 0. Use
—> A A
t = r * F , where r = (i - 2J)m .
Find the scalar product of (A + B) and (2 A - 3 B).
Find the vector product of (2 A+ B) and (3 A- 4 B).
Find the unit vector perpendicular to r\ = i-j + k and r2 = -i +2j-i.
Find a unit vector that points from the origin towards a position half way between the tips of two
A * —► A A *
position vectors rj = 4/ + 2k and r2 = -f + 3j + 2k .
Find the unit vector that bisects the angle between the vectors j + 2k and 3/ - j + k .
Show that A* (B* C) = B(A-C)-C(A- B).
Show that A* (B- C) = AB*C is the volume of the parallelepiped determined by the vectors
A, B and C .
1 — - 2 ]'dt’ — = 3 at an instant 
dt
2m
g = 10m/s2
Find the angle between A and B using dot and cross product.
Find the length of projection of A = / - j + 2k along the vector B = -i + 2j + k.
—► A A *
Find the length of the perpendicular dropped from the tip of a vector A=-i + J — 2k onto 
another vector B = i - j + k when the vectors act at a point.
Find the unit vector of the displacement vector when a particle undergoes a displacement from 
the point A (2, 3, -2) to the point B (1, 2, 5).
Determine the component of the vector A =1 —3j + 2k along the line which passes through the 
points P (-1, 2, 3) and Q (1, -1, 2).
Vector Calculus (Differentiation and Integration)
27. Find the derivative of a vector A = xi - y2 j + z2k when 
when, x = l,y = 1 and z = 1.
49
28.
29.
30.
31.
32.
2 7.33.
34.
35.
■dx.
36.
1.
2.
3.
4.
5.
6.
Prove that 4m- B) = ^ ■ fi+ J ~ . 
dt dt dt
Vectors
If A = 2ti-t2J + t3k and B = t2i-tj + 4t k, find ^AA-B). 
dt
Prove that ^(a~B) = ?A 5. 
dt dt
—►
Prove that y(A x B) = x B + A x . 
dt dt dt
Verify the identities given in Q.29 and Q. 30. by taking A = 1 - j + k and ~B = -i + j-k.
If r =2cosl0r/ + 2sinl0/ j, find . 
dt dt2
If r = /tcosa t i + Bsina t j, prove that ^~- = -a 
dt2 
—►
If a vector A rotates with an angular speed co, find the time derivative of that vector.
If a force F = lai acts on a particle, find the work done IF by the force in displacing the particle 
from x = 0 to x = xt by using W =
If a force F = xi + y2j + Jzk acts on a particle which moves from P(l, 2, -3) to £?(-!, 1, 2), 
find the work done by the force from P to Q.
Problems
If A-i - j, B = j-k and C = k-i, find:
(a) 7 x(B?) (b) A x(B + C)
(c) Z(B-C) (d) Ax(BxC)
_» _> -> A B C
If A + B + C = 0, prove that —— = = , where a, B and Y are the angles betweensin a sinp siny
(B, (?),(/?, respectively.
If -2k and B = -3i +j + k, find the: 
—> —> ' —> _
(a) angle between A and B (b) projection of A along B
(c) projection of B along A (d) angle between rf+B and A-B
If n is a fixed vector, show that A = (A -n)n + (n^A)xn.
t AR
If $+ \A x B | = find the value of 0.
The resultant of two orthogonal vectors makes an angle $ with their difference. If magnitude of 
one is double that of the other, find .
8.
9.
i + 2 j-i and C = i + j + k represent the side of a parallelepiped, find10.
11.
12.
13.
14.
ANSWERS
20. (c)
4. d
2. (a, b) 3. (c) 4. (c)
3. (d) 4. (a) 5. (d)
3. (F) 4. (F) 5. (F) 6. (F) 9. (F)7. (T) 8. (T)
G.R.B. Understanding Physics MECHANICS (Part-A)
A particle is situated at the comer of a cube. If three forces of magnitudes a, b and c act on the 
particle along the diagonals of the faces of the cube, find the magnitude of another force to 
equilibriate the particle.
2. (a)
10. (b)
18. (d)
8. (d)
16. (c)
24. (d)
50
7.
4. (a,b,c,d) 5. (a, b)
12. (a,b,c) 13. (c)
21. (a,b)
If X=2ti + 3t2j-Pk, find the angle between 1 and ^~,dXt = I.
Prove that, the time derivative of any rotating vector of constant magnitude is perpendicular to 
that vector.
If X>=2i -j + k, B 
its volume.
Two vectors in which one has magnitude twice that of the other, act on a particle. Find the angle 
between them, if their resultant is perpendicular to the first vector.
If five consecutive sides of a regular hexagon represent five unit vectors acting in the same 
sense, find their resultant vector.
By vector method prove that, if the diagonals of a parallelogram intersect perpendicularly, then 
the parallelogram is a rhombus.
By using the concept of scalar product prove that
| A + B | = J|A\2+| B|2 +2| J || J|cos6
Multiple Choice Questions
3. (b)
11. (b)
19. (b)
1. (b)
9. (a)
17. (a)
25. (c)
Assertion-Reason Type Questions
1. e 2. a 3. c
Match the Column
1. a - p, q, s; b - p, q; c - p, r and q; d - r
2. a - p; b - r; c - s; d - p
3. a - r, s; b - p; c - q; d - p
Comprehensions
Passage-1
1. (a)
Passage-2
1. (c,d) 2. (d)
True or False
1. (F) 2. (F)
6. (b, c) 7. (b)
14. (b) 15. (d)
22. (a) 23. (d)
51
4. orthogonal 6. (2 x area of the triangle)null 5.3.
5.(i) i-2j + k (ii) -i + 2j-k
9.-3 watts
13. (5/+3/ + A)10-3 kg m/s11. -2/-2V3 j units
23. 25.19.17.
27. i — 4j + 4k 35.
(d)(c) 01.
(c) 75 (d) cos(b)3.
105°5.
♦ ♦♦
I
76
1 -» -A*B 
2
(~j + W
372
-Z + 3k 
TIo
i-j + lk
751
(b) 0
5_
Ti
4 
729 .
Problems
(a)0
Vectors
Fill in the Blanks
1. (73)/ 2. zero 
Introductory Exercise
1. (ai + bj),(bj + cA) and (ai + ck) 3. -j + 5k
k 2
2 1
* * * 2 3 t7. v =2/ +3y + 4A; / = -^=,= -^== andn =
15. 2A2-3B1-A-B
(a) cosl(~7rA
7. ^(a2 + b2 + c2 + ab + bc + ca)
3
729
General Kinematics of 
A Particle
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.1 1
2.12
chapter2
StudyQ^Points
INTRODUCTION
Position, displacement and distance 
Trajectory of a particle 
relation between elementary distance and elementary displacement 
Velocity and speed
Displacement and average velocity from instantaneous velocity 
Distance and average speed from instantaneous velocity 
Uniform and non-uniform motion 
Acceleration
Uniform and non-uniform acceleration 
Four General Equations
Use of polar coordinate system in curvilinear motion
§ 2.1 Introduction
Kinematics deals with the study of motion of an object ignoring the agents causing the motion. 
We have three types of motion; translational, rotational and vibrational. For instance, sliding of an object 
is translational motion, spinning of a cricket ball is rotational and to and fro (periodic) motion of a simple 
pendulum is oscillatory (or vibrational). All objects ranging from electrons to galaxies are in (relative) 
motion.
An element of an object is called particle which is mathematically expressed as a point. When an 
object undergoes any type of motion, all its particles (points) move in arbitrary paths. Hence, to analyse 
the motion of an object, we need to understand the motion of any point (particle) of that object. The 
study of motion of a particle moving in an arbitrary path is called curvilinear kinematics, or three 
dimensional motion, or motion in space, or kinematics of a particle in curved path. 1 call it “general 
kinematics of a particle” for the sake of simplicity.
In this chapter we will talk about the motion of a particle in a curve. First of all we will define the 
location of a particle by a vector r called position vector. Then we will find the time rate at which r 
changes to define velocity 7 of the particle. Finally we measured the time rate of change in velocity to
y
XX r*/M
QN
z
(c) Locating a particle, position vector : After plac­
ing the observer at the origin of the coordinate system, let us 
now enquire about the location of a particle P, say “moon”. 
Let us consider a baby as our observer. When we ask him 
where the moon is, he promptly replies by pointing his 
finger towards the moon. Even though the baby is igno­
rant of physics, by his instinct he conveys the idea that 
“location of moon is associated with a direction” by point­
ing his finger towards moon. Furthermore, the distance 
of the moon from the baby is also required to specify the 
location of moon completely.
From the above explanation we understand that since 
the position of a particle requires both magnitude (distance 
of separation between the object and observer) and 
direction for its complete specification, it is characterised 
by a vector quantity ~r , termed as “position vector” of 
the particle with respect to the origin (point of reference).
Hence, reference frame is geometrically defined as a 
coordinate system.
t 
y
z
Fig. 2.1 The observer (baby) locates the 
position of a particle P(moon) from the 
reference frame (origin 0 of the coordi­
nate system x-y-z)
Fig. 2.2 In cartesian coordinate system 
position vector r Is given as 
7 = xi + yj + zk
^0
The “particle” is a physical concept represented by a mathematical concept “point”.
(b) Frame of Reference: In particle kinematics we 
measure the displacement, velocity and acceleration of a 
particle. For this, an observer is taken as a reference point. 
Generally we consider the observer at the origin of a 
coordinate system.The place where the observer (reference 
point) is placed (fixed) is known as “reference frame”.
General Kinematics of a Particle 53
define acceleration of the particle. These three vectors ~r , 7 and o are sufficient to understand 
the nature of motion of a particle in space.
We will use the general kinematics to discuss different types of motion of particles such as, 
straight line motion, circular motion, projectile (parabolic) motion, planetary (elliptical) motion, simple 
harmonic motion, spiral motion and helical motion etc, encountered all along the physics.
§ 2.2 Position, Displacement and Distance
(a) Particle : Kinematics deals with the geometrical property of motion of a particle. A particle, 
ideally means a point mass. However in practice, a particle need not be a tiny (point) object. For 
instance, an electron can be regarded as a point object (particle) compared to an atom. An atom can be 
considered as a particle when compared with a grain of matter. The celestial bodies like stars and planets 
etc., can be practically assumed as particles when viewed from earth as they occupy negligible volumes 
compared to the space of the universe.
G.R.B. Understanding Physics MECHANICS (Part-A)54
Sol.
0.4m
4m
x
,8 m
Fig. 2.3
GO fab + bc + ca(c)
(c) >5oxlOm(b) 50 m
It is the first fundamental vector quantity in physics.
In cartesian coordinate system ~r can be expressed as
r = xi + y j + zk
where x, y and z are the components (projections) of ~r along x, y and z-axes respectively. For this, you 
may retrospect the section “component method” in chapter-1.
cm.
y
Ex. 1. Find the position of the centre of the wheel of a ceiling fan hanging at the centre of a room of 
dimensions 10mx8mx4m. The length of the rod {shaft) of the fan is 40 
For the sake of simplicity, let us consider one comer 0 at the 
bottom of the room such that length, height and breadth are 
oriented along +x,+y and +z-axes respectively. Since, the tip P 
of the centre of the fan is at a distance of 5 m along x-axis, 
3.6 m along y-axis and 4 m along z-axis, the coordinates of the 
point P are given as Ps(x = 5, y = 3.6 and z = 4). Generally 
this is written as P(5, 3.6, 4). Substituting the coordinates in 
r = xi + yj + zk, we have
The distance between object and observer gives the magnitude of position vector r . The 
position vector is directed from observer to the object. The direction of position vector r is 
given by F = —.
|r| 
5mIX, |Q i
I|/ppjx,y.z) |
5m
10m—*
(d) VJTxlOm
Ans. (d)
(d) Rest and motion : When a particle does not change its position, we say that the particle is 
stationary. In other words, when neither the magnitude nor the direction of position vector of a particle 
changes relative to an observer, it is said to be at rest relative to that observer.
Ans. z7 = (5z +3.6/ + 4£)m
Student Taskf
> The maximum length of a straight rigid rod that can stay inside an enclosed room of dimen­
sions axbxc is: a^c 
GO a+b+c (Z>) VaW+c2 GO
Ans. (b)
> The bottom of an electric pole of height 20 m is situated at a distance of 30 VJ w from you 
due 30° north of east. The magnitude of the position vector of a crow sitting on the top of the 
pole relative to you is: 
(a) 30 m
55
y
r2
X
'0
r\ = X|f + yj + zxk and r^ = x2i + y2j + z2k , we have
a7 = ^-^ = (x2-X|)Z+(y2-y1)/ + (z2-Z|)fr
Sol.
t = t'
Fig. 2.6
CT
(b) In circular motion, if we consider the centre of the circle 0 as the 
origin, the magnitude of r which is equal to the radius of the 
circle remains constant whereas the direction of 7 changes be­ 
cause r rotates about 0.
Fig. 2.5 
(y
"70’
General Kinematics of a Particle
Let us now discuss the motion of a particle.
(e) Displacement: Let the particle move from position 1 
to position 2 in an arbitrary curved path during a time interval 
Ar. The position vectors of the particle at the beginning and end 
of the time interval are known as initial position vector zf and 
final position vector r2 respectively. During the motion of par­
ticle, the position vector changes from r( to r2 during time Ar. 
Hence the motion of the particle is associated with a change in 
position vector which is given as
Ar = r2 -
z
Fig. 2.4 The particle^undergoes a 
displacement s (= A r ) while mov­
ing from position 1 to 2 changing its 
position vector r[tor2;A r = r2- r,
2^2(x2. y2. zz)
It is worth noting that, during the motion of a particle, either magnitude or direction or both 
magnitude and direction of its position vector change. It means, motion of a particle is defined 
as the change in its position vector.
Let us discuss how position vector changes during different types of motion.
Ex. 2. Describe the case where r changes the {a) magnitude only (5) direction only (c) both magni­
tude and direction. 01 = 0
(a) When we drop a stone from rest relative to its point of release O, the 
magnitude of position vector r increases whereas the direction of r - JI 
remains vertically downward; I
That means, r2 = rj + A r ; By following the triangle law of vector addition we observe that 
A r is directed from initial position I to final position 2. As position vector changes because of the 
displacement of the particle from point I to point 2, the change in position vector can also be termed as 
“displacement vector” denoted as s; A r = s . T
56
0
o
Fig. 2.8
y
Sol.
Q
0, x
Fig. 2.9
Ans.
A
In planetary motion, when we choose the centre of the 
sun as the origin, ~r changes in both magnitude and di­
rection.
G.R.B. Understanding Physics MECHANICS (Part-A)
’ f
----- -
p, 0. 0)
Fig. 2.7
t= Os*
y
(0. 2, 0)
TR 
5? 
rp
> An insect crawls from A to B where B is the centre of the 
rectangular slant face. Find the displacement of the insect.
Ans. V15m B \
z/(0,0,1)
Fig. 2.10
Ex. 3. In the Example I, let us consider a fly sitting at the centre of the 
fan. When it flies to the farthest point of the room relative to the 
origin, find the displacement of the fly.
We can approach this in two ways.
Method-1. Coordinate geometry tells us that, the point which is 
farthest apart from the origin O is Q. Its coordinates are given as 
Q = (10,4,8). Since, the fly flies from the point P s (5,3.6,4) the 
displacement of the fly can be given as
? = A r = (xq - xP )i + (yQ - yP )j + (zQ-zP )k, where
xq = 10, yQ = 4, Zq = 8, xP = 5, yP = 3.6 and zP = 4
This yields s*= (5i + 0.4/ + 4k) m . Ans.
Method-2. Alternately express the position vectors of P and Q as
ry> = (5/ + 3.6 j + 4k) m and = (1Or + 4j + 3k) m
Then substituting r^ and ry> in $(= Ar) = /g-rP and simplifying, we have 
? = (5i + 0.4 j + 4k) m.
Student Taskf
(c) In projectile motion, when the point of projection is taken as 
the origin, ~r changes in both magnitude and direction.
General Kinematics of a Particle 57
1
o .2
where
Then, we have
Sol.
Ans.
Student Task
(path)
0
> Find the magnitude of displacement of the tip of the second's hand of length 10 cm during
10 seconds. Ans. 10 cm
*2
Fig. 2.11 When r, and r2 are known, 
| A 71= 7ri2 +rz + 2r/2cos0 . where 
0 Is the angle between and r2
I Ar | tells us how far the particle moves from the initial position (point). Its magnitude may 
increase or decrease with time.
When the particle undergoes a displacement from position I to 
position 2, let its the position vectors /j and r2 make an angle 0 between * 
them. Then the magnitude of displacement can be given as
I?I=|a7|
I a7 |=|rj-n I
7 = 7r]2 + r2 ~ ^r\r2 cos®
Ex. 4. Find the magnitude of displacement of a particle moving in a circular path of radius R, when 
it revolves through an angle 0 relative to the centre of the circular path.
Let us consider the centre of the circular path as the origin O. Then
directly apply the above formula
| &~r | = ^r2 + r2 -2r(r2cos0
and substitute r}-r2-R. This yields
A
|Ar i=2/?sin^
O’
Fig. 2.13 Relative to two different fixed origins, 
r may be different but Ar must be equal
If we consider two different fixed reference points 
0 and O', the initial and final position vectors may be 
different but the change in position vector relative to 
each reference point remains the same. Following this 
vector diagram, we have
. -> —> —> 
Ar = r2-/j ='2“rl
If the particle undergoes the consecutive 
displacements s[, s^,---s^ during time intervals
Arbitrary 
path
Ar^xX
Fig. 2.12
Relative to the centre of the circle, magnitude of position vector ~r remains constant; | ~r |= R. 
Hence A|7|(=Ar) = 0. Since the direction of 7changes, |a7|*0.
t = 0
At2
=7
s = s1 + s2 + s3Ans.
z
Sol.
P
—♦
0 y (East)7
x (South)
Fig. 2.15
Ans.r =
(f) Distance : When we speak of distance, we should not mistake 
it as the displacement. These two are entirely different quantities. As 
explained earlier, displacement is a vector quantity. Its magnitude is equal 
to the “shortest distance” between the initial position I and final position 
2 of the particle (as the particle moves from position 1 to position 2 
following an arbitrary curved path as shown in the Fig. 2.16).
On the other hand, the “distance D" is the length of the actual path 
traversed by the particle. It is a scalar quantity.
Fig. 2.16 The particle moves 
from position 1 to 2 in a curve
\/45°
5 ?
—r=i + 
V2
■s3\\At3
/
58 G.R.B. Understanding Physics MECHANICS (Part-A)
A/|,Ar2, •••Ar„ respectively, the displacement 7of the 
particle during the time interval Ar = (At] + Ar2 + • • • + Ar„), 
-+ ,=n-> is equal to the vector sum of all displacements; $ = £ s( 
i=l
Then, the final position vector of the particle is 
7=^+7: substituting 7 we have
-+10 | j + 20k m
V2 J
Let us now apply the above concept.
Ex. 5. If you walk 1 Om due east, then 5m due south of east, finally climb a pillar of height 20m, find 
your position following a cartesian coordinate system. Take south in +x-direction and east in 
+y-direction.
Final position is given as 7=^+7, where = 0 
[because the particle (you) was at the origin] and 
total displacement 7 = ^+s^+s3, where 
-> --> Find the ratio — when the particle revolves the circle 2.5 times, referring the above illustration.
D 2
Ans. 7—Sn
RQ: That means D = RQ.
0
s = AB = 2/?sin —
2
e
2sin-
2
RQ
2/?sin-
2
If a particle moves unidirectionally, the distance covered by it will be equal to the magnitude 
of its displacement. That means D = \ s | for unidirectional motion, which is a special —>
case of rectilinear motion. However, in other cases D>\ s |. Then, more generally, 
we have
plane
§ 2.3 Trajectory of a Particle
(a) Definition: When we write by a pen (or pencil) on a paper, 
the tip of the pen leaves a clear mark on the paper. The mark appearing 
as a line (curve or straight) represents the path followed by the tip of 
the pen. This is what we call the trajectory of the moving point (tip of 
the pen).
Sometimes when we look at the sky we find a long thin line of // 
smoke. If we curiously try to trace the object causing the smoke track, // 
we can see a point like object, that is, a jet plane at the end of the track 
of smoke. It means, the points of space through which the jet plane 
passes are traced by the smoke of condensed water vapour. That is 
called “trajectory” of the jet.
Even though neither the jet plane nor the tip of the pen is ideally a point, following the assumption 
as discussed in the beginning of this chapter, we can practically treat them as points.
G.R.B. Understanding Physics MECHANICS (Part-A)60
X-
X
tX
Sol.
y y
X t0
\ p(x. y)
/ ---------
Z i
r y
i
0
y = f(x) represents the trajectory
0
x »f(t) represents position-time graph
When displacement of a particle alongy-axis is expressed as the function of displacement of the 
particle along x-axis we call it “locus equation” or “equation of trajectory”.
y = must not be confused with x (or y) = /(/). The former is the equation of trajectory 
and the latter is the equation showing the variation of position with time, i.e., position-time graph.
y x
y4
y3 
y? 
yi 
o
Fig. 2.20
Ex. 7. Explain the difference between locus equation and displacement-time equation of a stone dropped 
from rest.
When we drop a stone from rest, it moves down in a straight line. Hence the locus is a straight 
line.
Now, we can define trajectory as the path traced by a moving point.
(b) Equation of trajectory : For the sake of simplicity let us 
consider a particle moving in x-y plane. If x coordinates of all points of 
the path traced by the particle are related with the corresponding y- 
coordinates by a definite relation (mathematically known as function) 
such as y = x2, y = sinxetc., we call it “equation of trajectory”. In 
general, the function y = /(x) represents the path of a particle moving 
in x-y plane.
In other words we can state that, the equation that relates the 
position vector or displacement (but not distance) of a particle along x a particle in x-y plane can be given 
and y-axes by a simple relation, that is, “function” is known as “trajec- as y " 
tory equation”.
It means,
Fig. 2.21
If you plot the displacement versus time you will get a parabola.We call it displacement (or 
position)-time y-t graph which shows the variation of position with time. Ans.
*1 l2 *314
0
Fig. 2.19 A curved path traced by
as y = f(x)
61
Sol.
(i>) circular (c) parabolic
D
P
dD=\d~r1
As the elementary distance being a minute portion of any curved path, it behaves as a straight 
line; hence it is equal to the length of an elementary displacement; dD =| d71.
If x = g(f) and y = h(t'), eliminating t from these equations, we have y = f(x\ which is known 
as equation of trajectory.
(d) hyperbolic
Ans. (c)
y_
2
Hence the particle follows a circular path having the centre at (0,2) with a radius of 2 units, in 
x-y plane. Ans.
Student Task
> The position vector of a kite varies with time as r = ati + 0/2J. The nature of trajectory of 
the kite is:
(a) elliptical
2
= 1
O x
Fig. 2.22 Elementary distance is equal to 
the magnitude of elementary displacement; 
dD = | ds'|
r2
X2
---+ I -4 I
General Kinematics of a Particle
(c) Finding trajectory equation : Let us find the equation of trajectory of a particle whose 
position vector is given as r = xi + yj, where x and y are the functions of time; x = g(r) and y = h(t). 
First of all choose the simpler equation, say x = g(/), and solve it for time /. Then substitute the 
obtained value of t (as the function of x) in the other equation y = h(t) to express y = f(x)as the locus 
equation of the particle.
Ex. 8. A particle moves in x-y plane such that its position vector varies with time as
~r = (2sin3t)r+ 2(1 -cos3t)j. Find the equation of trajectory of the particle.
Comparing ~r = (2sin3/)z +2(I-cos3f)/ with r = xi + yj, we have x = 2sin3r and
y = 2(1 -cos3/). This gives sin3f = ^ and cos3t = l-—. Eliminating t by squaring and adding 
the above terms, we have
rj + dr
§ 2.4 Relation between Elementary Distance and Elementary Displacement
As discussed earlier, the magnitude of finite displace- dDj
ment is less than the corresponding distance so long as the 
particle moves in a curved path; | s | -+ r
j = Ar = lim E Ar( = 
A/f-»O,=1
On the other hand, the distance D (length of 
the path followed) covered by the particle between 
the points 1 and 2 can be measured by adding all 
elementary distances dD. Since dD =| d r |, the total 
distance D is given as
D= lim £ IA^|= 
Azf—>0 /=1
Following the above logic, we can split any length of 
path traced by a particle into elementary segments. Each 
elementary segment is represented by an elementary displace­
ment vector d? as a tiny arrow as shown in the Fig. 2.23. 
If we add all elementary displacements vectorially between 
any two points 1 and 2, we will obtain the total (net) dis­
placement ? between these points;
which tends to one. That means, dr -> dD .
§ 2.5 Velocity and Speed
(a) Introduction : When you need to reach the college quickly you drive your bike faster. When 
you press the accelerator of your car, the car moves faster. If you apply the brake, your car slows 
down. You can sense the fastness of your car if you see any outside fixed objects. How fast does your 
vehicle move away or towards (or around) a fixed point, can be described by “velocity” of the vehicle.
J'Jr 2
Fig. 2.23 Vectorial addition of elementary 
displacements dr results a net 
displacement s = Ar = jdr
dr—
.... D- • •
1 X.
• 0
| Ar | \d~r\ S,n2 
lim------- =-------- = lim —-—
o->o D dD o->o 0
2
Ex. 9. In circular motion, show that \ d r \-dD.
„ • 02sin-
—from the previous section, when both Ar and D are
zu
>2
The average velocity of a particle for any time interval is defined as the ratio of change in its 
position vector (displacement) in that time interval and the magnitude of the time taken.
^=«2
General Kinematics of a Particle 63
When you travel in space, generally you cannot see any outside objects; in that case we can see 
the speedometer of the aeroplane to know how fast the plane is moving. Hence, the fastness of an object 
is related with both “space and time”.
In general, to define the fastness of motion, we need to find how fast the space coordinates 
(position vector) of a particle relative to a fixed point changes with time. We call it velocity. Apart from 
this, we need to know the rate at which the particle changes the length of the path which gives us speed 
of the particle.
We will describe the velocity as the state of motion and speed as the measure of motion. We will 
use these terms to define many important physical quantities such as acceleration, momentum, force, 
kinetic energy, power, pressure, electric current, magnetic field, electromagnetic field energy etc. Hence, 
velocity and speed are the two most important quantities in physics. We will categorise the velocity and 
speed and define them as following:
(b) Average velocity : It is a general tendency of human beings to enquire about the fastness or 
frequency or rapidity of occurring of any event (or change of anything) in nature. How fast do we eat, 
we move, we run? How fast do temperature, pressure, volume or anything change with time? To find 
the fastness of any event, generally we divide the “total number of occurrences” with “total time”. This 
gives an “average value” which tells us something about the quickness of the event. For instance, in fifty 
overs of cricket, India scores 250 runs; on the average, the score is 5 runs per over. As an over is 
proportional to time we can call this average score as “run rate”. This average time rate scoring of 5 
runs per over helps us to analyse the game at the first glance. That means, the opposite team must play 
more rapidly to maintain an “average score (run) rate”,more than “five” to win the game. It is a well 
known fact that, the average run rate may not be equal to the actual run rate per over. Hence the word 
“average time rate” (velocity) is just a mathematical quantity, which gives us a rough idea about the 
fastness of a moving particle.
Let the particle P move through an arbitrary curve 
between two positions 1 and 2 at times t = and t = t2 
respectively. During the time interval A/(=r2 -fj) the 
position vector of the particle changes from to by 
a7(=t£-^). Hence, on the average, the position vec­
tor changes with time at a rate Al . FJa- 2.25_Theparticle undergoes a displacement
A/ s = Ar = r2 - r, during a time Interval At»t2 -1|
This is what we call the “average time rate of after covering a distance D In an arbitrary curve 
change in position vector”. This can also be termed as 
“average velocity" of the particle which is denoted as vav .
t = t, path
-» Ar r2-r,
VaV A/ '2~'l
G.R.B. Understanding Physics MECHANICS (Part-A)64
is parallel to A r*(= s), average velocity points in the direction of displacement.
t = 1st=o
Sol. t = 2s
t = 3s
t = 4s
t = 5s
Fig. 2.26
For any time interval,
Sol.
and D = RS
Thus, we define the average speed of a particle as the ratio of its total distance covered (but not 
displacement) and the total time taken by it.
—> —> —► —► —>
Let S], S2, S3, S4 and S5 be the displacement of the particle 
during 1st, 2nd, 3rd, 4th and 5th second of the time interval of 
At = 5 s. These displacements neither need be equal to each
D
Ex. 10. Let us assume that a particle undergoes a displacement S during A t = 5 seconds. Then the 
average velocity over 5 s is given as
S
A/ 5
Then, can you define the average velocity as the displacement of the particle (/) per second (ii) 
in each second? Explain.
other nor be equal to £. In otherwords,
The average velocity over five seconds need not be equal to 
the average velocity over each second.
(c) Average speed : If the particle is going a distance D during
D
the time Ar(= t2 - /|), the average rate of distance covered by it is — for this time interval. We can call 
this ratio, “average speed” denoted by uav.
As
^av _ D
I Vav I I A r |
Since D >| Ar |, we have uav > | vav |.
Ex. 11. Find the ratio of average speed and magnitude of average velocity of the tip of a rotating 
blade of a ceiling fan when it moves through an angle 0.
When the tip of the blade rotates through an angle 0
-> fl A
| A r | = AB = 20/1sin — = 22? sin —
2 2
jZ - A.
“ A,
Distance covered wav =--------------------
Time elapsed
General Kinematics of a Particle
Hence zee
This yields
Ans.
Fig. 2.27
tangent
V
Q" ft
P'r op • 0
i , , . 2/?sin-lvl = IALl =2 
uav D
(d) Instantaneous velocity :
(i) Purpose: Following the above idea of average velocity, when we say that a car moves with an 
average velocity of magnitude 60 km/hr over three hours, does it mean that the car moves for the entire 
three hours? Of course, it may not be true. In between, at some places the car may speed up, sometimes 
it slows down, sometimes stays at rest. In this way, the car covers a total distance
D = Average speed x time = 180 km
It means, the average speed (and velocity) cannot tell 
about the details of the actual motion of the car. This is 
possible only when we will record the speed of the car at 
each point of its path or each instant during three hours. 
Hence, we need to know the instantaneous value of the 
velocity and speed of a particle to understand its motion.
(ii) Meaning of an instant: The word “instant” lit­
erally means a short time interval. How short? Shorter is the 
time interval, more accurate the meaning of instant will be. 
For this, we can write Ar tends to zero; (A/ -> 0). In other 
words, “instant” may mean an infinitesimal time “dt" which 
replaces the term “ At —> 0 ”. This idea of “instant” was in- fhe ,"ltlal P°5l,l°n p- Eventually v, tends
. . , . , ■ , . to point along the tangent drawn at P In
troduced by Newton to find the instantaneous velocity of a the form of instantaneous velocity v 
particle. Let us proceed with this idea.
(iii) Definition: To find the velocity at any time t, first of all let us allow the particle to move for 
an additional time Ar and record the displacement A r from a point P. Then we will go on reducing the 
time interval Ar to zero. While doing so, we observe the displacement $*(= A ~r = PQ) gradually leaning 
towards the tangent drawn at P As we know, the average velocity v£v points in the direction of dis­
placement A 7 . Hence v?v tends to be tangential when Ar tends to zero. Eventually v^v assumes a 
limiting value (of magnitude and direction) which is known as “instantaneous velocity” at the point P as 
shown in the Fig. 2.28.
.......
o r n+
Fig. 2.28 As ths time Interval M tends to zero, 
the final position Q tends to coincide with 
the Initial position P. Eventually v^„ tends 
to point along the tangent drawn at P In
0
. 0
2
66
Hence
dr 
dt
G.R.B. Understanding Physics MECHANICS (Part-A)
Hence we can express instantaneous velocity "v as
r(r + AQ- r(r) 
Ar
dy 
any function y = f{t) we have &
This was adopted by both Leibnitz and Newton at the same time, but independently. Newton 
used the derivative to define the instantaneous velocity and acceleration while analysing the 
motion of a particle. This was the beginning of new branch of mathematics called “differential 
calculus’’. Newtonian notation of derivative of y is y, whereas Leibnitz expressed it as . We 
d , dt
can mention the derivative of any function f(t) as f\t); ~z/(0 = f (0>’ similarly time deriva- 
_> d -» -> a
tiveof r(t) can be given as —[r (r)] = r'(r).dt
With the help of differential calculus, instantaneous velocity can be defined as a rate of change 
in position vector.
-» d~r
V ~~dT
.. Ar r2-r}v = lim vflv = lim —— = lim ——L 
A/-»0 A/-+0 Ar ^2 — h
Then, we define the instantaneous velocity, that is, velocity at an instant as
The limiting value of average velocity v?v at that instant (Ar -* 0).
(iv) Velocity as derivative: If we have the records of all positions of the particle in terms of 
time, this will give us a function, r = f(t). Then we can mention = r*(r) and r2 = r(r + Ar). Now 
the above expression can be given as:
dr ->The new term —, is popularly known as derivative of r with respect to time r. However, for
/(r + Ar)-/(r) 
lim----------------------
A/—>0
br .. hm —— = lim 
a/-»o Ar A/->0
-> .. A 7 ~r (t + Ar) -~r(r)
v = hm —— = hm —------/-----—
az-»o Ar a/->o Ar
When Ar 0, Ar -» 0. For vanishingly small (infinitesimal or infinitely small) values of Ar and 
A r , they can be expressed as dt and d~r respectively. The terms dt and d r are known as differen­
tials of time r and position vector r respectively. Then we can write
67
7(t + At)
C
A7(t)
t0
1.
2.
Since
we have
dD 
dt ’
General Kinematics of a Particle
(v) Velocity as slope of r-t graph: We can see in the veloc­
ity-time graph as shown in the Fig. 2.29 that, for any time interval
dD u = -r- dt
(f) Relation between ~v and u : As discussed earlier, elementary distance and elementary dis­
placement are equal in magnitude; dD =| d s |. x—
Dividing both sides by dt, we have
Slope of the tangent drawn in r-t graph at any instant gives (defines) the velocity at that 
instant. In other words, instantaneous velocity is equal to the slope of velocity-time graph.
The above statement must be matched with the fact as discussed earlier that, instanta­
neous velocity is tangential to the path that is given by trajectory y = f(x) equation.
dD _\d~s\ 
dt ~ dt
Ar the average velocity is represented by tan 0. Hence
the slope of the line AB gives us the average velocity in r-t graph as 
shown in Fig. 2.29.
When we reduce the time interval gradually to zero, the straight 
line AB tendsto be a tangent at the time t. That means,
Ar -» dr hm —— = v = — 
A/-»o Ar dt 
of the tangent drawn at the time “r” under consideration.
(e) Instantaneous speed : As explained earlier, the distance covered by a particle increases 
gradually with time. Following this fact, during a small time interval, a particle covers a small distance.
Definition: If the time interval is infinitesimal, the distance covered by the particle will be also 
infinitesimal. The ratio of infinitesimal distance and infinitesimal time interval is a finite quantity. We call 
it “instantaneous speed” denoted by the letter u.
Speed as derivative: Instantaneous speed is given as
r AD*
u- hm ——
A/-»0 A/
.. adI™.— 
A/->0
At \
t t + At
can be represented by tan 0, that is, the slope Fig. 2.29 7 = lim
At-+# At at
■ slope of the tangent AC-tanO
G.R.B. Understanding Physics MECHANICS (Part-A)68
\ i t= | v |, we can write
m = | v |
Sol.
we have
Ans.
fl/
Instantaneous speed is equal to the magnitude of instantaneous velocity. It gives us an idea how 
fast the particle is moving. The speedometer of a vehicle measures the instantaneous speed.
7 = lim 4.9- 
Ar-»0
dt
n. dDSince —— = u and dt
d . . . d 1
—(sin0 At
and substituting r(t) = 4.9t2 and r(t + Ar) = 4.9(/ + At)2
(r + Ar)2-/2
Ar
4 (cos cor) = -eosin cat • ^-(ea') = ae‘ 
at dt
The derivative of various functions derived by using the basic formula
lim
dx Ax->0 Ax
is the origin of a new branch of mathematics known as “differential calculus”. The derivative 
of few functions as derived by the first principle are given below for your immediate applica­
tion in physics.
„ A .. At2 + 2rAr
= 4.9 lim --------------
Al-»0 At
= 4.9 lim (Ar + 2t) 
A/->0
= 4.9(0+ 2r)
= 9.8r
Then substituting t=2 we have v = 19.6 m/s
69
Sol.
At t = 0,
Ans.
where
0 = 7t/2
Ans.
(c)
.-1 25
' 7>/l4
Hence
That means,
Student Taskf
> A particle moves such that its position vector ~r varies with time according to the relation 
~r =2ti -3t2j + t3k.
(a) find the velocity at the end of first second.
(i) what is the angle between 7 and 7 at the end of first second?
Ans. (a) 2i-6j + 3k (b) cos
-> d? ,v = ——, we have dt
d * d *v = A—(cos co/) f +/—(sincor)y
= -/cosin coz i +/cocos cor j
v — -/cosin Qi + /co cos 0 j
= /co j
(b) The angle 0 between ~r and 7 can be given as
0 = COS-1_L^_
|7||7|’
7-7 = (/coscori +/sin coz j)-(-/eosin cor i +/co cos cor j)
= co/2 (- cos cor sin cor + sin cor cos cor)
= 0
0 = cos
General Kinematics of a Particle
Let us now use the above derivatives directly in the following examples.
Ex. 13. / particle moves so that its position vector varies with time as r = A cos cot i + A sin at f. Find 
the
(а) initial velocity of the particle.
(б) angle between the position vector and velocity of the particle at any time.
(c) speed at any instant.
(a) Substituting 7 = A cos cor / + A sin cor j in
v =| 71= ^(-/cosincoz)2 + (/cocos cor)2 = /co Ans.
As proved earlier, the trajectory of the particle is circular. As r is always perpendicular to 
7, we conclude that the origin (reference point) is the centre of the circular path. But be 
careful to conclude that, when 717 at any instant, the motion is circular. When we consider 
some other origin except the centre of the circle, 7- 7 cannot be zero.
G.R.B. Understanding Physics MECHANICS (Part-A)70
We summarize the above section as following:
Substituting d~r =~v dt, we have
$*= a7
Then, the average velocity can be given as v*v =
v v
t t0 ti
Fig. 2.30
The magnitude of instantaneous velocity is a finite quantity because ratio of magnitude of 
infinitesimal displacement to the infinitesimal time interval is finite. The instantaneous velocity is 
a measure of motion of a particle which tells us “how fast” and “in which direction” the 
particle moves. The instantaneous velocity is always tangential to the path followed by the 
particle.
0
Area under v-t graph (curve) gives 
the displacement s = j"vdt = A
s*(= a7) 
A/
= dt
! M I
-Idt I*- t2 
d7» vdt ■ dA
s* = a7 =
§ 2.6 Displacement and Average Velocity from Instantaneous Velocity
After defining the average and instantaneous velocity, let us see how they are related and how to 
find distance, displacement and average velocity from instantaneous velocity.
Let the particle undergo a tiny (elementary) displacement d~r during an elementary time dt.
Since v = , we have
dt
\‘^dt=2i!___
Z2 ~zl
d~r =~v dt
Displacement as integral : Summing up (integrating) all tiny displacements, we have the net 
displacement
Let us plot the variation of velocity as the function of time. Let the velocity of the particle be v at 
time t. Then during time dt, the elementary displacement vdt represents an elementary area dA of the 
thin rectangular shaded strip of v-t graph as shown in Fig. 2.30.
dA = v dt
71General Kinematics of a Particle
Sol.
Ans.we have
(b) Since
Ans.we have
W(a)
tt
Fig. 2.31
Then, j^'vdt can be represented as the summation of all elementary areas between and t2 
which is equal to A, say. Then, we have
Area under v-t curve 
Time interval
s* = 2 v dt = jdA = A
Ans.
Let us now check the fundamental concept of motion by verifying the term “variation of 
velocity with time”.
Ex. 15. The velocity of a particle varies with time as shown in the following diagrams {graphs').
v v
Ex. 14. The initial coordinates of a particle are 1, -2 and 3. If the velocity of the particle varies as 
~v = (2i -3tj)m/s, find the:
{a) displacement of the particle during third second of its motion.
{b) position vector of the particle at the end of third second of its motion.
(c) average velocity of the particle over third second of its motion.
(a) Directly applying the formula s* = p v dt and substituting 
J,i
v = 2/ - 3(/,Z] = 2 s, t2 = 3 s,
7(= a7) = 2^dt i ~3^t dt j = (2i - 7.5j)m
) ) A A *
Ar=r2-rt and rt = (i -2j + 3£)m ,
r^ = 7\ + A~r ={i - 2j + 3k) + (2i -7.5J) = (3/ -9.5j + 3£)m
f2 v dt
'2~'l
We note that s*= dt = area under v-t curve and i£v =
Substituting s*
(c) In substituting ?=(2f-7.5/)m and A/ = ls,wehave 
v£v = 2i. - 7 • 5 j m/s
in v«v = T-. we have8V At
P v dt
-» Jr,
G.R.B. Understanding Physics MECHANICS (Part-A)72
Sol.
vV
*2
tt tl ‘2t
Student Task^
Distance as integral
> The velocity of a particle varies with time as ~v = 3/ + (4 - 5t)j m/s, Find the average veloc­
ity oj the particle for a time interval between t = 0 and a time when the speed of the particle 
becomes minimum. Ans. (3/ + 2y)m/s
Give the comments on the above graphs.
In graph (a), at any time t we have two velocities which is not possible. Hence, this graph is 
invalid.
Fig. 2.32
In graph (b), at two times /, and /2 we have same velocity v except at the point P. Hence, this 
graph is valid.
We cannot have two velocities at an instant. That means a particle cannot 
move in two different directions with two different speeds simultaneously. 
However, we can have same velocity at two different instants, i.e., an oscillat­
ing particle can have same velocity after a time equal to the time period of 
oscillation Tas shown in the Fig. 2.33.
t = 0
—->■- v 
t = T
Fig. 2.33
£>=
Then average speed for the time interval Ar = t2 -is
__2!_____
t2 -/j
W,v =TAr
§ 2.7 Distance and Average Speed from Instantaneous Velocity
Referring to the previous section, the elementary distance covered by the particle during time dt 
is dD=fv\dt.
Then the total distance covered by the particle between time rj and t2 is the summation (integra­
tion) of the elementary distances.
D= pDjsince dD = \d71 = 17\dt
v (b)
General Kinematics of a Particle 73
Sol.
D
Then,
DThis yields Ans.
Ans.
vav = v for uniform velocity
At = t2 - tv Then Av" = v"2-v".
t(b) Changein velocity : Is it true that when a particle 
moves in one direction, its velocity will be constant? Not neces­
sarily true. The magnitude of velocity may change. For instance, 
a body released from rest in gravity speeds up while moving 
downwards.
When the particle moves in a curved path (such as riding
2 J
Ex. 16. The position vector of a particle varies with time as r =
l + /24fr.
----- |2
If both magnitude and direction of velocity of a particle remain constant, the particle is said to 
be moving with uniform velocity.
If the velocity of the car averaged over any instant is kept constant, obviously its velocity aver­
aged over any time interval during its motion must be uniform (constant), i.e., 7 = i£v. Hence the 
particle must move unidirectionally.
Fig. 2.34 A particle changes its velocity 
a vehicle in a curve, flying of birds and insects), the direction of from V, to V2 during a time interval 
motion must change even though the speed of the particle can 
be kept constant.
r2
'o
= ||r7biy+ln(t + 7i77)| =|[2>/5 + ln(>/5+2)]m
(b) Substituting the value of D in “av = , we have
2x/5 + ln(V5+2) .
«av=---------4---------m/s
§ 2.8 Uniform and Non-uniform Motion
(a) Uniform velocity : While driving a car if you maintain the speedometer reading constant we 
can say that the car is moving with uniform speed. Can you tell in which direction the car is moving by 
just looking at the speedometer? Obviously no. That means, the direction of motion is controlled by 
turning the steering wheel. If you do not turn the steering wheel, the car would move unidirectionally. In 
this situation we can declare that, the car is moving with constant velocity.
- t2/J --^j' w. Find the (a) distance 
covered by the particle during first two seconds, (b) speed of the particle averaged over first 
two seconds.
(a) The distance covered by the particle is given by
= P 171 dt where 7 = — = — | ti - 
dt dt ( 
t| = 0 s and /2 = 2 s.
D= Jo2|O*-(/)|^ =
Av = v2 - V]
(b)
Fig. 2.35 Two ways of subtracting v, from v2 by bringing the tails of v1 and v2 to a point. In Fig. 
(a) we join the tips of v, and v2 and in Fig. (b) we add -v, with v2 to obtain Av
(ii) Process of finding A 7: In the case as shown in the Fig. 2.35 we notice that, the tails of 
and are not situated at the same spot because points I and 2 are different. Then, how do you
subtract v^ from ? Recalling the process of vector subtraction, we have two methods. In first 
method, we bring the tails of the vectors vf and to the point I, by sliding While sliding, we must 
keep the orientation (direction) of v^ unchanged. Then, by joining the heads of vf and we obtain 
A7 as shown in the Fig. 2.35(a). In the second method we can subtract vj from by adding -vf 
(negative vector of v^) with v2 (by using parallelogram law of vector addition), as shown in the Fig 
2.35(b). Hence, the change in velocity, i.e., &v is basically a vector subtraction; A7 = i^-v^ where 
and are I*16 velocities at the beginning and end of the time interval Az respectively.
(iii) Magnitude and direction of A"v: The magnitude of A~v is given as
74 G.R.B. Understanding Physics MECHANICS (Part-A)
(i) Definition: When we throw a stone horizontally, it moves in a curve (parabola). We can see 
that the body speeds up with continuously changing its line of motion. This is an examples of variable 
velocity. Let us conclude as following:
If the velocity of a particle changes its magnitude or direction or both, it is said to be moving 
with non-uniform (or variable) velocity.
Now let us find the change in velocity of a particle.
Let us suppose that the particle changes its velocity from v^ to v^ during a time interval Az = z2 - z(. 
The change in velocity in this time interval is
| a7| = | vj-vf | = 7v]2 + v22 -2v]V2 cos6 » 
where 0 = angle between and v^. Following the above ideas we understand that, A "v is directed in 
the concave side of the curve and points in the direction of Vp
Let us summarize the above concept.
75General Kinematics of a Particle
Sol.
Q
Fig. 2.36
Fig. 2.37
Ans. (d)
0
'0 = 6O°i
Av = Vv,2 + v22 -2V]V2 cos0
*2
Ex. 17. A particle slides with a speed ofim/s at P. When it reaches Q, it 
acquires a speed of 4 m/s after describing an angle of 60° at O 
as shown in the Fig. 2.36. Find the change in velocity of the 
particle between P and Q. Assume that the path followed by the 
particle is circular from P to Q.
As the particle moves from P to Q, its speed increases from 
=3m/s tov2=4m/s and the angle described by the veloc­
ity vector is equal to 0 = 60°. Then the change in velocity is 
given as
When the velocity of a particle changes from v^ to v^, the change in velocity is A v v2 V]. 
This is directed in the concave side of the path followed by the particle.
| A71= Jv/ + v22 - 2vjv2 cos0, where 0 = angle between and you should not write 
I Av|=| v£ |-| i? I;rather it is |^-v^|.
p
Fig. 2.38
Substituting V| = 3 m/s, v2 = 4 m/s and 0 = 60° in the above expression, we have
Av = V32 +42 -2-3-4cos60°
This yields Av = V13 m/s and the change in velocity is directed as shown in the figure. Ans.
Student Taskf
> A particle is moved with constant speed v. Find the change in its 
velocity at the junction of two surfaces at P 
(a) 0 (b) v(l-cos0)
a s 0(c) 2vsin^ Ifr) 2vcos-j
G.R.B. Understanding Physics MECHANICS (Part-A)
Then,
Sol.
Ans.
Fig. 2.41
Fig. 2.40 The average acceleration a^v is directed 
in the direction of Av (=V2 -v^) towards the concave 
side of the trajectory of the particle
76
§ 2.9 Acceleration
After finding the change in velocity, let us enquire how the velocity changes in order to analyse 
the nature of motion of a particle. If we want to find how rapidly the velocity is changing, we have to 
adopt two terms, i.e., “rate of change in velocity over a finite and infinitesimal time interval”.
(a) Average Acceleration :
Definition: When the velocity of a particle changes, it is said to be accelerating. If a particle 
changes its velocity by A"v(= v^- v^) during a time interval A/, on the average, the particle changes its
velocity at a rate of This is what we call “average acceleration” denoted as .
Fig. 2.39 Velocity changes at 
, Av -♦ an average rate of — = alv
| A"v | = yjv2 +V22 -2VjV2 COS0
= vv2 + v2 -2vvcos0
= 2vsin^
2
-> Av _ vz-vf 
°8V Af t2-t}
We define the average acceleration of a particle between any two points as the ratio of change 
in its velocity and the time of its motion between these points.
Since a^ is parallel to A 7, average acceleration points in the direction of change in velocity.
Let us find the average acceleration when the direction of velocity of the particle keeps on 
changing while maintaining its magnitude constant.
Ex. 18. A motorcyclist describes an angle 9 in a circular path with a constant speed v. Find the (a) 
change in velocity of the motorcyclist, (d) average acceleration of the motorcyclist during the 
motion in the curve (circle).
(a) As the motorcyclist moves from P to Q, the velocity turns 
through an angle 0-. Then,
77General Kinematics of a Particle
Substituting | A~v | and A/, we have I aw I- Ans.
Student Task^
DA
(b) The time of motion is Ar = —. Then, average acceleration is
2vsin®
v
> In a square cut, the speed of the cricket ball changes from 30 m/s to 40 m/s during the time of 
its contact Ar = 0.08 s with the bat. If the ball is deflected by the bat through an angle of 
9 = 90°, find the magnitude of average acceleration of the ball during the square cut.
Ans. 102 m/s2
o
t? = lim o Af
2v2 . 0 
= ^es,n2
(b) Instantaneous acceleration :
(i) Purpose: If the particle has same velocity at any 
two points A and B, say, in its path, the change in its velocity 
is zero. Then the average acceleration of the particle over the 
time interval Ar is zero. Does it (I request all my readers to draw my attention to any kind of errors which will 
help me to revise the next edition. 1 will remain indebted to you for any sincere suggestion and positive 
criticism to improve the quality of the book.
PREFACE
It goes without saying that a teacher is worth of millions of books. The teachers are the evolutes of 
the mental energy of potential students. A teacher without students is lame and a student without teachers 
• is blind. The teachers and students are interrelated, interconnected and interdependent in the world of 
knowledge. The association of teachers and students chum the ocean of knowledge to explore the 
riddles of nature which forms the basis of our lives. It is not always expected to enjoy the personal 
presence of a teacher. In that time, an ideal book which personifies thousands of the brains of students 
and enviable experience and expertise of many teachers takes the credit of impersonal guidance to the 
students. Let me present my effort in this regard as following :
Expressing my deepest gratitude to my revered teachers and the directors of the organizations for 
giving me the chance to cultivate my knowledge, extending my love and sincere thanks to all my 
students, unhesitatingly I can proclaim that by the raft of unswerving devotion to teaching and philosophy, 
the unalloyed motivation and support of my teachers, students, colleagues and well wishers. I am able 
to present this text named as Understanding Physics. It has four parts; Part-1: Particle Mechanics; Part- 
2: Rigid Body Mechanics, Fluid Mechanics, Gravitation, Properties of Matter and Simple Harmonic 
Motion; Part-3: Waves, Optics, Heat and Thermodynamics; Part-4: Electromagnetism and Modem 
Physics. Each part has nearly ten chapters. Thus, the total General Physics is divided into forty chapters 
approximately.
Understanding Physics Mechanics Vol.-I (Part A), has ten chapters covering the fundamentals of 
particle mechanics. Chapter-1 deals with the Vectors. In Chapter-2, using the vector ideas, I explained 
the kinematics of particles moving in arbitrary curves. Using the ideas of general kinematics described 
in Chapter-2, four new chapters are developed such as Straight Line Motion, Relative Motion, Circular 
Motion and Projectile Motion. After completing particle kinematics in Chapter-6, 
I introduced the concept of Force in Chapter-7 and explained the applications of Newton’s laws for a 
particle in various physical systems. In Chapter-8, I described the properties of frictional force and 
extended the applications .of Newton’s laws for a particle. In Chapter-9, roles of different forces in 
circular motion of particles are explained. Finally, Chapter-10 describes the concepts of work, energy, 
power, work-energy theorem and law of conservation of energy, in particle mechanics.
Each chapter of this book starts with a lucid introduction justifying the significance of the chapter 
in practical life and its relation with the previous chapters. The contents of each chapter are divided into 
well defined sections. Each section builds a new concept from the scratch. The student friendly discussion 
and labelled diagrams in the section make the thing easy to grasp. After the explanation in each section, 
the main points are highlighted for your quick reference. The mathematical formulae related with physical 
conc.ept are derived from the basics. I emphasized on the physical interpretations of each derived result 
(formula) rather than just getting the answer. Sometimes micro-interpretations are provided to understand 
the subtlety of macroscopic phenomena. At the end of an example, I ask few questions as the student 
task for deep understanding of the example. Furthermore, some finer points of the examples are tabulated 
by critically analyzing them which may help the students in broadening their understanding. After studying 
all sections of a chapter, you are expected to learn different tricks and techniques of problem solving, 
with a suitable level of analytical ability.
(viii)
Pradeep Kumar Sharma20 June, 2008 
Hyderabad
Then you are advised to refine your understanding by doing the exercises. To make the book 
balanced with theories and problems, I have tried to put varieties of thought provoking questions in 
different forms such as discussion questions, MCQ, fill in the blanks, match the columns, true or false, 
assertion and reason, comprehensions and numerical problems. While doing the exercises, you may find 
some difficult ideas conveyed repeatedly through different types of questions. This may help you to 
remember them longer and understand them deeper. Some difficult sections may seem to be bit elaborative 
for the sake of understanding. Thus, an average student will enjoy a methodical self study without an 
extra labor in browsing different sources.
At last, I strongly suggest the students to read the book without skipping intermediate sections 
because all sections are interrelated and continuous. You should not try to design your own method of 
solution without mastering the concepts and methods discussed in the book. Always try to understand 
the underlying concepts rather than just finding an answer which is the prime basis of this book. You 
should not directly attack the problems without referring to the relevant sections. You may lose the 
confidence if you will get wrong answer. Always try to get an answer qualitatively in terms of the given 
parameters; then put the given values to get the numerical answer. In this process, not only you can save 
your time without repeated numerical calculations but also can check the relevant steps more easily. 
Even though some students are deprived of getting proper guidance, sincere reading of this book may 
guide them step by step to improve their analytical skills to the highest level. The prime motto of this 
book is to make the General Physics more interesting and accessible to those who think that physics is 
tough and rigorous. If my effort will stimulate your interest in General Physics, I think it as the greatest 
reward for me.
I extend my sincere thanks to Mr. Prakash Chand Bathla, Proprietor Prakash Publications for 
undertaking this publishing work. Furthermore, I am grateful to my colleague Mr. S.K. Singh to review 
the matter scrupulously. I request all my readers to draw my attention to any kind of errors which will 
help me to revise the next edition. I will remain indebted to you for any sincere suggestion and positive 
criticism to improve the quality of the book.
CONTENTS
Chapter -1
1-51
70
87 •
Chapter-3
102-153
102
Chapter -2
102
52-101
103
GENERAL KINEMATICS OF A 
PARTICLE
38
40
42
43
43
46
46
46
49
50
28
30
30
35
61
62
104
105
72
73
76
84
86
2.1 Introduction 52
2.2 Position, Displacement and Distance 53
2.3 Trajectory of a Particle 59
1
2
3
4
7
8
20
22
24
91
92
94
95
95
98
98
98
99
100
VECTORS
1.1 Introduction
1.2 Definition of Vectors
1.3 Representation of Vectors
1.4 Types of Vectors
1.5 Multiplication of a Vector by a Scalar
1.6 Addition of Vectors
1.7 Subtraction of Vectors
1.8 Scalar Product of Two Vectors
1.9 Vector Product of Two Vectors
1.10 Some Fundamental Laws of Vector 
Algebra
1.11 Condition for Coplanar Vectors
1.12 Derivative (or Differentiation) of a 
Vector (Optional)
1.13 Integration of a Vector (Optional)
Assignments
O Discussion Type Questions
O Multiple Choice Questions
O Assertion-Reason Type Questions
O Match the Column
O Comprehensions
O True or False
O Fill in the Blanks
O Introductory Exercise
O Problems
O Answers
STRAIGHT LINE MOTION
3.1 Introduction
3.2 Sign Conventions for Straight Line 
Motion
3.3 Kinematical Equations for Straight 
Line Motion
3.4 Dependence of Acceleration with 
Distance, Time and Velocity
3.5 Use of Kinematical Equations
2.4 Relation between Elementary Distance 
and Elementary Displacement
2.5 Velocity and Speed
2.6 Displacement and Average Velocity 
from Instantaneous Velocity
2.7 Distance and Averageand the corresponding time interval, changes from time to time between A and b 
Hence aw cannot explain the change (variation) of velocity 
of the particle during any instant (short time) in the given 
finite time interval. For this, we need to establish the idea of instantaneous acceleration.
(ii) Definition: Let us assume that we know the variation of velocity with time as the function 
v = /(/). Hence we can write = "v(r) and = "v(r +Ar). Then, the average acceleration is
-> "v (r + Ar) - 7 (r) 
flav" Ar " Ar
If we want to find the rate of change in velocity over a short time (an instant), we must reduce 
the time interval gradually to zero, (Ar -> 0). In consequence, A ~v tends to zero (A 7 -> 0). Then, the 
ratio “ 2*m0^7' ” atta'ns a limiting magnitude and direction. We can term it as “instantaneous acceleration” 
denoted as a.
B(t + At)
v
Fig. 2.42 The velocity of the particle
but the net change in velocity between A 
and B is zero
78
Since
(iv) Graphical significance of ~a : Referring to the v-t graph, average acceleration
B
v
t0
Fig. 2.43
d~v 
dt
The instantaneous acceleration (or acceleration at any instant) is defined as the rate of change 
in velocity with respect to time at that instant.
I 
I 
I 
I
/ /•// / 1 \
-> dt P i
I 
I
I
I*-At—I
-» _ A~v
I
1
G.R.B. Understanding Physics MECHANICS (Part-A)
If we write lim AV = d~v (infinitesimal change in velocity) and Ar->0 as dt (infinitesimal 
M-»0 .
time), then an awkward looking term lim =^~ can be expressed as a simple term That means, 
a/-»o Ar dt
“ jLl ” can be termed as instantaneous rate of change in velocity. Hence,
Fig. 2.44 Instantaneous acceleration 
is equal to the slope of the tangent of 
v-t graph, a = 4^-= tan0 = slope of 
dt 
the tangent
lim ^1 = 
A/->0 Ar
Let us now find the magnitude and direction of instantaneous acceleration a.
(iii) Magnitude of : The instantaneous acceleration is given as the ratio of two infinitesimal 
(elementary) quantities d~v and dt. Hence its magnitude is finite. It means, a= lim —attains a 
a/-»o Ar
limiting (finite) value which can be expressed as
dt
lim dt a/->0 A/
Hence we define the instantaneous acceleration a as the limiting value of average acceleration 
at an instant (Ar -> 0). Thus, acceleration at an instant is defined as the derivative of 
velocity with respect to time, at that instant. Mathematically, we have
c/ = lim c?av = lim - ~~
At->o a/-»o Ar dt 
79
acceleration a
Physical concept 
particle 
velocity ~y
“ ” is known as
dt
derivative or differentiation of velocity with time. The idea of derivative (or differentiation) was invented
General Kinematics of a Particle
which is equal to tan 4>, that is, the slope of the straight line AB as shown in Fig. 2.43. If we go on 
reducing the time interval Ar gradually to zero, the straight line AB tends to be a tangent AC at the point 
of time / (under consideration). Hence = = tan® w^'ch *s the slope of the tangent
AC.
The rate of change in velocity at an instant, that is, “instantaneous acceleration” is given by the
slope of velocity-time graph. In general, slope = = tan 0 is a finite quantity which tells us
j \ dt J
that I a I is finite. The slope can be +ve, 0 and -ve.
Now, we have the following basic equation to find the slope of v-t graph.
ni dv i- A^ !• v(r + Ar)—v(r) tan 0 = slope=-V-= hm —— = lim —------—---- —
dt az->o Ar az-»o Ar
The slope of v-t graph gives us the rate of change in velocity. The term
and rate of change in velocity
moving termed as “velocity” V 
termed as “acceleration ~a ”,
Recapitulating, we have the following points:
1. Mathematical concept------- > Represents
(a) - (point) ♦
(b) Derivative of ~r ; ♦
dt
(c) Derivative of "v; ♦dt
2. Derivative of a vector is also a vector.
3. In general, we have derivative of y = f(x) with x is given as
— = lim — = lim which was invented by both Newton and
dx ax-»o Ax ax->o Ax
Leibnitz independently, to find the change of a quantity with respect to another quantity.
(v) Direction of : From the previous discussion we understand that a is equal to the slope of 
the tangent of v-t graph. Does it mean that ~a is tangential to the path? Let us see. To define the 
instantaneous acceleration when we go on reducing the time interval, we notice that the change in
by Newton to study the motion of a particle to measure the rate of change in position vector ? , i.e.,
v , i.e., d-Y_(='a). explains how fast the particle is
J —> 
and =-¥- explains how fast the velocity of the particle is changing 
dt
a*
P
o
This tells us:
1.
2. 0
? -v
Fig. 2.47
Alternate procedure:
Alternately, we can explain the above as follows :• 
As 7 is tangential, we can write
When both magnitude and direction of velocity of a particle change 
continuously, its acceleration at any position will not be tangential. 
However, in straight line motion a± will vanish and in uniform cur­
vilinear motion fl|| will vanish.
P
T(t)/
7(t +At)
7 =| v | x >
a'= af,
Fig. 2.46
Dividing both sides by the time interval At, we have
-> _A v _Avr Ayj
" At " At + At
(vi) Component of ~a : Since the time interval At tends to zero gradually, we have
a = lim = lim —- + lim
A/->0 Ar->0 At A/->0 At
Substituting lim = a. and lim —— = a,, we have
A/->0 At " A/-»0 At
7 = ®|i+
80 G.R.B. Understanding Physics MECHANICS (Part-A)
velocity, A 7 attains a limiting direction which is not tangential as shown in the Fig. 2.45. Hence, the 
instantaneous acceleration is not tangential to the curved path. If we consider a very short time interval 
At, we can imagine the points P and Q very close to each other. This idea helps us to just join the tips 
of the velocities 7(/) and ^(r + A/) in order to obtain the change in velocity A 7. We can observe that 
A 7 is not tangential (directed in the line of motion) at P. Hence A 7 must have a component Avy 
along the tangent (line of motion) and another component Av^ along the normal (Ito the line of 
motion). Then, we can write A v = Avy+Av^.
7 (t) a7
atX
0
Fig. 2.45 When At-»0 the limiting direction of change in velocity a7is neither tangential 
nor normal to the line of motion. It is directed to the concave side of curvilinear motion.
Hence, a = lim -tt- is not tangential in a curved path 
At-»o “
^7
81
substituting V =| v | i, we have
dO
e
tA
Hence dt
7
0"
A/->0
Ans.
dv 
dt ’
din
a*
The direction of a
Z di
Ex. 19. Find the instantaneous acceleration in uniform circular motion.
Sol. From the previous example, we have
General Kinematics of a Particle
where t is the unit vector of ~v .
(vii) Normal and Tangential acceleration. Hence a =-^- 
—X k
a = —(| v |t). This yields
3=2121
di 
dt
Substituting | dx | = | t| d& we have —■ =| i | = 0, 0 -> 0.
■ 9 - 2 • 6sin^ 2v sin- 2
Since Iim —= 1, we have ~a = lim af = lim — .
0.0 6 zv->o av a/->o RQ R
2 ? 0-*°r 
is radially inward because a = is perpendicular to ~v .
If the line of motion 7(orr) of the particle turns an 
elementary angle dQ during an elementary line dt, the change 
in unit vector, that is, dx is shown in the Fig. 2.48. Here we 
observe that dx1 i. Since, t is tangential, dx must be directed Fig. 2.48 dt is perpendicular to v (or x) 
normal to the line of motion. and directed normally (radially) inward
0
2
82
-*
(•) (ii)P
y»
x
Sol.
zThis gives Kid
Then Fig. 2.51
Ans.
G.R.B. Understanding Physics MECHANICS (Part-A)
Hence a is known as “centripetal acceleration”. We will discuss more about it in chapter-5. 
Let us tabulate the following points in the foregoing example:
I 
I
I 
I
I 
I
~a = -0.4n2 sin-^f -0.4k2 cos^ j2
At time f = ys,
After simplification we have a = (-3.464/ +2J) m/s2.
The tails of v, v' and 7" are brought to the point 
P a7 tends to be normal to 7 when At -» 0
1. As Ar -> 0 ; A 7 attains a limiting angle; so that it will be perpendicular to ~v . That means, 
lim ^-(= 7) ± 7.
A/->0 Ar
2. Since 7 is tangential, 7 must be normal and directed inward (towards the centre 0 of 
curvature). Hence ~a is known as “centripetal acceleration”.
3. The tails of the velocity vectors 7,7' and 7’ are brought to a point P. We see that, when 
Af -> 0, Av tends to be perpendicular to 7. As 7 is tangential, A 7 tends to be normal 
and radially inward.
Ex. 20. In an amusement park ride a kid slides down the helical track. The position vector of the kid 
changes with time as
~r = (0.4 sin nt i + 0.4 cos nt j - 0.3nt) m
Determine the velocity and acceleration of the kid after t = ^s. 
The velocity of the kid is given by
7 = ^~ = ^(0.4sin nt i +0.4cos7t/ j-Q.3tk).
7 = 0.4k cos k// - 0.4ksin ntj - 0.3fc
a = ~ = -0.4k2 sin nti - 0.4k2 cos ntj.
^7 v 
v" 7
Three different velocities 7, 7 and 7" 
of the particle moving in the circular path
Fig. 2.50
2
3
83
Ans. 1.3 m/s
(i) Since ~a = aav =Sol.
, where kt{ - aj and kt2 = a^This gives
Ans.Then,
(ii) Since a = ic,
Ans.This gives
k sin co/ where T = —. It gives a[v =^~ 
co n
Ans.(iii) T
(Z>) In which direction does the kid accelerate relative to its line of motion at t = 0 ?
Ans. Kid moves perpendicular to acceleration.
General Kinematics of a Particle
StudentTaskf
('2-a)
(c) Calculation of aav from a :
Let the variation of acceleration as the function of time be given as a = /(/).
During an elementary time dt, the change in velocity of the particle is
d~v = lx dt
Summing up (integrating) both sides, the change in velocity (a7 ) du.ing time interval 
A/ = (/2 - /|) can be given as
r(/22-/,2) 
2(/2-/,)
= f'2 ci dt
aav
A"v = pcf"v 
JV|
A v
A/ ’ —
> Referring to the above example’.
{a) What is the average speed of the kid till he descends through a distance of 20 ml
Substituting A"v in aL = , we have
A/ -----------------
f > \‘2adt 
-> J/,
aav . .
_____z2 ~'l
Ex. 21. Find the average acceleration of a particle during a time measured from starting (t = 0), if 
(i) a=lct (0) a = lc
(iii) a - k sin tot, where k is a constant vector.
PaL Jfi____
Z2 -/l
£/,+£/, 
2
• 4=^
^iv=~tLT 
2
aav = k = a
G.R.B. Understanding Physics MECHANICS (Part-A)
? \[t. The average acceleration of the particle
(d) zero
Ans. (c)
~adt
T
t t
The area under a-t curve gives us the change in velocity (but not velocity).
84
Student Task^
§ 2.10 Uniform and Non-uniform Acceleration
(a) Uniform acceleration : If both magnitude and direction of acceleration of a particle remain 
unchanged, we can say that the particle is moving with uniform or constant acceleration.
When a particle moves with constant acceleration, its velocity changes linearly with time during 
any time interval (finite and infinitesimal). Hence, average acceleration and instantaneous acceleration of 
a particle are same in uniformly accelerated motion.
= = A
2flmax
7t _____________________________________
Now, let us discuss the graphical significance of j'adt.
(d) Graphical representation of j a dt: Since, a dt represents the area dA of the elementary 
strip (shaded), fadt will represent the total area A under a-t curve between the time interval ^t-t2-t\.
Av = |
Fig. 2.52 Area under a-t curve gives the change In velocity; A 7 = Jl?dt = A
(0) 4
2
_ f_
— t —H dt h—
> A particle moves with an acceleration ~a = 
during a time interval starting from the start (t = 0) is :
(A) A (c)
We can derive the following points from the above example :
1. For constant acceleration a,aay/ = a for any time interval.
a>+at
2. For an acceleration varying linearly with time, aav = —r-2-, but we can not use this for any 
other function.
3. For sinusoidal (sin or cos) variation, aiv =
General Kinematics of a Particle 85
Sol.
Then,
•••(')
The average acceleration
where
-(ii)This gives
From eqs. (i) and (ii), we have Ans.
a - aw for uniformly accelerated motion. During uniformly accelerated motion of a particle, 
velocity changes with constant rate in one direction for any time interval. However, the particle 
need not move in a straight line.
The following example resembles the above fact.
? 5-5
The velocity of the particle is 7 =
v[ = ai + (6-c/])j, v^-ai +(b-ctf)j
aav =—:—~ =
*2
-> —>a = aav •
Even though the particle moves in a curve y = — -^7 which is a parabola, its acceleration 
a 2a
remains constant; "a = 
= ^-v-.That means, a is the first 
dt2
, we have a = ~Substituting "v
in both sides, we have
dvii
v
Fig. 2.53
Then,
in a =
dv^ 
->dV
§ 2.11 Four General Equations
Let us recast the general kinematical equation.
dv 
dt
dr* 
di
Since r dt =
-» d v a = ~vdt
dt
This gives ^dt-d^v
Executing a scalar product with v
(because^ ± dv^)
ad~r = v-dv
= v(dv||+ dvjJ
= ~v-dv\\ + ~vdvy
= |7||v 
d r , we have ~ad~r = ~v-d^v
87
(ii)
(iv) vdv = a d r
yi
x0
ng. 2.54
Sol. As per the given condition,
Since
•X
Fig. 2.55
If you write
Ex. 23. A particle moves with a tangential acceleration at = a- r, 
where a is a constant vector in +x-direction and x = —.v
If the speed of the particle is zero at x = 0,ftnd v = /(x).
This will give you correct answer but the procedure is wrong because a-dr is in scalar 
product. Remember that we cannot write |a|=|o>|=a because at = tangential acceleration 
and a = constant vector.
General Kinematics of a Particle
Tabulating the general kinematical equations, we have
§ 2.12 Use of Polar Coordinate System in Curvilinear Motion (Optional)
Let a particle move in a plane in any arbitrary curve. We can use polar-coordinates (cylindrical 
coordinates) to analyse the nature of motion of the particle more comfortably.
= adr cos 0
= adx
ai = a-t
= at cos 9
= a(l)cos0
= acosO
vdv = a■ d r = adrcos0
-* d v a = — dt
= £ adx = a £ dx (because a = constant)
Ans.
vdv = ~a - d~r
Then, vdv
This gives
= a, dr (because a cos 9 = a,)
Substituting at = acos© in vdv = atdr, we have 
vdv = (a cos Q)dr 
= a(dr cos'B) 
= adx
v = flax
dt
..... d2~r
(in a =^-~ 
dt2
G.R.B. Understanding Physics MECHANICS (Part-A)88
7
y
p—♦
X
0
Then,
a + IA[ (00
where A is any vector, we have
dr
Substituting
dt
Since
A
' V “♦
'7/fv
/y i
v I 
i
J v
-> dv, a =—r~
Z =| r | r 
where r - unit vector of r
V=Vr + 70 and 
—* —• —» —♦ —♦ 
a =ar+ aa = at + an
Fig. 2.56
aB
v0 U--A 
an
where ~^ = vr (radial component of v ) and rco = v0 (transverse component of
Acceleration: The acceleration of the particle is
-» d v* a —----
dt
v = vrr + ru)0, we have 
-» dfy.r) d , Ax a =—r~^ + ^(rco0)
Velocity; The velocity of the particle can be 
obtained by taking the time derivative of ~r given as
~v =^- where r*=| r*| r
dt
v^(\ r\r)
dt
Using the relation = = ^111
dt dt dt
dB
dt
89
e
* fo d^ar + 2v.co + r—r- 0,
I dt )
a =
2where
. vr tells us at what1.
Acceleration of the particle is given as2.
3. dt
4.
5.
0-ar
(ii)(i)
6.
]
Fig. 2.61
r
Fig. 2.60
dr 
dt
d0 
dt
^16 
dt
Fig. 2.59
°4
\ a r 2 
a =ar+ag =an+at, where ar = —y-r© 
\dt
dvr _ d2r
,2 •
Ex. 24. An insect crawls along the rod with a velocity u relative to the rod. 
If the rod rotates in the horizontal plane with a constant angular 
speed a^ shown in the Fig 2.61, find the (i) velocity, (ii) accel­
eration of the insect as the function of distance r of the insect 
from the axis 0 of rotation of the rod.
d2r
—y-r©'dt2
particle.
Summarizing the above ideas, we note the following points.
( d2Q I dvr
and ae = 2vfco + r—- ;ar can be positive, zero and negative; ar * —
V dr J at
v2 , , d\v\ vdv
an = — where p = radius of curvature and at = - .
v2 2
acp =—,(but notrw ).
r and 0 are the perpendicular rotating unit vectors in polar coordinate system; ~ = ©0
. dr . - dr , . . . ono dQ and — lr;— leads r by 90°; — d( dt ~ 7 dt
and |10; J leads 0 by 90°.
Any particle can not have uniform accelera­
tion while moving in a curve with constant speed.
( In the above example, (/) find the tangential and normal acceleration (ii) Draw the trajectory 
of the insect.
vfr/ 
' ve
co2r
+ r2co2
r2 Since co -— constant,
dlr n 
~v = ° and —X- dt2 dt2
dill Q ,lfO ko.fA 
Jr r'
General Kinematics of a Particle 91
Assignments
1.
2.
3.
4.
5.
6.
2 . Can you interrelate between these two formulae?a =
10.
11.
15.
16.
12.
13.
14.
7.
8.
9.
Discussion Type Questions
It is always true to state that speed is the magnitude of velocity. Following this logic, can you 
define the average speed as the magnitude of average velocity?
In curvilinear motion, is it possible to have average velocity during a time / (from starting) 
pointed in same direction of instantaneous velocity?
Can a particle moving in any arbitrary curve maintain (i) a constant 
acceleration (ii) a constant value of acceleration (iii) a constant value of 
tangential or radial acceleration? Explain all the above cases with suitable 
example.
While moving in a curve, a particle’s velocity is always tangential to the path. That means, 
velocity has no component along the normal direction, whereas its acceleration can have one 
normal and one tangential component (if | v | varies). Sometimes, we talk about the radial and 
transverse components of velocity of a particle. Are they different from the normal and tangential 
components? Explain. Mark the difference between radial, normal and centripetal acceleration, 
with an example. '• • '
Let us present some terms regarding velocity, i.e., resultant velocity, change in velocity, average 
velocity and relative velocity. Distinguish between these terms.
d2r
We know that, a = —r-; but sometimes we come across a strange looking formula 
dt2
v2
d2r—-rco
{dt2
Why does a particle accelerate towards the concave side of its path?
Are the terms “moving” and “accelerating” different? Explain.
Can a particle (i) move while not accelerating (ii) accelerate while not moving? Support with 
suitable examples.
Can a particle (i) accelerate (ii) decelerate, while moving with a constant (a) speed (b) veldcity? 
Support your answer with examples.
Is it possible to move in a curve with zero acceleration, constant acceleration, constant velocity 
and constant speed?
A particle moves in a curve with negative (total) acceleration. Is this statement correct? Explain.
Can a curvilinear motion be periodic? Explain.
Can we apply the formulae v2 =u2 +2^~s , vdv = ads for a particle moving in a curved path ?
In the formulae, what does 'a' signify? Is it at or an or otota| ?
What is the physical significance of change in acceleration?
Can a particle move with an acceleration of constant magnitude while retarding at a constant 
rate?
av
-t0
21
1.
2.
(c) o = (d) a =(a) v =
3.
(d) straight line
4.
5.
6.
(a) vav (b) vav = (c) Vav _ V|5| + V2S2
St + s2
18.
19.
92
17.
dr 
dt
d~v 
~dt~ dt
V1+V2 
d\lv\+d2l v2
d\ r | 
dt
If 7 = J sin co// + 5cosw/ j + cP'k, the path of the particle is:
(b) spiral (c) helical
vxi + vyj, which of the following conditions is/are correct if the particle
‘ 0
Can a speedometer detect the direction of a vehicle?
In which case acceleration is greater; (i) putting the accelerator on (ii) pressing the brakes 
(iii) turning in a curve?
When a particle slows down or speedsup, it is said to be accelerating. Is it always true?
fa, dt, j | a | dt, | j^ dt |, j^ dt, j| a„ | dt.
(a) circular
If ~r = xi + yj and ~v = 
does not accelerate?
■ (a) vx > 0, vy 0
(c) xvy + yvx >0 (d) vx 0
Which of the following statement/s is/are correct?
(a) If a point always accelerates parallel to a fixed line, it must move in a straight line
(b) If a particle always accelerates towards a fixed point, v- a =0
(c) If a particle always heads towards a fixed point, 7 and are collinear, where r is the 
position vector of the particle relative to the fixed point
(d) If a particle always heads towards a fixed point, it must move in a straight line
Which of the following formula/e is/are correct for average velocity, if V] and v2 are the magni­
tudes of average velocities during time r( and t2 covering distances d\ and d2 and displace­
ments j] and s2 respectively?
V^2
2
20.
. Physically interpret the terms jo dt,
Multiple Choice Questions
Which of the following represent/s that the particle is moving with a constant velocity?
(a) x = A sin a>t,y = /cos cor (b) x = /sincor,_y = Bcoswt
(c) x = e',y = e~' (d) x = At,y = Bl
Which of the following expression/s is/are correct?
G.R.B. Understanding Physics MECHANICS (Part-A)
Let us represent the motion of a particle in a curve by v-t or a-t graph. What should we 
understand from this?
(b) | "v | =
5] +51
V, v2
93
7.
(d) 7x7 + raj2 sin 4>, where ) 
vr
dr 
dt
(d) vr and v
any position (r,9) and instant t, the tangential and 
normal component of v respectively are:
dr(a) v and 0 (b) and rw (c) rco and v
An insect P crawls along a horizontal rod with a constant 
velocity vr relative to the rod, away from the end 0. If the rod 
rotates with an constant angular velocity ^(= °)> the velocity 
of the insect is:
(a) vr
dVa .
(c) —and 2vrco9 (d) zero and 2v,co9
The centripetal acceleration and component of acceleration tangential to the rod are : 
(a) -r(02r and zero
(a) “9 (b)
at dt
The particle turns around the point 0 with a velocity:
(a) (b) r J (c) v0
The particle moves away from O with a velocity:
l
, -rco2r
(a) parabolic (b) r = roe*°
The total acceleration'of the insect is:
(c)
(c) -rco2n and -rco2r
(d) 2mvrQ and -rco2r
The trajectory of the particle (insect) is:
vr
CD
G.R.B. Understanding Physics MECHANICS (Part-A)
11.
1.
5.
6.
1.
2.
3.
4.
5.
6.
The position vector of a particle is given as 7 = 2t2i - e'j + sin 1 Or £. Find its (i) initial velocity, 
(ii) acceleration at t = n.
2.
3.
4.
7.
8.
9.
10.
1.
2.
3.
4.
Introductory Exercise
Position, velocity and acceleration
A person moves 100 m in the direction of east, then heads due north through a distance 200 m, 
finally moves due N-E through a distance of 200-^2 m. Find the final position (displacement) of 
the person.
Find the change in position vector of the tip of a second hand of a watch of diameter 4 cm during 
20 s. Assume that the second hand was lying on x-axis at t = 0.
In problem 2, find the (i) average velocity and average acceleration of the tip of the second hand 
(ii) acceleration of the tip of the second hand.
- x“, where w and A are positive constants. 
than | f v dt |.
When a particle retards, the value of 7- 7 is
Fill In The Blanks
When a particle moves with a constant speed in a curve, its total acceleration is equal to its 
acceleration.
In automobiles, the speedometer records
Average velocity is equal to instantaneous velocity. Then, the motion of the particle is 
The positions at which a and v are maximum are respectively, when 
v = o^2" ”2 
f| v\dt is
98
True Or False
In a curvilinear motion, total acceleration cannot be a constant.
Magnitude of average velocity is called average speed.
Magnitude of instantaneous velocity is known as instantaneous speed.
v-dv = v dv
The average velocity of a particle moving in a curve is given as y J I v I dt.
A particle moves in x-y plane. The tangent to y-x graph gives the direction of motion of the 
particle.
| f ~v dt | = f | ~vdt | in a curvilinear motion.
When a body has zero velocity, it must have zero acceleration.
Average velocity of a particle is equal to the displacement of the particle in one second.
Average velocity has a magnitude lesser than the magnitude of instantaneous velocity in curvilin­
ear motion of the particle.
In curvilinear motion, particle will be farthest from the reference point (origin), when = 0-
General Kinematics of a Particle 99
5.
KaVfl during6.
time interval is given as Kav =
7.
.Ar„ respectively, prove that the average speed over the total time is “av =A/|,A/2
8.
9.
10.
11.
12.
1.
2.
3. The end of a rigid rod of length / is moved with a constant velocity v. 
Find the velocity and acceleration of the midpoint of the rod.
Mav„
Trajectory
The position of a particle changes with time as. r = Jti + t2j. Find the equation of trajectory.
The velocity of a particle changes with time as V =asina)// + 6cosGraphs and Graphical Solutions
3.1
3.2
3.3
3.4
3.5
3.6
3.7
§ 3.1 Introduction
When we drive a car in a straight road we feel more comfortable than driving the car in a zig-zag 
road. It signifies physically that one dimensional (straight line) motion is simplest type of motion. Here 
we need to apply all general kinematical equations derived in the last chapter.
In this chapter we will modify the equations of general kinematics with proper sign convention, 
for straight line motion.
Then we derive a set of kinematical equations for constant acceleration. At the end of the chapter 
we will solve kinematical problems by using graphs.
We can apply the straight line kinematics in analysing various straight line motion such as oscil­
latory motion, motion of a particle in viscous medium (motion of rain drops), motion of particles 
(vehicles) in straight roads etc.
§ 3.2 Sign Conventions for Straight Line Motion
While moving in a straight line we have two possible directions for acceleration, velocity and 
displacement of the particle. Hence, for the sake of simplicity, we need not mention them with their 
usual vector symbols. Rather we prefer to write them like scalars a,v,s and r and use plus (+) and 
minus (-) to treat them as vectors keeping their senses (directions) in our minds.
_^chapter3
103
v
+s
+s
0
Sol.
and o- ds = 7- d~v, we have the following equations for one straight line motion, i.e.,
Sometimes a particle may accelerate opposite to the direction of its 
motion and sometimes it may accelerate in the direction of its motion. 
That means, “motion.” and “acceleration” are not the same thing. 
Motion is characterised by “velocity” of the particle whereas “accele­
ration” is defined as the “rate of change in velocity of the particle”.
At A, 
MB, 
AtC, 
At D,
v is +ve, 
v is -ve, 
v is -ve, 
v is +ve,
s is +ve
s is +ve 
s is -ve
s is -ve Ans.
0-----------*s
s is positive, v and a are negative
-?-T 
D a
Fig. 3.3
I pMMWW] — 
wzzzzzzzz/z//z/////////// —>
the direction of its acceleration (or accelerates in the direction of its motion), we say a favours v; 
hence 7 = +—. When the particle moves against the direction of its acceleration, 7 opposes 7;
dvhence a = . In other words, when a particle speeds up, it is said to be accelerating, hence we can
Jv
write — = +a. When the particlb slows down, we call the particle decelerating (retarding), hence 
dv — = -a.
dt
d . ...
Fig. 3.5 Since the block 
experiences a spring force 
F = -kx, its acceleration is 
position dependent; a = fix)
G.R.B. Understanding Physics MECHANICS (Part-A)
When s increases, v = +— and vice-versa 
dt
-> dvWhen v increases, o = +— and vice-versa 
dt
a = — and vice-versa, 
ds
Then, how to use these equations? Let us discuss.
Generally, in rectilinear kinematics the acceleration of the particle is given in the question and we 
are asked to find the nature of motion, that is, r(ors) = /(/). Then, by successively differentiating 
s = f(t) with respect to time, we find the velocity and acceleration of the particle as the function of time.
§ 3.4 Dependence of Acceleration with Distance, Time and Velocity
Furthermore, we have to find the variation of acceleration of the particle from the statement of 
the question. The acceleration may vary with distance, velocity and time because a force which causes 
an acceleration according to Newton’s 2nd law may vary with distance, time and velocity.
Gravitational
(i) a = f(r): For instance, gravitational force acting on a 
particle varies inversely with the distance r between the 
particle and the centre of earth according to “Newton’s 
law of universal gravitation”. Hence, gravitational 
acceleration of a particle depends upon its position. In 
other words, acceleration due to gravity is space 
(distance) dependent; a = f(r).
Spring-particle system
A block connected with a spring experiences a force which 
linearly varies with deformation “x” of the spring or dis­
placement of the block from the unstretched position of the 
spring, according to Hooke’s law. Hence, acceleration of 
the block varies linearly with the displacement s(or x).
Straight Line Motion 105
v
(iii) a = f(v)
Fig. 3.6 The body moving in 
a viscous medium experiences 
a viscous force F n av2 say. 
Hence, its acceleration is 
velocity dependent
Viscous Drag
(ii) a = f(v): When a body moves in a viscous liquid it 
experiences a “viscous force” which opposes its motion 
(velocity). Since, viscous force depends upon the velocity 
of the body, the acceleration of the body in a viscous medium 
depends upon its velocity.
(iii) a = f(t): When we try to slide an object (a table) say, we 
cannot do it immediately. It takes sometime to make it to 
slide. This becomes possible when we continuously increase the magnitude of muscular 
push (force) on the table till it overcomes the limiting friction between the ground and table. 
This signifies a force which varies (increases) with time. If we keep on increasing the 
force, the table will move with a time varying acceleration. That means, acceleration of the 
table increases with time.
§ 3.5 Use of Kinematical Equations
dv
(a) Use of a = — : This formula relates acceleration with velocity and time. When acceleration 
is given as the function of time and velocity, we can use the above formula to find the 
velocity and displacement at any instant.
Case I; fl = /(f): When acceleration, as the function of time is given, i.e., a = f(t),
/ I | mmmmr
Fig. 3.7 As we pull the table with more and more force, the table 
will eventually move with increasing acceleration after sometime
In kinematics, we “ignore the cause of acceleration”, that is, force. However, we cannot understand 
the variation (dependence) of acceleration with (distance), time and velocity without referring to “force”because force causes acceleration according to Newton’s 2nd law of motion which will be dealt in the 
chapter-7.
Ultimately, the knowledge of force is obvious to set a kinematical question. However, we do not 
require a prior knowledge of force to solve the kinematical problems when acceleration is directly given 
as the function of distance (position), time and velocity depending upon the respective sources and 
nature of the forces involved in the problems.
Then how to apply the simplified kinematical equations when acceleration varies as the function 
of time, distance and velocity such as
(i)a = /(f) (ii) a = /(s) and
We will discuss it in the next section.
dv 
substituting this in the above equation, we have — = /(f).
di
106
5 =
S -
V =
dv —
If the velocity changes from v0 to v during time / from starting, integrating both sides, we 
have
= say- 
dsSubstituting v = g(s) in di = — and integrating, we obtain the time 
v
This yields the velocity as the function of time
v = v° + = say.
Again integrating v with respect to time’we have
j0 vdt = £ g(t)dt = h(t), say.
Case II; a = /(v): When acceleration is given as the function of velocity similar to the 
dvfirst case, substituting a = f(y) in a = —, we have
, dtdv \
— = /(*) dt
Then rearranging the terms, we have
dv 
dt =-----
/(v)
When velocity changes from v0 to v during time /, integrating both sides, we have
f' j fv dv
Jo Jv0 /(v)
This yields velocity as the function of time; v = g(r), say.
Then, integrating velocity, we have displacement
vdt Jo
107Straight Line Motion
I Force)
a =f(v)
jwfv = J adst
-»| Velocity, v
= [_
J av = ^adt
, vdvds -------
/(v)
If the velocity of the particle changes from v0 to v when it undergoes a displacement s, 
integrating both sides, we have
f'A=r—Jo
This gives velocity as the function of displacement, i.e., v = g(s).
Then, executing the previous procedure, we have
rs ds ts dst = — = Jq ——, which gives us s = h(t).
Let us tabulate the program to find the displacement from the given force (or acceleration).
,= f-= f —Jo Jo v Jo g(s)
This yields the displacement as the function of time.
Case II; a = f(v): When acceleration is given as the function of velocity as a = /(v).
Substituting this in vdv = ads, we have
vdv- f(v) ds.
Rearranging the terms, we have
108
Ex. 2. A particle slows down from a speed v^while moving in a straight line. If the deceleration is
Sol.
Then,
This gives In— = -at. Hence v = voe‘
t
dt =v0The displacement s =
o
Ans.
Then, we have
Since
ij
atThis gives
Hence Ans.)
v0
directly proportional to the speed of the particle, find its displacement as the function of time, 
dvMethod 1. a = — .where a = -avdt
G.R.B. Understanding Physics MECHANICS (Part-A)
Let us use the above ideas in the following problems
—In a
I as1-— 
vo
/
/
/
!i
/
This yields
dsv - —dt
dv-av = — dt
Rearranging the terms, we have — = -adt.
When the particle slows down from v0 to v let it take a time t. Hence, we have
rv dv rt— = -a dt• Vq V Jo
-az
e-°‘
-af'vdrJo
J = 2!0(l_e-“' 
a
This gives s = —(l-eaz)a
Method 2. vdv = ads where a--av
Then, we have vdv = (-av)Js
This gives dv = -ads
When the particle slows down from v0 to v, it undergoes a displacement s.
[ dv = -a f ds
Jv0 Jo
v = v0 - as
, we have' = Substituting v = v0 - as, we have
= -“ ln(v0-as) , 
aL JoJo v0 - as
v
vo
109Straight Line Motion
v
vo
t
s
s
The average velocity vav = - =Sol.
Hence
This gives Ans.s =
3. a-s and v-s graph : In this type of motion, both speed and acceleration 
decrease linearly with distance which can be given as v = v0 - as
and a = av = a(v0 - as), as shown in the Fig. 3.8.
After a long time, both velocity and acceleration will be zero. Hence the 
body cannot retrace its path and the motion is unidirectional.
tt 
av
T‘ 
sm
a = -aVv in
1. v-t graph : The velocity (or the motion) of a body decays exponentially
as v = vQe~al, when t -> oo, v -> 0. It means, after a long time (/ -> oo), it 
/ vostops (v = 0). _£
dv 
Q —------.
dt 
r~ dv dvHence, -avv = -r. This can be written as -t= = -adt 
dt yj y
If the particle takes a time / .to stop, we have 
f° f' j-7= = -a \ dt 
»v0 \Jv *0
—t 
)
Use of a = ~ : Now we will find the time of motion by substituting 
dt
2. s-t graph : From the equation s = -^-(l-e we understand that, 
when t -> oo, 5 _> . It means, displacement builds up exponentially.
a
vn = v()e-“'
(a)
(b)
(c)
Fig. 3.8
Ex. 3. A particle retards from a velocity vQ while moving in a straight line, if the magnitude of 
deceleration is directly proportional to the square root of the speed of the particle, find its 
average velocity for the total time of its motion.
. . s Total displacement
■ t Total time
Use of v dv = a ds: Let us calculate the displacement by substituting 
a = -a>/v (for retardation) in vdv = a-ds. Then, we have v dv = (-aVv) ds.
This gives Jvdv = -ads
When the particle slows down from v0 to 0, it covers a distance s.
f yfvdv = - f a ds.
Jv0 Jo
2v03/2
3a
G.R B- Understanding Physics MECHANICS (Part-A)110
This yields
Ans.
Sol.
V X
Putting these values in
we have
Ans.This yields
Student Task
+Vo
-vo
a-t graph
Fig. 3.10
-»t 
v-t graph
x
+xo
— t 
x-t graph
t = t 
—o—
> In the above example draw x-t, v-t and a-t graphs.
Ans.
t = 0
___
s
Fig. 3.9dv 
a — — ♦ dt
When velocity changes from v0 to v during time t, integrating both sides with respect to time, 
we have
r1 we find a is +ve which is given by 
o = +6(/-l).
Let the particle has velocity +v as shown in the Fig. 3.9.
Here, s = +x, v = +v, a = +6(7 -1).
Straight Line Motion HI
(='2>
Fig. 3.11 (locus-diagram)
v = v0 +
-(i)v = v0 + at
we have
...(H)This yields
...(iii)
Following points are noted from the previous example:
Putting x = 0 in x = t (t - I) (/ - 2) we have t = 0, I and 2. That means, the particle crosses the 
origin twice at t = I and t = 2. After t = 2, x is positive. Hence its displacement points in positive 
x-direction and goes on increasing its magnitude.
Since v = 3t2-6t + 2, putting . 
say). Hence the particle remains stationary at / = /j and t2 ■ 
That signifies to and fro motion of the particle during first 
two seconds as it starts retracing its path at t = t] and 
t = li­
lt - 0 
It ="l
v = 0, we have r = l--^(=/b say), and t =
Then, substituting v from eq (i) in $ JvcZr,
J = £(v0 + at)dt
At last, using V
l+-£
V3.
43
1 2 s = vot + -at
v2 = vq + 2a» ''
Let us tabulate the above equations for constant accelerated motion, as following:
v = v0 + at 
’ 1 25 = vQt +—at
v2 = Vq + 2as
The sign convention for the above equations is same as that described for non-uniform 
acceleration.
(c) Kinematical equations for constant accelerationa = c : Let us now apply the previously 
obtained general kinematical equations such as
£ a dt, s = £ v dt, v2 = Vq + 2 £ a ds
for a particle moving with constant acceleration a.
Since, a is constant, taking “a” out of the integral in v = v0 + £a dt, we have
,2 = vq + 2 £ a ds for constant acceleration, we obtain
Sol.
we have t = t
- Oh s
Fig. 3.12
t s
Ans.Then, t = 4 s
Student Task
Sol.
•••(>)
..(ii)
From the Fig 3.13, we have
...(iii)
m
t = 0 
......... »■ a
Ans. (i) 4 m/s, (ii) 12 m/s.
t = 0
777/ '////////////////f///////////z
•*—12 m
.............
______ Xc_
xm
Fig. 3.13
—> v
—I
t = 0
u = 20 m/s
> In the above example, find the:
(i) speed of the car just after the braking.
(ii) average velocity of the car during the braking.
 -(-10) ± 7(-10)2 -4(24)(1)
2
xm=vt
During the time t, the car moves through a distance 
1 2 xc — — nt
Substituting x(
This gives t2 -10/ + 24 = 0
Solving the quadratic equation, we have
xm xc 12 
from eq. (i) and xc from eq. (ii) in eq. (iii), we have
1 ■> _ 
vt-—at =12
Then substituting v = 8 m/s and a = 2 m/s2, we have
/2-8/ + 12 = 0
j I2 G.R.B. Understanding Physics MECHANICS (Part-A)
Ex. 5. A driver applies the brakes of his car while moving in a straight path through a distance o/48 
m. If the speed of the car before applying the brakes is 72 km/hr, what is the time span of the 
braking? Assuming the retardation due to braking is 4 m/s1.
Let the time of braking be t. Putting s = +48 m, a = - 4 m/s* and
1 2 u = 72 km/hr = 20 m/s in the formula s = ut + -at ,
48 = 20/+ |(-4)/2
Ex. 6. A motorcyclist at t = 0 , situated at a distance s = 12 m 
behind a car, moves with a velocity v = 8 m/s. If the car •)
starts acceleratingfrom rest with a = 2 m/s~ at t = 0, when 
and where do they meet?
Let they meet after a time /. During this time the motorcycle 
covers a distance xm , say.
113
Ans.
where
Then, we have
motorcycle
t0
v
t
y
X
xo[
o
During first two seconds the motorcycle 
approaches the car.
At / = 2s , the motorcycle overtakes the 
car. Then it diverges (moves away from 
the car) till the car acquires a velocity 
equal to that of motorcycle at / = 4s . 
Then the distance of separation will 
gradually decrease as the car moves 
faster than the motorcycle after 4s. In 
consequence, at t = 6 s the car over­
takes the motorcycle. After that, the car 
will take a lead leaving the motorcycle 
behind it.
■car
•motorcycle
— t
0
x
§ 3.6 Kinematical Equations for Freely Falling Bodies
When a body moves under the action of earth’s gravitational 
field alone, we say that the body is falling freely. The acceleration of a 
freely falling body varies inversely with the square of the distance of 
the body and the centre of earth. When a body falls from a height 
much smaller than the radius of earth (Re = 6400 km), the acceleration 
of the body can be approximated as a constant.
| | | 'a = -9.8 j m/s2
Fig. 3.15 Near earth’s surface 
the bodies accelerate uniformly 
and vertically with 9.8 m/s2
; ; ;
i2 •" i6
i i II J4 L 
TT?
■ 2 ;4 |6 
I I 1
! •
Fig. 3.14
(i) 0 6 the car takes a lead moving away from the 
motorcycle.
Straight Line Motion
This yields two real values of time. i.e., t = 2s and 6 s.
The distance of separation between the car and motorcycle is
x = xcm = xc~ xm,
I 2 xc=x0 + -at anaxm-vt;
x' = x0+^at2 -vt
Substituting a = 2, v = 8 and Xq = 12.
we have x' = f2-8/ + l2.
x' is minimum, when — = 0
dl
This yields t = 4s. Hence x^in = -4 m .
Since x - 0 at the time of their meeting,
We have f2-8/+ 12 = 0. This yields 
t = 2s and 6s. Ans.
The following conclusion can be made 
from the above mathematical proceedings:
114
s
Fig. 3.16
------ t = 1sSol.
t = 2s
s*3
V2
t = 3s
s4Ans.
Student Taskf
t = 4s
Fig. 3.17
Ans. (i) I : 3 : 5, (ii) 1:4:9
s-t graph is parabolic, whereas the trajectory (y-x graph) is a straight line.
S?
> In the above example find the ratio of distances covered 
(/) in Is/, 2nd, 3rd second and so on.
(ii) during Is/, 2nd, 3rd second and so on.
Ex. 7. A multiflash stroboscopic camera photographs different snap 
shots of a ball which is falling from the top of a cliff. Draw 
the snapshots after each second of time interval.
If we magnify the photograph of different snapshots of the 
ball to its natur! size using the proper scale, we can find 
that the distances fallen after one, two, three and four sec­
ond etc. can be given as; st = 4.9, s2 = 4.9 x 4, s5 = 4.9 x 9, 
s4 = 4.9 x 16 and so on. If we plot s-t graph we can get a 
parabola given as,
- Op w
0 :
!|a = g £
zz/zzz/zzzzzzz 
Q
ForP 
s + ve 
v0 + ve 
a - ve 
For Q 
s -ve 
v0 + ve 
a - ve
rfr
I a = g
zzzzzz/'zzzz7z
P
ForP 
s - ve 
v0-ve 
a - ve
| Ozz'^zl////7z> ’z'Z/ " t = Os
a, = g |
v v,
a2 = g |
G.R.B. Understanding Physics MECHANICS (Part-A)
In this section, we will consider the straight line motion of a particle 
near earth’s surface. It is experimentally verified that ail bodies accelerate 
towards the centre of earth with an acceleration of 9.8 m/s2 “near earth’s 
surface”. When viewed near the “earth”, its surface appears as flat. Hence 
any particle appears to accelerate vertically downwards with an 
acceleration of 9.8 m/s2.
For vertical motion of an object, first of all we choose the point of 
projection as the origin of the coordinate system. If the particle is projected 
up, its velocity is considered positive. For downward projection, velocity 
of projection is assumed as negative. If the final point is above the point 
of projection (origin or reference point), the displacement is upward. 
Hence we consider “s” as positive. If the final point is below the point of 
projection, “s” is assumed as negative. Since the acceleration of the particle 
is directed vertically downwards, a = -g. (You may also choose the 
downward as positive and upward as negative. However, you will get the 
same answer).
j = 4.9 r2
Hence the acceleration of the ball is constant, that is, 9.8 
m/s2 directed downward.
The speed of the ball after a time t can be given as
v = ^ = 9.8/
dt
a3 = g | o— 
vv3
a4 = g | O—
v v4
115Straight Line Motion
Sol.
x
Vo
V
y*
tO
'7777777777
Fig. 3.18
.2
Student Taskf
Ans.
> In the above example, find the time after which the particle reaches the ground which is 
situated at a depth h = 20 m from the top of the cliff. Assume vq = 5 m/s and g = 10 m/s2.
y..h
y
Ivo
I
Vo 2vJ\ 
g g
2 vo
At t = ~0 is 
the origin of the coordinate system. For upward 
projection
Sol.
Substituting the above values in
t = 0
Ans.
Student Task
> Referring to the above example, find the
(a) velocity of the ball at the time of passing through the point P.
under consideration. , Ans. (i)
8j\ ttime /| and t2 respectively.
Sol.
116
Ex. 9.
, = til 1 
P
zzzM9zzzzzzzz
0
Fig. 3.19
6 ?t = t2 
vo!
Ans. |(/2-'t)
(b) maximum height reached by the ball relative to the (i) point of projection O (if) point P 
f0i+^2)2,(ii) f('2"'l)2 
o o
Ans. y[tf^
Alternately we can solve the above example by solving the equation h = vt-^gt2- This yields a 
quadratic equation t2 - — + — = 0. If this equation is satisfied for two roots /j and t2, we have 
8 8
t]t2 = — and t] +12 - — (because in the quadratic equation Ax2 + Bx + C = 0, x(x2 = — and s s
Xj +x2 = -%-).
G.R.B. Understanding Physics MECHANICS (Part-A)
A ball is projected vertically up such that it passes through a fixed point after a time and t2 
respectively. Find the height at which the point is located with respect to the point of projection 
and speed of projection of the ball.
Let the ball be projected up with a velocity v0 = v. It passes through the point 
/’at t = /| during its ascent and at t = t2 during its descent.
For the motion of the ball from O to P
s = +h, vQ =v, a = -g and t = and t2.
I 2 s = vot + —at ,
, , I 2 I 2we have h = W, - - gt{ = W2 - - gt$
Solving the above equation, we have
and v=g(,|+';)
2 2
(c) time offall of a body when released from rest from the fixed point P.
(d) average speed and average velocity of the ball during the motion from 0 to P for the
Ans. |^4^2
2 2 4r2 2
Ex.10. An elevator of height h moves up with an upward acceleration a. At certain instant during the 
motion of the elevator, a bolt loses contact with the ceiling of the elevator. Find the time offall 
of the bolt till it strikes the base of the elevator.
Let us assume that, the elevator has a velocity v when the bolt loses contact with the elevator at 
t = t\. Let the bolt strike the base of the elevator after a time t = T, below the point of projection
117
y
in the equation
(ii) se
This yields Ans.T =
Student Taskf
Ans. (c)
“-a” in T =
I
ia
' 2h 
g + a
2(1 + -1I g)
b
t = ti + T
2h
g + a
of the elevator.
4. When we look from the ground frame, the bolt moves up and accelerates down with 
a = 9.8 m/s2 . This is because the bolt moves up with a velocity equal to the velocity of the 
elevator at the time of its losing contact with the elevator; then the bolt falls freely under 
gravity and attains its highest position before coming down. When observed from the 
elevator, the bolt appears to move down with a downward acceleration (g + a) from rest 
which will be explained in the chapter of “Relative motion”.
sb
(c)
Til 
ib
t = 0
Fig. 3.20
> In the above example, if the bolt strikes the floor of the elevator at the point of its losing 
contact, the velocity of the elevator at the time of losing the bolt is:
agh
2(g + o)
releasing the bolt, given as T =
3. If the elevator accelerates down, the time of fall of the bolt can be given by putting a as
I 2A. This gives T = ----- .If a > g, the bolt can never touch the floor
Ng-a
/»=|(g+a)r2 
! 2/i 
g + a
Straight Line Motion
of the bolt. Hence by substituting sb = -y, v0 = v, a = -g and
/= Tin s = v0t + ^at2,
From the previous example we note the following points:
1. For suitable value of V|, the bolt may strike the base of the elevator above, below or at the 
point of its release.
2. The time of fall of the bolt does not depend upon the velocity of the elevator at the time of
we have y = ^gT2-vT ...(i)
Likewise, for the elevator, substituting se = +y' = h-y,
v0 = v, a = +a and t = T 
s = v0/ + -a/2, 
we have h-y = vT + ^aT2
Adding eq. (i) and (ii), we have
Sol.
(i)Then,
®2
For 2nd particle,
Then, ..(ii)
(iii)
/ = Ans.
Student Taskf
Ans.
> In the above example, if the bodies meet between the top and bottom of the cliff, find the time 
of meeting.
t = 0
Fig. 3.21
v2
It 
>=y 
isi
t
Vii t = oA
^2=v2/~2^
Referring to the Fig. 3.21, s2 - = h
Substituting j] from eq. (i), s2 from eq. (ii) in eq. (iii), we have 
h
(v2 -v,)
dr
V = —= dt
s = sv v0 = v},a = -g
1 2
S = s2, v0 = v2, a = -g
GR.B. Understanding Physics MECHANICS (Part-A)
1 1 o
Ex. 11. Two bodies 1 and 2 are projected simultaneously with velocities V| and v2 respectively. The 
body I is projected vertically up from the top of a cliff of height h and the body 2 is projected 
vertically up from the bottom of the cliff If the bodies meet above the top of the cliff, find the 
time of meeting of the bodies.
Let the particles meet after a time t. First of all we choose the point of 
collision above the top of the cliff.
For 1st particle,
h 
V2~V|
The time of meeting of the bodies for simultaneous projection of the two bodies is given as 
( h h
~ I v2 T vj | ” | vre| |
If the bodies are projected vertically in same direction use I vrei I = I v2 ~ vi I • ^e bodies are 
projected vertically in opposite directions use | vre( | = (vj + v2). We will discuss the relative 
velocity in the chapter-4.
§ 3.7 Graphs and Graphical Solutions
Introduction: There are four main parameters in rectilinear kinematics such as displacement (s), 
velocity (v), acceleration (a) and time (/). One can be plotted as the function of the other. We can plot 
them in six possible ways, i.e., s-t, v-t, a-t, v-s, v^-s and a-s.
J J
The time derivative of position vector or displacement gives velocity; v = — = —. The time 
derivative of velocity gives acceleration; a = Following the reverse process, integration of acceleration 
with respect to time gives us change in velocity; Av = $adt. Similarly, integrating velocity with respect 
to time we find the displacement (change in position vector); s = Ar = jve/r.
V
,2
y
si
X
t
Jad^s
1. The slope of the straight line joining any two points in s-t graph gives the average velocity 
between these points of time. Symbolically, vav = — = slope of the line joining the points in 
s-t graph. If slope is positive, vav is positive; ifslope is negative, vav is negative; if slope is 
zero, vav is zero.
2. In the process of graphical solution, we should not mistake the displacement-time graph as 
equation of trajectory.
Straight Line Motion 119
Sometimes, the velocity is given as the function of displacement. In that case, we find acceleration 
vdv 1 dv“ ® , •by using the formula a = — = ■ This may also be written as — = —■ That means, derivative of
‘v ’ with displacement yields the ratio of acceleration and speed.
When acceleration is given as the function of displacement, integrating acceleration with respect 
to displacement, we have
I
*1 t2
Fig. 3.22 The slope of a straight line connecting any 
two points in s-t graph gives the velocity averaged 
over the time interval (t2 -1,)
= f vc/v = |(v2 -Vq)
J v0 1
From the above explanations we understand that, sometimes derivative of one parameter leads to 
the other, and in some other cases, integration of one parameter results the other. As we know, derivative 
of a function signifies its slope and its integration gives the area bounded by the function. In the present 
context, we will discuss the slope and area bounded by the kinematical functions. We will develop and 
use the idea of graphs in solving the kinematical problems.
s
(a) Displacement-time graph
Average Velocity: Let us consider an arbi- s2
trary s-t curve; s = /(/). Please remember that 
s-t curve and the trajectory (path) equation of a 
particle are different concepts.
As derived earlier, average velocity of a 
particle over a time interval Ar can be given as
Av
Vav" A/’
Av
Graphically, — is represented as tan (slope of AB) where is the angle of inclination of the 
Ar
straight line AB with positive /-axis. In other words, tan is equal to the slope of the straight line AB
that connects the coordinates ($,/) under consideration on the s-t graph. If slopeis positive, the
Av
average velocity is positive. The average velocity is negative for a negative slope. This is the 
graphical significance of average velocity.
A.y
vav = — = slope of the line joining the points in s-t graph
____ Ar____________________________________
From the above analysis, we conclude the following points:
G.R B- Understanding Physics MECHANICS (Part-A)
s
t
Sol.
s
-tO
Student Task
Ans. 5 m/s
(ii) Find the equation of s versus t graph.
find the significance of instantaneous velocity in the
s2
S1
Since
0 tand
we have v = slope of s-t graph (= tan 9)
2s
Fig. 3.23(a)
T 
AS 1
yl p
Instantaneous Velocity : Let us now 
graphs. To find the instantaneous velocity, we have to gradually 
decrease the time interval Ar to zero. When Ar tends to 
zero, Aa also will tend to zero. Inconsequence, the straight 
line AB tends to coincide with the tangent drawn at A as 
shown in the figure. That means, approaches 0. Eventually 
tan Referring to the above example:
(f) V find the average speed of the particle during 2 s.
ds The slope of s-t graph, that is, tan6 = —
at any point gives the instantaneous 
velocity at that point
v= lim tand> = tan0
*->0
lim tandi = tan0 
♦->e
120
Ex. 12. A particle moves rectilinearly possessing a parabolic s-t graph. ^Find 
the average velocity of the particle over a time interval from 1 = ~s to 
t = ].5s.
Ans. s = 10r-5r2
,B 
4
I yB'
tangent
/ * 1 • \f I I I \
1/2S 1s 1 5s 2s
♦At*
Fig. 3.23(b)
2
As s-t graph is a parabola it can be given as 5 = k\t - k2t , where Aj and 
k2 are constants. Since it is symmetrical about r = Is, we have equal 
1 3displacements at r = —s and r = - s. That tells us that, the line A B joining 
two coordinates is parallel to r-axis. Hence, slope of AB is zero. This 
implies that the average velocity during the time interval Ar = Is is zero.
Ar /(r + Ar)-/(r) ds v = hm —= lim —----- -—=—,
ai-»o Ar At-»o Ar dt
//
Straight Line Motion 121
s
tO
Sol.
s
As = s (= 2m)
3s)-**,Q ^-At = to
t
It gives Ans.
v
v
s =
t0
dl
elementary time dt
Fig. 3.26
are equal, the straight line QP must be tangential at P. Hence the point P has
3
2
1
Vav
I
i 
k
As vav = v = — 
A/
1 2 3
Fig. 3.25
/R 
z/fp(‘i. »1>
I»i!
— t-
2m „ ,
— = 0.67 m/s.
3s
Now we conclude that, in s-t graph the velocity at any instant is numerically equal to the slope 
of s-t graph measured at that instant (time) which is given by
ds
v = — = tan 0 = slope of s-t graph 
dt
If the slope is positive, v is positive (the particle moves along positive axes) and when the slope 
is negative, v is negative (directed along negative axes).
Ex. 13. A cockroach moves recti linearly such that, after sometime t0 let its 
(instantaneous) velocity be equal to its average velocity over that time. 
Referring to the s-t graph as shown in the Fig.3.25 for the motion of 
the cockroach, find the time t0 and the average velocity of the cockroach 
over the time t0.
Since vav = v (given), the “slope of QP = slope of the tangent 
PR”. Hence, QPR is a straight line. i
Suppose at point ”P” of s-t graph, both vav and ~v are equal. 
The average velocity during time t is given by the slope of the 
line QP . That means, vav = ——— by referring to the s-t 
l° C
graph. On the other hand, the velocity v of cockroach at the | 
time t is equal to the slope of s-t graph at that instant so
-> ds n
(point P)\ v = — = slope of the tangent PReA P. u
dt
Since vav and ~v 
coordinates (3,3).
(b) Velocity-time graph
Displacement: When the velocity of a particle is given as the function 
of time, i.e., v = f(t), we find the total displacement ~s by summing up 
(integrating) the elementary displacements ds*;
jV eft
The elementary displacement d s = v dt can be given as the area 
of the shaded elementary strip. Summing up all the elementary areas between Fig. 3.27 Area of the 
' 1 = 'l and / = /2, we have the total area under v-t graph enclosed within represent el?me"nteJy” 
the time interval which gives the total displacement s of the particle during displacement during 
.. . elementary time dt
the time interval Ar(= z2 - fj).
G.R.B. Understanding Physics MECHANICS (Part-A)122
dt = A (= Area under v-t graph)
vv
t0
t
‘1 l2
v2
V1
 .Since = tan —> tan 0, where tan 0 = —-.
Since tan 0 is equal to the slope of v-t curve (graph) at the point A as shown in the Fig. 3.31, the
—> (/y
instantaneous acceleration a = — at any point of v-t graph can be represented by the slope of v-t 
. . . dlgraph at that point.
—> d y
a = -y s tan 0 = slope of v-t graph
Fig. 3.33
Z B ,/k
If the slope of v-t graph is positive at any point of time on the graph, the acceleration of the 
particle is positive at that time. If the slope of v-t graph is zero, it signifies that the acceleration is zero. 
When the slope of v-t graph is negative, it means a negative acceleration (or retardation).
20 2Ex. 14. A diwali rocket moves vertically up with a constant acceleration ai =~ m's • After sometime 
■J 2
its fuel get exhausted and then it falls freely with an acceleration a2 = 10 m/s . If the maximum 
height attained by the diwali rocket is h, using graphical methodfind its speed when the fuel is 
Just exhausted. Assume h = 50 m.
Let the rocket acquire a speed v at the time when all its fuel get 
exhausted. Referring to the v-t graph in Fig. 3.33 the slope of OP 
gives the acceleration a{.
v
The slope of OP gives the acceleration fl] = tan 0] = —rt
—vThe slope of Pg gives the acceleration (-a2) = tan02 =— ,..(ii) 
r2
From the equations (i) and (ii), we have v = = a2r2 ...(iii)
The area under v-t graph = Area of the tsOPQ = ^OQ • PM = s
: ,Atir:/ / /
/'/' t
w
G.R.B. Understanding Physics MECHANICS (Part-A)124
••■(iv)
al
Finally, substituting
v =we have
This yields . v = 20m/s Ans.
Student Taskf
Sol.
V
•••(>) V
t
—»i—rThen, ...(ii)
This gives ...(hi)
-----12-------- 1
— T------
Fig. 3.34
> ReferringSpeed from 
Instantaneous Velocity
2.8 Uniform and Non-uniform Motion
2.9 Acceleration
2.10 Uniform and Non-uniform
Acceleration
2.11 Four General Equations
2.12 Use of Polar Coordinate System in 
Curvilinear Motion (Optional)
Assignments
O Discussion Type Questions
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O Match the Column
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O True or False
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O Introductory Exercise
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(X)
Chapter-5
189-217
189
198
Chapter -4
RELATIVE MOTION 154-188
Chapter -6
174 218-262
220
2 233
236
131
132
113
118
168
169
173
176
177
180
180
182
183
183
185
186
239
240
241
207
207
209
209
210
211
215
200
201
203
204
190
190
196
218
219
159
165
139
140
143
143
143
147
152
3.6 Kinematical Equations for Freely 
Falling Bodies
3.7 Graphs and Graphical Solutions
Assignments
O Discussion Type Questions 
O Multiple Choice Questions
O Assertion-Reason Type Questions 138 
O Match the Column 
O Comprehensions 
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Assignments
O Discussion Type Questions
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Q Comprehensions
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Q Problems
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KINEMATICS OF CIRCULAR 
MOTION
5.1 Introduction
5.2 Uniform and Non-uniform Circular 
Motion
5.3 Linear Quantities
5.4 Angular Quantities
5.5 Relation between Linear and
Angular Quantities
5.6 Use of Polar Coordinates in 
Circular Motion
5.7 Radius of Curvature
4.1 Introduction 154
4.2 Galilean Transformation of Velocities 155
4.3 Motion of Swimmer (or Boat) in
Water
4.4 Relative Angular Velocity
4.5 Equation of Trajectory of One Particle 
Relative to the other (Optional)
4.6 Time of Meeting of the Particles
4.7 Relative Acceleration
4.8 Application of Kinematical Formulae 
in a Moving Reference Frames
Assignments
O Discussion Type Questions
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O Assertion-Reason Type Questions 179
O Match the Column
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PROJECTILE MOTION
6.1 Introduction
6.2 Trajectory Equation
6.3 Use of Trajectory Equation 
(Locus Equation)
6.4 Use of 5 = ut + - at
2
6.5 Projection onto an Inclined Plane
6.6 Application of u = w + a t in 
Projectile Motion
6.7 Application of u2-u2 = 2as
6.8 Horizontal Projection
(xi)
O Answers 355
243 Chapter -8
FRICTION 359-406
Chapter-7
263-358
263
264
Chapter -9
407-454
320
246
247
327
329
337
338
342
343
343
349
369
371
377
378
379
407
408
431
431
385
385
389
389
390
394
402
265
266
271
274
286
293
312
359
359
362
366
409
414
419
425
250
251
254
255
255
258
261
NEWTON’S LAWS OF MOTION 
OF A PARTICLE
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O Discussion Type Questions
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□ Match the Column
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6.9 Projectile Motion Relative to a 
Moving Reference Frame
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■ O Introductory Exercise
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7.1 Introduction
7.2 Interaction between Mass and 
Environment; Force
7.3 Newton’s First Law of Motion;
Law of Inertia
7.4 Newton’s Second Law of Motion
7.5 Newton’s Third Law of Motion
7.6 Application of Newton’s Laws
7.7 Spring-Particle System
7.8 String-Particle System
7.9 Wedge-Particle System
7.10 Application of Newton’s law in 
Non-inertial Reference frame
DYNAMICS OF CIRCULAR 
MOTION
8.1 Introduction
8.2 Static Friction: Definition
8.3 Laws of Static Friction
8.4 Angle of Friction
8.5 Static Friction on Accelerating 
Surfaces (Optional)
8.6 Kinetic Friction
8.7 Role of Friction
9.1 Introduction
9.2 Centripetal Force
9.3 Centrifugal Force and Rotating
Frame
9.4 Horizontal Circular Motion
9.5 Role of Friction in Circular Motion
9.6 Vertical Circular Motion
Assignments
O Discussion Type Questions
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O Match the Column 434
(xii)
455-554
469
O Comprehensions
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Q Problems
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Chapter-10
436
438
439
439
442
450
455
456
458
471
472
476
481
485
506
509
494
495
512
514
488
489
519
520
523
524
524
533
548
Assignments
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O Match the Column
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O Answers
10.10 Relation between Force and 
Potential Energy
10.11 Stability
10.12 Subtle Difference between 
Conservative and Non-conservative 
Forces
10.13 Conservation of Energy
10.14 Work Done by Non-conservative 
Forces and Mechanical Energy
10.15 Power
WORK-ENERGY-POWER
10.1 Introduction
10.2 Definition of Work
10.3 Work Done by Different Forces
10.4 Work Done by a Force Relative 
to Different Reference Frames
10.5 Work Done by Many Forces
Acting on a Particle
10.6 Energy
10.7 Work-Kinetic Energy Theorem
10.8 Conservative and Non-conservative 
Forces
10.9 Potential Energy
VECTORS
§1.1 Introduction
Physics deals with the study of nature as the basis of observation and experiments. Mathemati­
cally, we express the hypothesis, rules, laws and principles by using physical quantities like mass, 
length, time, volume, pressure, density, energy, velocity, acceleration, force etc. Some of the physical 
quantities have directions apart from their magnitudes. We call them vectors; others are called scalars. 
Mass, length, time etc. are scalars and velocity, acceleration etc. are vectors. In Greek language the 
word “vector” means “carrier” which signifies a directional nature.
After defining vectors, we name and represent them. Then we develop the mathematics (algebra 
+ calculus) of vectors. Algebra of vectors deals with addition, subtraction and multiplication of vectors. 
Calculus of vectors deals with differentiation (derivative) and integration of vectors. You should remem­
ber that, mathematics of vectors is not abstract; vector algebra and calculus are invented as we need 
them while formulating the outcomes of analysis of properties of nature.
Let me tell you the necessity of “vector mathematics” in brief. While we find the acceleration, we 
need the “net force” which means the “vector addition” of forces. While describing the displacement,
1.1
1.2
1.3
1.4
1 .5
1 .6
1 .7
1.8
1 .9
1.10
1.11
1.12
1.13
CHAPTER 1
Study^^Points
INTRODUCTION
Definition of Vectors
Representation of Vectors
Types of Vectors
Multiplication of a Vector by a Scalar 
Addition of Vectors
Subtraction of Vectors
Scalar Product of Two Vectors
Vector Product of Two Vectors
Some Fundamental Laws of Vector Algebra 
Condition for Coplanar Vectors
derivative (differentiation) of a Vector 
Integration of a Vector
= ^BdA,
where
2 G.R.B. Understanding Physics MECHANICS (Part-A)
we subtract final position vector from initial position vector. Relative velocity and change in velocity are
- expressed as the subtraction of two velocities. We define the scalar quantity “work” as the dot product 
of two vectors, i.e., force F and displacement s . The rotational effect of force, that is, torque is
—> —1►
expressed as the cross product of two vectors r and F . Thus, using the idea of product of vectors 
we can express many physical quantities such as power, energy, densities, angular momentum, etc. 
While describing the motion of a particle in kinematics, we need to express the velocity as the time 
derivative of position vector and acceleration as the time derivative of velocity. In reverse way,to the previous example'.
(a) Find the time taken by the diwali rocket to attain the maximum height. Ans. 5 s
(i) What is the average (/) speed (ii) velocity of the diwali rocket over the total time of its
motion in air? Ans. (i) 20(5 - VT0)/3 (ii) zero
a2
v s = —
Substituting Og’ = /1+/2 and PM = v
v v
'1+') = a + p
Area under v-t graph gives the total displacement 
s = Area of the trapezium
2s T
q is + ve
A_—Av2 is-ve
A = jdA = | a dt
Straight Line Motion
Substituting q + /3 from eq. (ii) and /2 fr°m ecl- (*•*) *n etL (D» we have
T = —a
T 5 
J = — ■
V
I
> A particle moves in a straight line obeying the v-t graph as shown in 
the Fig. 3.35. The average velocity of the particle over the time T is:
v 2s + ------- TP v
, I 0 
M“+pJ *
(0^
(c) Acceleration-time graph
Let the acceleration be given as the function oftime, i.e., a = /(/). 
Then the elementary change in velocity during a time dt is:
d v = a dt
We can observe that in a-t graph, a dt is equal to the area of the 
shaded elementary strip (rectangle of sides a and dt). Summing up all 
the elementary areas dA, we have the total area L
A 7 = fa dt = A = area under a-t graph
That means, the change in velocity a 7 over any time interval is represented by the algebraic 
sum of all positive and negative areas which we call “area under a-t graph” over that time interval; 
A = £A+ + ZA_. For positive area, write 
(+J) and for negative area, write (-A). 
If the area is positive, A 7 is positive 
which means that A 7 is directed in 
positive direction of the coordinate axes. 
If the area is negative, A v is negative; 
A 7 is directed in negative direction of 
coordinate axes. When area is zero, 
a7 = 0 which means no change in Fig-3.37 Av=Ja 
velocity.
a
0
Fig. 3.36 a"dt is graphically 
represented as the area of the 
elementary strip of sides “a” 
and “dt” in a-t graph
G.R.B. Understanding Physics MECHANICS (Part-A)126
5
t(s)Sol. 0
Fig. 3.39
...(H)
5
0
-10
Fig. 3.40
This yields
, 1in A2 = — x x x 5 ,Substituting
and ‘3
we have
+5 J-7.5/= -2.5 j Ans.
a (m/s2)
a (m/s2)
When a 7 is positive, you should not immediately accept it as A | v | > 0, because Av is a 
vector quantity. Hence positive area means a positive A 7 but not positive A171 (t Av).
Ex. 16. A particle moves vertically with an upward initial speed 
v0 = 5 m/s. If its acceleration varies with time as shown in 
a-t graph in the Fig. 3.39, find the velocity of the particle 
at t = 4 s.
The velocity of the particle at t = 4 s can be given as
74=vo+a7, ...(i)
where A 7 ■ A (= area under a-t graph during first four 
seconds)
Referring to a-t graph (Fig. 3.40) we have,
A = At + A2 - Aj - A4
where A] =5x1 = 5, A2 = -^ x x x 5
=^x(l-x)xl0 and A4 =|x2xl0 = 10
We can find x as following :
Using properties of similar triangles, we have
x 1 -x
5--io~
1 x = —
3
x = l i ’ 
3 2
A3 =|(1-x)x10, 
4 5 . , 10A2 = — and Aj = —
Then substituting AhA2,A3 and A4 in eq. (ii), we have
A =-7.5
Negative area tells us that A 7 = -7.5 j m/s. Hence substituting v^ = +5j and A 7 = -7.5 j in 
eq- (i) 
we have
127
v2
v —
9
s s
It gives
1.
2.
dv 
ds
The slopes can be positive, negative or zero which signify positive, negative or zero 
accelerations.
I// dv --At.
I \
h-a—H
Using the formula vdv = ads, the slope of the 
a 
v
;'dv2 
ds]
v2-s graph.
acceleration = a = --7—7- = slope of v2-s graph
2 ds 2
Summarizing the above facts, we conclude the following points:
Acceleration = speed x slope of v-s graph
= yxslope of v2
0
Fig. 3.41 The slope of v-e graph at any 
point gives the ratio of acceleration 
and speed at that point;
slope = — r ds v
T 
V2 
1 
O|—s —|
Fig. 3.42 The slope of v2-s graph at any 
point is equal to twice the acceleration 
that corresponds to that point 
slope = ^ = 2a
. ■ . dv . j
v-s graph, that is — can be expressed as
acceleration = a = = (speed) (slope of v-s graph)ds
(d) Velocity-displacement graph
When v is given as the function of s, we can find the acceleration at any position by using the v-s 
graph as described below:
v
The above formula can also be written as a - That means, half the slope of v2-s graph
2 ds
can give the acceleration at any point of the graph. Hence, we can write
Straight Line Motion
Student Task f
> Referring to the previous example, the particle starts moving down at t which is equal to:
(a) 3 s (A) 3.5 s (c) 4/3 s ( Find a = f(s) and draw
v (m/s)
0
; 100 60 \we have v = —m/s.
c- v dv ,Since a = —— and 
ds
’ I
100 200 250
Fig. 3.43
2 k »k ♦ • ^(v2)
v-s graph, that is, —-—- 
ds
40 , 2= —3- m/s , obtained from the given as 
tZ(v2) _ , i/(v2)
4r2 = 2o, and —— 
ds ds
dv 1
-r = TFf°r 100 Find the time required to cover 100 m mark.
> Find the distance covered during first two seconds.
a
a
v
vo--
s
s
2400
1600
800
-----s ——{ds H—
Fig. 3.48 The elementary 
area dA = fds is equal to vdv
I 
a
v = 20>/2 m/s
I 
I
I 
200
(b)
_ A
_£\p(s, v2)trr
80 100 200
(a)
(l-M
I N13 J
Ans. (2>/260 - 1) m
a-s graph in a displacement ($2 -$|), we can write
v2-s
40 (260-s) as plotted in the Fig. 3.47.
v2
(e) Acceleration-displacement graph
If the acceleration of a 
particle is given as a function of 
displacement in the form of a-s 
graph, how do we find the 
(instantaneous) velocity of the 
particle?
By using the formula 
vdv= a-ds and integrating both 
sides, we have vdv= {'za-d~s.
J»’o ■'•’1
It gives the speed
= + 2p a-ds
As P a -d~s represents the area under 
Js, 
the speed v at any point of a-s graph as
v = y vq + 2 (Area under a-s graph)
S1 s2
Fig. 3.49 Area 'A' under a-s graph 
gives the speed as per the formula : 
v»7*o2*2A wher® A=J^’d«
Fig. 3.47 
v2-2400 -40 . ..
-- 7^’ =-r-. This yields200-80 3 3
(c) Considering a point P(s,v2) on the 
v2 -2400 40 „ . ..---------- = . It yields v = s-80----- 3----- 7
Student Task^
.a
Sol. s
••• (i) a
ao
s
...(H)
It gives
Then,
Since the area A under
This yields v = Ans.s
Ans.
130
Ex. 19.
— 8..........
Fig. 3.50
a0 =
Now substituting a0 in eq. (ii), we have
A =
.2
substituting v0 = 0, we have
G.R B. Understanding Physics MECHANICS (Part-A) 
a
Substituting s( in the above equation — = cota, we have 
tan a tan 0s 
tan a + tan 0
Student Task^
> Find the velocity of the motor boat when it attains maximum acceleration.
a-s graph is equal to
s2 tan a tan2 0 
(tan a + tan 0)2
A racing motor boat speeds up in a straight line in a lake, from 
rest. Referring to the acceleration-displacement graph for the 
speeding boat, find its speed when it passes a raft at a distance 
of s from the starting point. (Neglect the initial speed in 
comparison to the speed when it passes the raft).
If the boat speeds up from v0 = 0 to v after covering a distance 
j then
y = =
—- S2”"
Fig. 3.51
—-H
s2 tan a tan 0 
2(tan a + tan 0)
v2-vq
2 
s2 tan a tan 0 
2(tan a + tan 0) 
tan a tan 0 
tan a + tan 0
Since — = cota and — = cot0 as obtained from the graph,
1 tanP 
s\/s2 =7-----1 L tana
S] tan0
S] + s2 tan a + tan 0 
~ JtanP 
tan a + tan 0
v2 =Vq + 2
From the a-s graph it is evident that its acceleration and 
displacement are both positive (unidirectional). Hence, 
|a-ds = jadr.
This gives the area A of a-s graph as 
A = v4| + A2
Cinna 21 
a0 
we have,
_£q£l +fo£2 _2o,_ . . aos ~ 2 + 2 ~ 2(1 2)-T
■22 =
*0
131Straight Line Motion
Assignments
Discussion Type Questions
i.
2.
t t t0 00 0
3.
t
4.
for a non-uniform acceleration? Will this expression satisfy for some5.
6.
9.
13.
10.
11.
12.
A particle starts from rest and moves with accelerations a, 2a for two equal consecutive time 
intervals covering a distance x. If it travels with accelerations 2a, a for same two consecutive 
time intervals it covers a distance greater than x. Is it true? Explain.
Describe the nature of motion of a particle whose x-t graph is shown.
x
7.
8.
Can we use vav =
cases? Discuss.
A particle when projected vertically up can attain a height h. If we disregard the air resistance, 
how high will it go during the same time of ascent assuming same velocity of projection? 
Does a “freely falling” body take less time to travel if it is heavier?
Can you give example of a motion where the speed increases whereas acceleration decreases. Is 
it possible for a falling body?
“Average speed is equal to the magnitude of average velocity”. This statement is meant for 
unidirectional motion. Is it true?
If v*v = 7, can we call it “uniform motion”? Can the reverse of this statement be true?
The clouds do not fall but appear to float, why?
Can a coin appear to move up when released from the base of a falling cage? If the cage is falling 
freely, how does the coin appear to move relative to a person inside the cage and on the ground? 
If a stone is released from a balloon ascending with a downward acceleration a, is the accelera­
tion of the balloon g or g + a or g -a? What will be the acceleration of the stone if it is thrown 
inside an elevator which moves up with an acceleration a?
Can you jump higher (i) on moon (ii) at the pole of earth?
In each second, a rabbit moves one-half the remaining distance from its nose to head of a lettuce. 
Does the rabbit ever get the lettuce? What is the limiting value of the rabbit’s average velocity? 
Draw the velocity-time and position-time graph.
14.
15.
vi+vf 
2
Is the formula =u + (2n-l)^, dimensionally incorrect? Explain.
Look at the following four v-t diagrams for a falling body and discuss the nature of motion, 
v v v v
and
1.
(c) (d)
2.
3.
4.
5.
6.
7.
a
132
16.
(d) 4:3 
a oc v . Then, which of the follow-
G.R.B. Understanding Physics MECHANICS (Part-A)
In one dimensional uniformly accelerated motion, verify the relation between distance and dis­
placement which can be given as
d = s = vqI + — at1 t Iq, where Iq
2
5m - (b) h5m
A boy moving with a velocity v is behind a stationary bus separated by a distance x, at t = 0. If 
the bus starts accelerating at t = 0 with constant acceleration a:
(a) the boy cannot touch the bus if v yj2ax and the:bus cannot cross the boy
(c) the distance between the boy and bus will be minimum when their relative velocity is zero
(d) when their relative velocity is maximum, their relative distance is minimum
The ratio of average speed and maximum speed of a particle whose acceleration varies linearly 
with displacement as a = -kx, is:
(a) 1:1 (b) 1 :2 (c) (d) 2:n
A stone is dropped from a certain height. The ratio of speeds averaged over the total distance and 
total time of fall is:
(a) 1 : 1 (b) 2 : 3 (c) I :2
A particle slows down from velocity v0 obeying the relation 
ing graph/s is/are correct?
Straight Line Motion 133
(b)(a)
(d) o(c)
8.
9.
10.
(a)
(d)(c)
11.
(C) tC - tD - yjtAtg
(d) lA >fC ~tD>tB
12.
13.
(c)as as
jB^v
V
(b)(a>£
Z1Z2 vlv2
2 ’N 2 /] +r2’vi +v2
The particles A, B and C are projected simultaneously from the top of a tower with equal speeds 
as shown in the figure, whereas the particle D is dropped from rest at the same time. Then: 
(a) all the particles hit the ground with equal speed y
(h) h=-gtAtBandv=^(jB-tA) |D;
h : V
A particle moves in a straight line obeying the relation x = (r - IX' - 2). Then:
(a) the particle moves in negative x-direction initially
(b) the particle moves with constant acceleration
(c) the average velocity and speed of the particle may be equal to the instantaneous speed
(d) the average speed and average velocity of the particle must be equal
9A
A stone is dropped under gravity from rest from a height h and it travels a distance ~ in the last 
second. The height h is:
(a) 140 m (b) 122.5 m (c)150m (d)l00m
A stone dropped from rest passes the top and bottom of a pole with velocities V| and v2 at times 
/| and t2 respectively. Then the time and velocity of passing the midpoint of the pole are (respec­
tively):
A +f2
2
'777777777777777777777777777
N stone is released from the top of a tower of height 15 m such that it covers 15 m during the 
third second. The average speed of the body (stone) during three seconds is:
25 . 20 ' .
(a) — m/s (b) — m/s (c) 5 m/s (d) 25 m/s
A train moves from rest with a uniform acceleration a. Attaining a maximum speed v it starts 
moving with uniform retardation a. Assuming s = total distance covered in the unidirectional 
motion of the train, its total time of journey and maximum speed are (respectively):
15;
16.
(a) (b)
(c) (d)
17.
18.
s
s
V
19.
v
X(b) yo
V V
*y y
134
14.
(c)
0
A glass ball is released from a point above 
graph/sis/are valid ?
v
G.R.B. Understanding Physics MECHANICS (Part-A) 
A bullet loses n'h fraction of its velocity while passing through a sand bag. The minimum number 
of sand bags required to arrest a bullet is:
1 1 , I /JX 1 + n2
(a) |-/j2 l-(l-l/n)2 n ( ) n
A particle moving in a straight line with a constant acceleration will be situated at a distance of 
12 m from a stationary point after four seconds from starting. It covers a distance of 6.5 m 
during 12th second. The average velocity of the particle during 5th second is:
(a) zero (b) | (c) > 1 (d) 2
Which of the following v-t graphs is not possible in practice?
(a) 
0
(b)
0
a
2(d)
0
a smooth horizontal floor. Which of the following
I
u 
---------- -t
Wo
A particle moves in a straight line so that t = Vx2 -1. Then a (acceleration) is equal to: 
1 1/2 /2 1(a) - (b) l-L. (c) (d) ij-L
x X x3 x3 x3 x2
Which of the following graphs representing one dimensional motion is/are valid? 
a
-----------»t
(d) o
.2
(d)
V
E B
A,
C
22.
23.
24.
25.
(b)
26.
(d) 60 s(c) 80 s
27.
28.
(b) v increases when 7-7 >0
(d) v decreases when | 7 | decreases
2a 
v
G/
I
135
+ 0x, where a and 0 are positive constants. The decel-
(c) 20v3
! ••
• • i » ••• 
! •' 
i • • 
i {• 
ilLt
*1 At1 l2At2
Water drops fall from rest from a nozzle from a height of 5 m at regular interval of time such that 
the first drop reaches the ground when sixth drop is released from the nozzle. The height of the 
fourth drop from the ground is: (g = 10 m/s2)
(a) 3.2 m (b) 0.8 m (c) l.8.m (d) 4.2 m
Which of the following option/s can be correct for a body having a zero acceleration?
(a) Velocity has same magnitude and direction
(b) Velocity changes its magnitude only
(c) Velocity changes its direction keeping the magnitude constant
(d) Force acting on the particle decreases
Which of the following option/s can be correct for a moving particle?
(a) a = 0, s * 0 (b) a * 0, v = 0
(c) o*0,s = 0 (d) s*0, v = 0
Three particles start from the origin simultaneously. The first particle moves with velocity V| 
along +ve x-axis, the second particle moves along the +y-axis with velocity v2 and the third 
particle moves along the line y = x. The velocity of the third particle so that all the three particles 
lie on the same line is:
(a) vt” (b) (c) (d) ^v>+v2
Two cars leave one after the other and travel with an acceleration of 0.4 m/s . Two minutes 
after the departure of the first, the distance between the cars becomes 1.9 km. The time interval 
between the departures of the cars is: 
(a) 190 s (b) 50 s
In rectilinear motion:
(a) v decreases when ~v- ~a 4, the velocity, displacement and acceleration 
are negative
A particle starts from rest whose a-t graph is given. The 
maximum speed of the particle (in m/s ) is: 
(a) 110 
(c) 550
When we project a stone:
v = +^ 
dt
d2y 
 —>
_vl + v2
2
Assertion-Reason Type Questions
(a) Both assertion A and reason R are true and R is the correct explanation of A.
(b) Both assertion and reason are true but R is not the correct explanation of A..
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
(e) Both assertion and reason are false.
A : The v-x graph of a particle is given in Fig.(a). Then, the a-x graph will be given as fig.(b).
a
(c) z>=|vf|/1+|v2|r2 
y
A: The point Q is pulled down with a speed v. The speed of the smooth ring ///// 
hanging from the thread is 2vsin—. 1 Rinya
R: The velocities each of magnitude v are added vectorically on the point P. \ / \Vr 
p v
dv a dv
R : -- = ~ and — is negative.dx v dx
G.R.B Understanding Physics MECHANICS (Part-A)
A man moves with a velocity during a time interval /]. Then he moves with a velocity 
£(J_ vf) during a time interval t2. Then:
(a) |a7| = 7v|2 +vj (b) s-IV] +v2/2 |
A particle moves in j'-axis as shown in cubic y-t graph:
(a) vA > vB (b) aA > aB
(c) vc = 0 (d) ac = 0
139
4.
5.
2A
t0
A
6. y
X
0 t = 0
7.
8.
v
9.
is negative and — is positive.R:
x
0
P
1.
Vav
over any time interval(q)(b) Uav
A : The path of a particle is given as y = /(x). The average veloc­
ity over first t seconds is parallel to the velocity of the particle 
at time t at P.
R: Displacement s is tangent at P to the path and always velocity 
is tangential to the path.
/p
A : A body can have acceleration while at rest.
R : The body is momentarily at rest while reversing the direction of its velocity.
A : Taking air resistance into account, time of ascent of a particle projected vertically up is 
greater then the time of its descent.
R : Air resistance favours g while ascending and opposes g while descending.
A : The acceleration at P is negative.
vdv . . . dv .
n — v I-'— — 
dx ’ ' dx
(^) ^ascent
(d) descent
Match The Columns
For a body projected vertically up with a velocity vq from the ground. Then,
(a) £ (p) Zero for round trip
-> -> 
vl + v2
2 
v0
(r) — over the total time of its flight
vo(s) 7
Straight Line Motion
A : A particle moves so that its coordinates vary with time as x = acosco/ and y = asin at. The 
trajectory of the particle is circular.
R : Two mutually perpendicular velocities when added, give rise to a circular motion.
A : v-t graph of a particle moving in a straight line is given. The v 
distance covered by the particle is 3A and displacement of the 
particle is A.
R : Distance = Sum of magnitude of positive and negative areas 
under v-t graph.
Displacement = Algebraic sum of positive and negative areas 
under v-t curve.
(vii) Slope of v-s graph s
,2
(viii) Algebraic sum of area under a-s graph =
I.
1
I
t(s)
0 1
2.
40 t(s)
i4
i
i 
i
G.R.B Understanding Physics MECHANICS (Part-A) 
initial velocity. Then, match the following terms:
(p) Speed
(q) Acceleration
(r) Area under v-t graph
(s) Velocity (w = 0)
140
2.
v (m/s)
1 ; -
A particle moves in a straight line with zero
(a) (slope of v2-s) graph
(b) Slope of v-t graph
(c) Slope of r-t graph
(d) y/2 (Area under a-s graph)
Comprehensions
Passage-1 (Graphs and Graphical Solutions)
Generally, we can solve the problems of straight line motion by using graphs, based on the 
following points:
(i) Slope of s-t (or r-t) graph = velocity
(ii) Algebraic sum of area under v-t graph = displacement
(iii) Slope of v-t graph = acceleration
(iv) Algebraic sum of area under a-t graph = change in velocity
(v) Sum of magnitudes of area under v-t graph = distance
(vi) Slopeof v2-s graph = 2 (acceleration)
acceleration
velocity
v2-u 
2
(ix) Slope of the line joining, any two points in s-t graph = vav
(x) Slope of the line joining, any two points in v-t graph = aav
Following the above points, answer the following questions :
The position-time graph of a particle is shown. If the particle was at x = 2 m:
(a) the particle was at rest for 2 s x (m)
(b) the particle moves forward with uniform velocity of 1 m/s in
+ x-axis and then moves in -x-axis with same speed
(c) the particle will be at distance of I m from the origin after 4 s
(d) the average speed of the particle is equal to its initial speed dur­
ing last two seconds.
The velocity-time graph of a particle is shown:
5
(a) the displacement of the particle is equal to ~ m
7
(b) the distance covered by the particle is equal to - m
(c) the average velocity of the particle is m/s over 4 s
(d) the motion is uniformly accelerated for 4 s
1 \
I
2 3
3.
v
*0
0
4.
t(s)0 J 32
5.
x(m)
0 !3
6. v (m/s)
2
x(m)
0
7. v2 (m/s)2
2
x(m)0 1
+ 
1
i 
i 
i 
i 
i
2
a (m/s2)
2
a (m/s2)
4 \
Straight Line Motion 141
The velocity-time curve is shown for a particle moving in a straight line. The particle moves with:
(a) uniformly decreasing acceleration
(b) unidirectional acceleration
• (c) gradually .decreasing acceleration
(d) unidirectional motion , '
The acceleration-time graph of a particle moving in x-axis with 
initial velocity = -3/ m/s is given. The velocity of the particle 
at t = 3s is:
(a) -/ m/s
(b) -2/ m/s
(c) i m/s
(d) zero
The acceleration - displacement graph of a particle moving in 
x-direction is shown. If the initial speed of the particle is 
u = -3i m/s, the velocity of the particle at s = 3 m is:
(a) J\3i m/s
(b) Vu i m/s
(c) —2>/3 i m/s
(d) 2>/3 i m/s
The velocity - displacement graph of a particle is shown. The 
acceleration of the particle when x = 0, is:
(a) -^-m/s2
2 ,2(b) “^m/s
(c) ym/s2
(d) none of the above
The v-x graph is given for a particle undergoing a straight line 
motion. The acceleration of the particle is:
(a) 2i m/s2
(b) li m/s2
(c) ~i m/s2
(d) -1/ m/s2
----------------- ► (
-1— 
1 2
s
CB
A
t0
when a = f(x)
when a = /(v)
when « = /(/)
I.
(a)
t
a
(c)
s
2.
(a) (b)
(c) (d)
with velocity as 
is/are correct?
v
142
8.
G.R.B. Understanding Physics MECHANICS (Part-A)
The displacement-time graph of a particle is given. Then:
(a) velocity is maximum at A
(b) the particle remains stationary at 0 and C
(c) the velocity of the particle at B is equal to its average velocity 
over the time taken from 0 to B.
(d) the particle moves in same direction
vdv use a = —
dx
dv vdvuse a-— or a = —
dt dx
dv use a = —
dt
Remembering the above points answer the following questions:
Which of the following graph is valid for a rain drop falling from a large height of 10 km, say:
a
(d) 
0
A particle moves with an initial velocity v0 in a straight line. The acceleration of the particle varies
a = -Kv, where K is a positive constant. Then which of the following graph/s
I
(b)
0
Passage-2 (Drawing Graphs using General Kinematical Equations)
As learnt in this chapter, we can find functions x = /(/),w = /(/), v = f(x),a = = /(*)
and a = f (v) by using the general kinematical equations modified for straight line motion followed by 
proper sign conventions. The main equations used are given below:
dxv = —
dt
dv vdva = — =----
dt dx
143Straight Line Motion
3.
(b)
x
v
(d)
t
1.
2.
3.
4.
5.
1.
2.
5.
6.
Displacement, Distance, Average Velocity and Average Speed
1. You move along + x-direction through a distance of 10 m and then move back through a distance 
of 4 m. If you repeat it four times during ten minutes, find the (a) total (i) distance (ii) displace­
ment (b) average (i) speed (ii) velocity.
3.
4.
a = Kv2 . Then, which of the followingA motor car moves in a straight track with a retardation 
graph/s is/are correct?
v
A car driver applies the brakes which retards the car at a rate of 8 m/s2. If the initial velocity of 
the car is 10 m/s, the speed of the car after 5 seconds will be
The ratio of magnitude of average speed and average velocity of a particle projected vertically up 
with a speed v0 = 10 m/s during 1.5 s from the instant of projection is respectively. 
The area under a-t graph gives
The time taken by a body to cover last metre of its path when it is released from rest from a large 
height h « R, where R = radius of earth, is 
The acceleration of a particle which moves in a straight line from rest is given as a = 6t. The 
average velocity of the particle over a time t measured from starting is
A stone is projected vertically down with a velocity v0. The magnitude of average velocity of the 
stone over the total time of falling a distance h is ........
Introductory Exercise
(c)
01
True Or False
The ratio of distance covered by a freely falling body released from rest during first and last half 
time of its descent is 1 : 1.
The average velocity of a particle moving in a straight line is equal to half of the vector sum of its, 
initial and final velocities.
The bodies of different masses are released from same spot. They will strike the ground simul­
taneously, if we ignore air resistance.
If the average and instantaneous velocities are equal, a particle is said to be moving with constant 
velocity.
Different bodies released from same position (or same altitudes) hit the horizontal ground with 
same speed ignoring air resistance.
Fill In The Blanks
(a)
o
I B
I
I
I
IA
3.
4. x(m)
t(s)0 10 20 30 40 50 60 70 80
5.
6.
7.
J. t(s)O
graph. Find the velocity of the particle at A, B, C 
and D.
40
30
20
10
D
4-
G.R.B. Understanding Physics MECHANICS (Part-A)
An insect crawls along the frame of wire AB with a constant speed v. Find the average velocity 
of the insect.
144
2.
Instantaneous Speed and Velocity
A particle oscillates obeying the relation x = A (1 - cos co/). Find the:
(i) velocity of the particle at t = 0,
(ii) maximum velocity of the particle,
(iii) displacement of the particle when its speed is maximum,
(iv) average speed over the time during which the particle acquires a maximum speed for the 
first time.
A particle moves in a straight line obeying the relation x = A sin cof + B cos co/.
(i) Describe the natureof motion of the particle.
(ii) Find the time after which the speed of the particle is maximum.
(iii) Find v0.
A particle moves in a straight line obeying the x-t X(m),,
A bird moves from the base of a tower of height 40 m due east during 1 minute covering a 
distance of 30 m. Then it flies in a straight line to the top of the tower with an average speed of 
5 m s. Find the average (i) speed (ii) velocity of the bird for its trip from the bottom to top of the 
tower.
A runner runs 40 m on a straight track in 30 s and 
then returns back to his initial position. If he takes 
a total time of 80 s for to and fro journey, find the 
average speed and average velocity for the total 
journey.
V
/Illi
1 2 3 4 5 6
145
8. y*
xOA B
9.
10.
11.
12.
x
tO
13.
14.
15.
16.
T 
Ax 
1
Straight Line Motion
A particle connected with a spring oscillates between A 
and B about the mean position O. Find the (i) velocity at 
t = |^s (ii) distance and displacement of the particle 
during t = |^s, if the motion of the particle is given as 
x = 20 cos 10/ cm.
Average and Instantaneous Acceleration
A car moving with a speed v = 12 m/s applies its brakes so that it stops within a time interval of 
A/ = 3.6 s. Find the average acceleration of the car. Can you use the relation v = u + at?
A raindrop falls obeying the relation v = v0(l-e-a') from a large height. Find the velocity and 
acceleration of the raindrop at (i) t = 0, (ii) t -> oo.
The position of a particle is given by x = At2 + Beat. The particle is at x = - 1 m, t = 0 moving 
with a speed v = 2 m/s.
(i) Find the value of B and a.
(ii) What should be the value of A so that the initial acceleration of the particle is zero?
A car starting from rest accelerates such that x-t graph is parabolic. Find the acceleration of the 
car.
A vertically moving glass bail loses n = 40% of its initial speed v0 = 20 m/s during the collision 
with the horizontal floor. If its average acceleration during the time of its contact A/ with the 
horizontal floor is 80 m/s2, find A/.
A body moves in a straight line obeying the relation a = A - Bv.
(i) Find v = /(r) assuming at/ = 0, v = vo-
(ii) Describe the nature of motion of the body.
(iii) Draw v = /(/).
A machine changes the speed of an object moving in a straight line according to the relation 
v = voea', where vq=5 m/s and a = 1 /s.
(a) What is the displacement of the object in first second?
(b) Find the initial acceleration of the object.
An object moves with an acceleration which depends on time as a = -ky[t, where k = 3/2. If at 
t - 0, the speed of the object is v0 = 8 m/s, after what time the object will stop?
ti t2
G.R.B. Understanding Physics MECHANICS (Part-A)
17.
s(m)
O 1 4
18.
V(m/s)
t(s)
19.
20.
21.
22.
23.
24.
A car moves in a straight track as per the v-t graph.
(i) Find the average velocity of the car over A/ = 8 s.
(ii) What is the average acceleration over the period from /] = 1 s to = 7s?
146.
Graphs and Graphical Solution
The a-s graph of a body moving in a straight line is given. If the speed of the body at t = 4 is 
4 m/s, find the initial speed of the body.
afm/s2) ■ ‘
3| = ^fl1a2 A/.
ti tg 6s
A sand bag is dropped from the top of a tower of height h = 80 m. An archer standing at the base 
of the tower shoots an arrow vertically up aiming at the sand bag, after 1 s the bag is dropped. If 
the initial velocity of the arrow is 32.5 m/s, when and where do the arrow hit the bag?
Problems
A particle of mass m is moving along positive x-axis under the action of a force F = —-, where 
2xl
A is a positive constant. If the particle is at a distance x0 at t = 0, find the (a) velocity when 
x = 2x0 (b) time taken to reach x = 2x0.
A car while moving with a constant acceleration from rest, covers the distances a, b and c during 
time x01, /b and 2th seconds respectively, from starting. Prove that
a(y-z) + 6(z-x) + c(x-y) = 0.
Two bodies are projected vertically up with a velocity v0 from a point at a time interval t. When 
and where do they meet? What are the velocities of the bodies just before meeting ? Take neces­
sary assumptions.
A particle moves such that its velocity is given as v = 3(/2-/). Find the average (a) speed
(b) velocity (c) acceleration, during two seconds from starting.
A sleeve slides along a smooth rotating horizontal rod such that its acceleration varies with the 
d2xposition x along the rod as —- = kx, where k is a positive constant. If the sleeve starts from 
J
t = 0
L_
B
t = 0
_L
A
G.R.B. Understanding Physics MECHANICS (Part-A)
A body when projected vertically up covers nQ1 fraction of total distance (during its ascent) in last 
second. Find the (a) maximum height attained (b) velocity of projection.
A stone is hanging by a thread from a balloon which is moving up with upward acceleration a. 
Find the total time of motion of the stone when the string is cut after a time /0 from starting of 
the balloon from rest.
A particle is projected vertically up from ground at t = 0 and reaches a height h at t = T. Find the 
maximum height attained by the particle.
A Diwali rocket ascends with an upward acceleration 20 m/s2 from rest at t = 0. The fuel gets 
exhausted at t = 10 s and it continues to move up. Find the:
(i) maximum height attained (ii) total time elapsed
(iii) total distance covered
Two stones are simultaneously thrown vertically up from the top and bottom of a pillar of height 
20 m, with speeds 10 m/s and 20 m/s respectively. When and where do the stones meet?
A coin is dropped from the roof of an elevator which moves with an upward deceleration a. If it 
strikes the base of the elevator after a time t0 with a velocity v0, find the average velocity and 
average speed of the coin.
A car applies its brakes when its speed is vq.. It stops after covering a distance x with a constant 
deceleration a. Find the average velocity of the car over a time interval between its application of 
brakes and stop.
A particle retards from v0 with an acceleration a = -kt, where k is a constant. Find the total 
distance covered.
A particle moves in x-axis such that x varies with t as at x = 4? - 2t + 5 . Describe its motion.
A stone is dropped from a bridge. Just after 3.63s from dropping the stone, a person on the 
bridge hear the splashing sound of the stone. Find the height of the 
bridge. •
A man at A starts accelerating from rest with a = 2 m/s2 due east, 
at t = 0. Another man at B situated at a distance of 12 m behind the
man A starts moving with v = 8m/s at / = 0. Find the distance p| {2m
2000Referring to the v-s diagram of a particle, find the displacement 
of the particle during last two seconds.
Straight Line Motion 149
22. 3
x(m)500
v (m/s)
23.
s(m)500
v (m/s)
24. 20 m/s
t(s)
25.
t (s)0 5 15
26.
10
5
s(m)0
■
150
30
25
Referring to v-t diagram, find:
(a) the time when the velocity of the particle becomes zero
(b) the time when the particle crosses the origin.
i i
i i 
■ i
200 300
Referring 
v0 =0.
a (m/s2)
a (m/s2)
O 
-10 m/s
Referring to a-t diagram, find the distance covered by the particle during first 10 s. Assume 
v0 = 0.
a (m/s2)
a-s diagram, find the time taken by the particle to travel 250 m from starting. Assume
6
Referring to a-x graph, find the velocity when the displacement of the 
particle is 100 m. Assume initial velocity as zero.
/ : \ 6070
-4—I—I I —I-
1O2O3Q4O5P?SJ
Referring to v-s diagram, find: 20
(a) acceleration of the particle when its velocity becomes half of the 
initial velocity
(b) total distance covered by the particle
(c) average velocity of the particle during its total time.
2
0
28.
s(m)
29.
v (m/s)
15
25
30.
10
s (m)
0 40
diagram, find the total time taken to cover first 200 m. Draw the a-s, v-t and a-t 
v (m/s)
G.R.B. Understanding Physics MECHANICS (Part-A)
Referring a-t diagram, find the time when the particle stops.
I50
27.
I 
I 
I 
I 
I 
I 
I 
I 
I 
I 
-4- 
250
50" 
40-• 
30-- 
20-- 
10"
/i
f i
i
i
i
i
i
i
-4-
100
40------- yr
i 
i
i 
i
i 
i
i 
i 
i
Referring to v-s 
cover first 500 m.
a (m/s2)
a (m/s2)
Referring v-s 
diagrams.
0
diagram, draw a-t and a-s diagrams. Find the time taken by the particle to
Referring to a-s 
v0 = -5 m/s.
0
diagram, find the velocity of the particle when the particle just covers 30 m.
I 
I 
I 
I 
I 
I 
I 
I 
I
-4—>s (m) 
100
151
31.
2
t(s)
50
32.
t(s)
20
33.
t(s)
0 1
34.
35.
36.
(a) Prove that
0
Referring to v-t diagram, find the:
(a) maximum speed (b) average speed of the particle
(c) accelerating time and retarding time of the particle.
v (m/s)
ft, = 30° 
-------- T0 
Referring to a-t diagram, find the distance covered by the particle during 2nd second of its 
motion. Assume at t = 0, v = 0.
a (m/s2)
Straight Line Motion
Referring to a-t diagram, find the velocity of the particle at t = 20 s. Assume that the particle was 
at the origin at t = 0 and was moving with a velocity of 10 m/s in -ve x-direction.
a (m/s2)
e2 = 60?
-3f
The acceleration of a particle varies with time as a = 2t+1t- 21. Find the speed of the particle at 
t = 5 s, if the particle was moving in -x-axis with a speed of 20 m/s.
Two cars start moving simultaneously from same spot with velocities V| and v2 and travel 
unidirectionally with uniform accelerations a\ and a2 respectively. Find the distance covered by 
each car before they meet again.
A particle moves in a straight line and travels through a distance d\ with constant velocity V| for 
a time interval /|. It then suddenly changes its velocity to v2 and moves with that velocity for a 
time interval /2 travelling through a distance d2.
V]t/| +V2 V]/| + v2/2 
t/| + d2 /] +t2
(b) Under which condition will there be an equality?
G.R.B. Understanding Physics MECHANICS (Part-A)152
ANSWERS
29. (d)
8. (e)7. (a)6. (d)5. (a)4. (c)2. (e)
8. (a,b,c,d)7. (b)6. (b)5. (c,d)3. (b,c,d) 4. (d)
3. (a,c)
5. (T)4. (T)3. (T)
3. Change in velocity2. 5:3
,2 6.5. t4.
(b) (i) 5.6 m/min, (ii) 2.4 m/min1.
3.
(iiiM(ii)co/4(i)05.
m/s, 0 m/s, tan (1), 07.
13,0.4 s
19. (>i)
1. (d)
9. (b)
4. (a)
12. (c,d)
20. (a)
11.
15.
17.
7. (a)
15. (b)
23. (a)
8. (a.b)
16. (a,d)
24. (a,b,c,d)
tan
(i)-l,-2(ii)2
(a) 5(e -1) m 
2 m/s
^2 gh + yjlgh' 
kt
(b) 5 m/s2
, 2cozl 
(,v) —
9. - 10/3 m/s2; No
-(vo + 7vo +2g/»)
Multiple Choice Questions
2. (c,d) 3. (b)
10. (c)
17. (a.b) 18. (c)
25. (b) 26. (b)
33. (a,b) 34. (b)
11. (b.c)
19. (a.d)
27. (a,b,c) 28. (a.b)
35. (a,b,d) 36. (b,c,d) 37. (a,b,c) 
Assertion-Reason Type Questions
1. (a) 2. (e) 3. (b)
9. (a)
Match The Columns
1. a-p, q; b — r; c — s; d — s
2. a - q; b - q; c - p, r, s; d - p
Comprehensions
Passage 1
1. (a,c,d) 2. (a,b) 
Passage 2
1. (a,b,c,d) 2.(a,c)
True or False
1. (F) 2. (F)
Fill in the Blanks
1. 0 m/s
g ■
V2s
Introductory Exercise
(a) (i) 56 m, (ii) 24 i m
(i) 1 m/s (ii) m/s
5. (a) 6. (d)
13. (b) 14. (b)
21. (a,b,c,d) 22. (d)
30. (a,b,c,d) 31. (a,b,c)32. (a,d)
38. (a,b,c)
Straight Line Motion 153
21.
(ii) 30 m/s25.
5. (c) 3
(b) (c)7.
9.
13. 2 s, top of the pillar. 15. —11.
17.
'6
19.
.2 (c)0(b) 50 m
29.43255 m approximately25.
22 m/s31. 35.
♦ ♦♦
21.
23.
' 27. (10 + 2V10)s
2(vi - v2) (viq2 ~ v2al) 
(«1 -«2)2
37
33. -m
(b) I
3yo 
n 
(b) gn
^ands = ^ 
a
x = -^sin>/Fr 
4k
2v • 
o
(i) 20 m
Problems
(a) 1.5
(a) ~^=
64k
(a) gV2w
(gT2+2h)2
8gT2
The particle moving in negative x-direction reverses its velocity at / = -!=• and moves in positive 
v6
x-direction, with an acceleration given by a = 24/
B overtakes A at / = 2s, A overtakes B at / = 6s
SB = 16 m, SA = 4 m during first 2 seconds
SB = 48 m, SA = 36 m during first 6 seconds
240 m
(a) -4 m/s'
15 , 8) — + ln- s 
.8 3;
23 a = 2^h~v2
2(H-h)
v0
2-
CHAPTER
relative Motion
kJ
Hence we can write
Introduction
Galilean transformation of velocities
Motion of swimmer (or boat) in water
Relative Angular Velocity
Equation of trajectory of one particle relative to the other
Time of meeting of the particles
Relative acceleration
Application of kinematical formulae in moving reference frames
Let us present the following familiar experiences of relative motion.
When you look outside a moving train, it appears as if the trees, electric poles, bushes etc., are 
moving behind. If you spin your body while standing, you can see the surrounding objects revolving 
around you opposite to the sense of your rotation. Similar to the last example is the “motion of sun when 
viewed from earth”. Sun moves while earth is stationary. This was,indeed, a common belief of the 
people before Galileo. It is evident from this fact that, even the scientists and philosophers were bewil­
dered by the relative motion between sun and earth. Then what to speak about the common man! Even 
though the relative motion is purely practical, its physical (psychological) effects very often deludes us
StudyQ^Points
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
§4.1 Introduction
In chapter-2, we defined the motion of an object relative to an observer as the change in its 
position vector relative to the observer. Let us assume the object as A and observer as B. When the 
position vector of A relative to B, that is, rAB changes, we say that, “A moves relative to B". The rate at
* t/r —>
which rAB changes, that is, *s defined as velocity of A relative to B which is denoted by vAB
M - d,
rAB
VABfAC
fBC
drAC
= velocity of A relative to C = v^c
Hence, the motion of an object relative to different observers is different so long as the observ­
ers move relative to each other (yBC *0). In the other words, same object appears to move 
with different velocities when viewed by different observers. Hence, the nature of motion of an 
object depends on the motion of the observer. With respect to one observer, the object may 
appear stationary, whereas with respect to another observer it seems to be moving. This con­
cludes that “motion of an object is relative”. Hence, when we talk about the motion of a particle, 
we must mention the reference frame (or observer). In this sense, motion of any object in the 
universe is not absolute.
Relative Motion 155
in our day to day lives. For instance, a passenger in “Delhi Express” asks another passenger, “Sir, when 
Delhi will come?” The second passenger answers “Delhi will come after one hour”. In fact, Delhi is not 
coming (relative to ground), rather train is going towards Delhi. Then, the question posed by the passen­
ger may also be correct relative to the train. When we drop a stone from a uniformly moving train, it 
appears to fall vertically straight (to us) whereas a stationary observer on the ground sees the stone 
moving in a parabola.
From the above examples it is evident that, the nature of motion of a particle depend upon the 
reference frame. Relative to different observers (reference frames) the motion of a particle may be 
different.
vAC=vAB+vBC
Jhe above expression tells us that, simultaneously different observers (B and C) will record 
different velocities ( and v^c ) of the object (A).
Substituting =
-> dt
dfnr —>
and = velocity of B relative to C = vBC, we have 
dt __________
4*
Fig. 4.1 The motion of an object
A viewed by the observers B and
C from the reference frames 
S' and S respectively.
‘s’*
§ 4.2 Galilean Transformation of Velocities
In last chapter we discussed the motion (velocity) of an 
object relative to a fixed reference frame say 5. Let us now find 
the velocity of a particle A observed from a moving reference 
frame say S'. Let us assume that the reference frames are only 
translating but not rotating (not changing the orientation of the 
coordinate axes). Furthermore we assume that, observers B and 
C are fixed with the reference frames S' and 5 respectively. Let 
the observers B and C measure the position vectors of A as ~?AB 
and 7^c respectively. If the position vector of B relative to C is 
r£c , following the triangle law of vectors, we have
—► —> —>
rAC = rAB+rBC
Differentiating the above expression with time we have 
drAC _drAB , drBC 
dt dt dt
velocity of/1 relative to B = vAB,— 
dt
/Vxj ... . " V*C
156
Sol.
x
Ans.
VAC
Fig. 4.3
Student Task
C will observe the ball as moving vertically upwards?
> Answer the following questions with reference to the previous example'.
(a) In which direction should the man B throw the ball relative to himself such that the man
Fig. 4.4
v 
/?7277ZZzZ^Z/%/
VAC ~ VAB + VBC
Substituting = uj and vgC = vi, we have
vxc =vi +uj.
Relative to B, motion of A is a straight line whereas relative to C, the motion of the ball is curved 
(parabolic). Initially C observes the line of motion of J inclined at an angle
= tan-1 y with horizontal. In the above example, we should not assume
that C is stationary with ground; vcc * 0.
Ans. 0 = sin ’ — with vertical in opposite direction of motion of the cart. 
u
(£>) If the man C moves relative to ground with a constant speed v' towards left, in'which 
direction should the ball move relative to the man C, such that an observer standing on 
the ground will see the ball moving vertically upward?
I E oAi -»■ a
Is >vbc = vi
........
Fig. 4.2
G.R.B. Understanding Physics MECHANICS (Part-A)
Let us discuss few examples to use the above law of Galilean transformation of velocities.
Ex. 1. A man B standing on a cart throws a ball A vertically up with a velocity u relative to himself 
while the cart is moving rightwards with a velocity v relative to another man C. What is the 
velocity of the ball relative to the man C?
As the cart moves relative to C with a velocity
~v towards right, ~vBC =vi. Since, the man 
B on the cart throws the ball vertically (with 
respect to himself) with a velocity w,
= uj, the velocity of the cart relative to 
the man C is given as
Ex. 2. A rat is moving down the slant of a wedge of angle of 
inclination 0, with a velocity v , as shown in the 
Ffg. 4.4. If the wedge moves towards left with a velocity 
u, find the (/) velocity of the rat relative to ground, 
(ii) value of 0, if the rat moves vertically downward 
relative to an observer G fixed with the ground.
Ans. 0 = sin"'4, (with vertical) 
v' 1
C a
S’ uuzzzwzzzzzzz/zzzzzz/zzzwzz/7/
157
Fig. 4.5
Ans.
Fig. 4.6
Sol.
rr^rr. 
Fig. 4.7
> Referring to the above example, if the wedge is pushed towards right with a constant velocity 
~u , find the magnitude and direction of velocity of the rat relative to ground; assume u = v.
o 0 0
Ans. 2vcos- at an angle of ~ with vertical.
Ex. 3. (Man in a rain) Applying the Galilean law of transformation of velocities, explain the varia­
tion of velocity of rain relative to a man moving relative to ground.
Let a ground observer G record the rain falling with a velocity ~v . When another observer 
(say the man M) moves with a velocity vmg relative to ground observer 
G, he will observe the rain falling with a velocity ~vrin = vrg-'v„g accord­
ing to the Galilean law of transformation of velocities. Since ~vyg is as­
sumed as a constant (for the rain falling at a constant rate), the man will 
record different velocities of the rain if he moves with different velocities 
(with respect to ground).
vrg 3 vrm, + vmfll 
Fig. 4.8
vRC = (v cos 0 - m)2 + (v sin 0)2
= fu2 +v2 -2wvcos0
(ii) If the rat appears to move vertically downward relative to G
VRGS = vcos 0-w = 0
This gives 0 = cos 
can find 0.
Student Taskjf
Relative Motion
Sol. (i) The velocity of rat R relative to the ground frame G is given as
vRG=vRB+vBG
according to Galilean transformation of velocities, where ~vRB = v and v^c = ~u (given).
Then, 'v'RG = ~u + ~v. B
Substituting v* = vcosG i - vsin0 j and
~v = -ui in the above expression, we have 
~vRG =(vcos0-w)i -vsin0y 
Hence
; 1 Y • Alternately, referring to the vector diagram we
Ans.
G.R.B. Understanding Physics MECHANICS (Part-A)158
vrm2
■vrm3
Sol.
This yields
Mdappear to fall vertically with v* =~vrg~ ^mg2 35 shown in Fig.
4.9. If the man increases his speed further, he will see the rain 
falling with a velocity
= tan 11 —
12
>Vmg?
^zzz/^zZ/Zzzzzzz/zzZ
vrffl=vr"vm-
zzz2 —>
vrmj = vrg “ vmgj
as shown in Fig. 4.10. From the above three cases we under­
stand that sometimes the man observes the rain falling forwards, 
sometimes vertically downwards and sometimes the rain ap­
pears to fall backwards. Hence, the velocity of rain relative to 
man depends upon the velocity of man relative to ground ac­
cording to Galileantransformation of velocities.
jh
vrm2
~vr = -2j and ~vm = 1 z, 
vrm =-G+2y)m/s
 = tan-11 —
I Vr
vmg3
vrg = vrm3 * vmfl3
Fig. 4.10
For different velocities of man, the rain appears to fall with different velocities (to the man). 
However, the velocity of rain (relative to ground ), that is, the vector sum of velocity of rain 
relative to man ~vrm and velocity of man relative to ground is a constant; that means, 
~b = bm, + S = \m2 + % = ^rm2 + bn, = constant, as shown in the figures 4.8, 4.9 
and 4.10.
* vmg3
* vmg2
—> —>
= bg “ bng|
Let us discuss more about the above facts through the following example:
Ex. 4. Imagine yourself in a rain steadily falling vertically with a speed of 2 m/s. If you start moving 
with I m/s due east, in which direction should you hold the umbrella to protect your self from the 
rain in best possible way?
Most practically, we hold the umbrella in the direction of motion of the rainfall jelative to the 
man (vrm). Theretbre, we need to find ~vrm which caii be given as
Substituting 
we have
Fig. 4.9
If the man speeds up, at a particular velocity , the rain will
Vrg=Vrm2
Relative Motion 159
vm
vrm Vf
Fig. 4.11
Then, the observer B fixed with the ground will
vc = u + vvcb = v
The umbrella should be held in the direction of rainfall relative to the man (running in the rain). 
In other words, we should hold the umbrella in the direction of 7^, = v^ -~v^ . When the man 
is stationary, ~vrm =\y, which is a special case.
That means the umbrella should be held at an angle $ = tan 
direction as shown in the fig. 4.12.
with vertical in forward
Aus.
_1_
2
Fig. 4.13 The observer A records the 
velocity of the cockroach w.r.t. the board 
7cb whereas the observer B fixed 
with ground records the velocity of 
the cockroach relative to the ground, 
vc = vcb + vb
Fig. 4.12
Student Taskf
> At 12 noon we hold the umbrella vertically when walking in a sunny {not rainy) day. If we run 
at certain speed, do we need to incline the umbrella? Explain.
Ans. No, the velocity of light does not depend on the reference frame according to 
Einstein’s special theory of relativity.
§ 4.3 Motion of Swimmer (or Boat) in Water
Let us assume a cockroach crawling on a stationary cardboard. Let an observer A fixed with the 
board measure the velocity of the cockroach as ~v . 
also record the same velocity ~v of the cockroach.
Let us now move the board with a constant velocity 
~vb (= ui, say). Even then the observer A records the same 
velocity ~v of the cockroach. That means, the velocity 
ofthe cockroach relative to the board v^b does not depend 
upon the motion of the board because it is the velocity of 
the cockroach measured by the observer A fixed with the 
cardboard. However, the observer B will record a different 
velocity of the cockroach. According to the Galilean 
transformation of velocity, the velocity of cockroach is 
given as vcg = vcb+vbg as shown in the Fig. 4.13 where 
vcb = V and = u .
From the above experiment we note that,
160
y
»x
we have
1. The velocities of boat relative to moving water and still water are same which is given as 
7^,, . It means, the person fixed with the water frame observes the boat moving with 
velocity vbw .
2. Since ~vb =7^,1,+Y,., when there is no relative velocity between the reference frame (water) 
and ground we have vb = vbw which is a special case. Hence vbw should not be confused
—>
with vb because water moves relative to ground.
3. Velocity of the boat will be different for different directions of heading of the boat. That 
means, the boat will cover different distances relative to ground in different directions, 
during a definite time interval in a running water.
4. Since the boat moves with a constant magnitude of velocity relative to water (vbw = 
constant),when viewed from water the boat seems to cover equal distances in equal time 
intervals, in any direction.
Let us use the above concepts in the following example:
b* % = %w * *w where 1%J = 
magnitude of the velocity of the 
= vwi in boat relative to water.
vbg = vbw
v’bw
Vb =^bg = (Vbw C0S 0 + Vw) 1 + vbw sin QJ
The above expression teaches us the following facts :
G.R.B. Understanding Physics MECHANICS (Part-A)
Velocity of cockroach relative to the board \.b is essentially different from velocity of cock­
roach (relative to ground ) v>cg when the board is moving. However, both ~\%b and v*cg are 
equal when the board is stationary relative to ground (v*cg =0). This may be a special case.
Hence, if we maintain v*cb as a constant ~?cg depends upon ~vbg .
Using the above ideas, let us now discuss the relative motion between the boat (swimmer) and 
water. For this purpose, let us imagine the “board as the surface of water” moving with a velocity 
(velocity of water relative to ground).
Furthermore, we can imagine the cockroach as a boat
rowing with a velocity ^bw relative to water. When the water 
moves with a velocity ~vw , the velocity of the boat relative to 
ground can be given as
="^M,+~?wg according to Galilean transformation of
......5
~vbg =~Vbw+\g according to Galilean transformation of 7 vw
velocities. V
Let the boat head at an angle 0 with the direction of water jjhe y^jocity of boat is given
boat relative to water flow. Then, ~vbw = vAw cosOz + vbw sin 9/.
Substituting = vbw cosQi + vbwsmQj and vwg
+
Sol.
Since
1 OA
where
and
Ans.Hence
Student Task
When we observe from the cork which does 
not move relative to water (reference frame 
fixed with water), the boat seems to move to 
and fro with equal speed. That means the boat
(a) Line of motion and line of heading of the boat:
Referring to the previous section, the velocity of the boat is
= (vAw cos 0 + v„) / + v6w sin 0 J-
----------------- d —
Fig. 4.16
B
O~*vw O-*vw
Fig. 4.15
?bw
man
1 
appears to cover equal distances «/ = v6h,/i in cork 
both upstream and downstream when viewed 
from the cork (water frame).
\boat4
k
jXboat, 
vbw
Relative Motion 161
Ex. 5. Let us consider a boat which moves with a velocity vbw=5 km/hr relative to water. At time 
t = 0, the boat passes through a piece of cork floating in water while moving downstream. If it 
turns back at time t = , when and where does the boat meet the cork again? Assume r( = 30 
min.
Since the velocity of boat relative to water has a constant magnitude, it takes equal times in 
moving from 0 to A in downstream and from A to B in upstream, before meeting the cork.
That means toA-lAB
tOA=h (given) we have tAB=t}
Then, after a time t0A + tAB = /j +t} =2tx) the boat 
will meet the cork.
During this time the boat moves a distance
D = OA + AB,
OA = (vbw+v„)tOA = (viw+vw)z1
M = (vbw - vw)tAB = (vbw -vw)t}.
30D = 2vbwt} = 2x5x— = 5 km
> In the above example if the cork moves a distance d = 2 km (/) what is the speed-of the cork 
(water)? (//) find the ratio of distances covered by the boat in upstream and downstream, (Hi) 
find the average speed and average velocity of the boat till it meets the cork.
7
Ans. (i) 2 km/hr (ii) - (iii) 5 km/hr and 2 km/hr respectively
From the above example we note the following point:
162
+ 2vbw v„ cosGvb
1 Water current
4> = tan 1
y
vb “*
d
x
vw
>---- X
as obtained earlier.
x -
d 
‘m = — 
____ vm
,2 w
d
vbyi>
Line of motion 
l/'B
(v6w, cosB + vJt/ 
v6w sin 0
This yields
The boat moves relative to ground along the line AB with a 
velocity as shown in the Fig. 4.17. Hence this line can be termed 
as “line of motion” of the boat. The boat moves in the line AB 
making an angle I with the direction of flow which can be given as A
Vbw sin 9 v
COs9 + V,J
(c) Drift of the boat: The distance covered by the boat along the bank 
of the river is known as “drift”. The component of velocity of the boat parallel 
to the bank, that is, x-component of is responsible for displacing the boat 
along the bank of the river. Hence, the drift that is, x-displacement of the boat is 
d 
vbw sin 9
x = vbxt where vfcx = vbw cos 9 + vw and t = 
This gives
G.R.B. UnderstandingPhysics MECHANICS (Part-A)
Line of heading
________ D/
vbw
vby = vbsin0 
= sin 0
Fig. 4.18
tf-—vbx
The boat is drifted with 
a speed vbx 
Fig. 4.19
Hence the boat moves with a speed
vb = 7(v6wcos9 + vh-)2+(vAi 
= \lvbw +v‘
,w sin 9)2 ■
vb = vbw + vw
Fig. 4.17
While moving along the line AB as seen from the ground, the body of the boat is oriented parallel 
to the line AD. In other words, the boat heads along the line AD passing through the length of the boat. 
Hence, we call the line AD the “line of heading of the boat” and the corresponding angle 9 is called 
“angle of heading”. In fact, the boat appears to move along the line AD when viewed from the frame of 
water. That means, the “line of heading” is the “line of motion of the boat relative to water.”
(b) Time of crossing: Since ~vb = (vbw cos9+vu,)/+v6h, sin 9 j, the 
y-component of ~vb, that is, vby is given as vby(= vbw sin 9). Since v^ >s 
responsible for taking the boat across the river, we can write d = vbyt, where 
d = width of the river and t = time taken by the boat to cross the river.
Then the time taken by the boat to cross the river is
d t =---------
Vbw sin 9
For minimum time of crossing the river, 9 = 90°. Then the boat will 
cross the river in minimum time given as 
163Relative Motion
vw): When the drift is zero, x = d = Q
Hence the angle of heading is
Substituting p = sin
water flow
Fig. 4.20
d. That means, the
COS94-Vw 
sin 8
This yields cos 0 = - —. Hence, the boat must head at an angle 
vbw _______________
v^sinS vb„
boat can not reach the right opposite point-. However, it can get the nearest point to the right opposite 
point on the opposite bank of the river.
-, we have x =Substituting 9 in the expression x =
„ n . .
0 = —+ sin
2
-11 VH- L —, we have
Case I (vbw
,->2k 
yiw
Case ll (vbw vw for zero drift
y
* w’^ direct>on flow.
.2 _„2
-——d
x_ m
2. (a) Forviw >vw, the boat moves with a drift (x-displace- 
ment) of magnitude zero when it heads at an angle of
{vbw)
iw > vw
of flow of water; for this,
Then, the time of crossing the river for the minimum 
(zero) drift is
d
Vb Jvbw-vl ----
Hence the boat will attain the right opposite point
Q if it starts from the point P.
(b) For vbw , its velocity relative to an observer 0 who is moving 
with velocity vq is given by
> Referring to the above example, find the average speed of the man in each case in terms of 
(vmw ~ vw ) 
vmw
AnS-
'mw + vw and when he moves in
'mw~vw Hence the total time for to and fro journey is 
d d . t d d ---- - =----- and /UD = — =--------- 
v/n, vmw+------------- vm2 vmw ~ vw
ffd fmw
Vrmv ~
Comparing the above expressions of 7j, T2 and 7} we have
T22=7jr3
Ex. 6. A man swims with a velocity vm. in still water. When the water moves with a velocity ( where /(jown =
166
PP
vo
relative to 0.
...(ii)We can write
®P0
Sol.
A
Then,
Fig. 4.23
Ans.
>vba(= v) 
BA1 (= v cosG)
P'
J
_ vPOl
R
The direction of angular velocity is given by the right hand thumb rule as discussed in the section 
5.4 with respect to 0.
Ex. 7. The angular velocity of a particle moving in a circle relative to the centre of the circle is equal 
to co. Find the angular velocity of the particle relative to a point on the circular path.
m = Vbai
BA AB ’
where vflyvpoVVpo-
~PO
; o/
d® _ vpol 
dt R
Eliminating from eq. (i) and eq. (ii), we have 
dt __________
Substituting = coB0, we have
R
0
P moves with vp0 (■ vp - v0) 
relative to O
(ii)
ri rPO
Fig. 4.22
Then, the particle P appears to turn around 0 with an angular velocity
G.R.B. Understanding Physics MECHANICS (Part-A)
That means, when observed from O, particle P seems to turn about O with a velocity vPOi_ 
which is the component of ^vP0 perpendicular to the line joining between P and O. In other words, 
vpoi J-
v v 
=~AC~2R
1 / 1 X
f ' 12
1
Ans.
1.
1
Fig. 4.26
2.
VI2, = vl,~%
= -V! sin 0| j - v2 sin 02 j
*2l/= *21 ) 
^2
' *17
1zy
Fig. 4.25
Relative Motion
Alternate procedure :
The point 1 seems to turn about the point 2 in anticlockwise 
sense as vjj is directed in negative ^-direction as shown in the 
figure. Similarly the point 2 seems to turn about the point 1 with 
same angular velocity as shown in this figure.
Hence, co^| =co^2 but = -vl2. However, you can use the 
expression co2i xr2i =V2I in general.
You may be tempted to write (i^2 = ©[ - co2 following the relation 
V|2 = V]-v2, which does not have any sense.
CO21
r21
. dQ where — = (0nj
dt UA
Substituting | vl2y |, we have
V| sin 0] + v-> sin 02 
°12-------------—-------
C0|2=—)
1. The angular velocity of a moving point relative to two different fixed points may be different 
whereas the linear velocity of a point relative to any stationary observer (point) remains 
constant.
t/0 dangle made by the position vector BA. Hence 
dQ ‘
— = rate of change in angular position (or magnitude of angular velocity) of the position 
vector of B relative to A(rBA).
Ex. 8. Two particles 1 and 2 move with velocities V| and v2 making the angles 0, and 7 
02 with the line Joining them, respectively. Find angular velocity of 2 relative to 1. 
Let us find the relative velocity between the particles perpendicular to the line 
joining them, the velocity of 1 relative to velocity of 2 is
?2 =^-^2
 = 20 
d$, , dQ
®BO = 2® BA
G.R.B. Understanding Physics MECHANICS (Part-A)
(o)
Fig. 4.27
(d) loj-ozl(C) Ans. (c)
path of A
VA
,C
path of B
VB
B
''abT'V/k
,..(i) B
...(ii)
168
Student Task^
B
The paths of A and B 
relative to ground and C is the 
point of collision of A and B
Fig. 4.28 (a)
A
Let us consider two particles A and B moving with constant 
speeds vx and vB in arbitrary curved paths. If they meet at C, how do 
you find the time of their meeting measured from the initial positions?
For the sake of simplicity, at t = 0, let the distance of separation 
between the particles be r0. After a time t, the relative distance of sepa- 
—> —> —> —> —>
ration between the particles is given as rAB = rA-rB where rA and rB 
are the position vectors of A and B respectively at time t. Differentiat­
ing both sides, we have
> Two satellites 1 and 2 are orbiting with angul ir velocities ©| and 
©2 relative to the centre of earth respectively, n the same plane. If 
the radii of their orbits are q and r2 respectively, the angular veloc­
ity of 1 relative to 2 is numerically equal to,
/it°|-j-/2t°2 (Z,) rlr2 I fl>2 ~ s component of ~v perpendicular to 
the line joining the particles, we have 
rdQ = v_l dt
Dividing eq. (i) by eq. (ii), we have
r vj.
^AB =VA-~VB
That means when observed from B, A seems to move with a 
velocity of magnitude vAB(= vsay) tracing an arbitrary curved path as 
shown in the Fig. 4.28 (b) and (c). In other words, A approaches B 
with a velocity reducing the distance of their separation by dr during rAB
an elementary time dt.
Hence
|©2r2 ~(Oiri I 
r2-n
§ 4.5 Equation of Trajectory of One Particle Relative to the Other (Optional) 
A
B at rest and A appears 
to move (relative to B) 
with a velocity vAB
Fig. 4.28 (b) 
A
/ / path traced by
/ / h relative to B
8 A approaches B with velocity 7
and turns about B with a velocity , where v ■ vAB 
Fig. 4.28 (c)
169Relative Motion
A
This yields
iv
0
Hence
B
we have, r0 e
This gives dQdt =
This yields ^0,
§ 4,6 Time of Meeting of the Particles
Substituting r in the expression r dti = v±dt 
fide
hi- fi After what time the distance between the insects becomes half of its initial value? Ans. —
> What are the average velocity and average speed till the insects meet?
> What is the distance covered by each insect till it meet the others?
21 r°°
>/3v Jo
21x = —
3v
In the above example,
> Find the equation of trajectory of one insect relative to the other.
r = t = 0 and r =
J Vj^
. 3v □ V3
r0=/, vll=-7 anc* vi =—v> we have2 2
dQ
Ans. r = w'S* 
... I 
3v
Viv Ans. ----
2
2, 
Ans. -I
Method-1:
After a time t, let the distance of separation between the insects be 
r. The velocity of approach (component of relative velocity vre) 
between the insects along the line of their separation) is
3v
vrcin = v+vcos 60° = y
The perpendicular component vre| is
Vreix =vsin60° = y v
Directly using the obtained expression of time of meeting,
3v Since (vrel)||cfr = -dr, substituting (vre( )y =—, we obtain
3v — dt = -dr
2
0 at t = t (time of meeting), integrating both sides, we have
3v rr , r0— dr =- dr 2 JO it
21 t = —
3v
171Relative Motion
I
v
Missile
Fig. 4.30
Sol.
Fig. 4.31
->
= Jo’(w-vcos0)+9 = ^. c/ = -d9
-drRM = (y-u cos 0) dt ...(i)
In horizontal direction, the relative elementary displacement dx^ =(yRM)x dt, where 
(Ivrm Dx =I%-X I- Substituting ~vR, = =v cos 0/ and dx^ = dxRMi,
we have dxRM = (u - v cos 0) dt .. .(ii)
Since at t = 0, rRM = /, xRM = 0 and at the time of meeting t = x, rRM = 0 and xRM = 0 
Substituting these values in the eqs. (i) and (ii), we have
"RM = Jq(v-h COS 6)dt
172
Fig. 4.33Hence
This gives
Ans.Then,
We can also find time of meeting
Ans.
,2
is real; meeting really takes place.1. When v > u, r =
Then,
•••(>)i-r= vT-v
v r 0 r 0= — cosec dQ -I cot 9 dQ u Jk/2 Jn/2
X«— .
v‘
where r = /(9)
StudenTTask^
In the above example,
-PM
G.R.B. Understanding Physics MECHANICS (Part-A)
t by substituting r in eq. (v), which can be given as 
r dQ 
u sin 0’
%
> Find the point of meeting.
> What are the total distances covered by the missile and rocket before meeting?
j uvl j j
Ans. d\ = —---- =- and a2 s-5---- 2
v -u v -u
> When the missile rotates through an angle 9 = 30°, find the distance of separation between
the missile and rocket? Ans. 2/(2 - V3)v/“
vl
v2-u2
2. When v = w, r = oo; they will never meet.
3. When v " sin 9]^2
V
/(cosec 9-cot 9)" r =----------------------
sin 9
4. If v = u, 
let us assume that the distance of separation between the rocket and missile be /'
- j/ drRM ~ jQr(v-v cos 9)dt- This yieldswe find 
displacement by integrating velocity, velocity by integrating acceleration, with time. We define work as 
the line integral of force, flux of any vector field can be given as the surface integral of that vector v. For
instance flux of vectors v,E,J and B can be given as
v = jv-dA, ) or cap (A) represents the direction of the vector and the letter “A” repre­
sents the magnitude of the vector.
(ii) Graphical representation: Let us 
represent a “force of magnitude 5 N” 
acting oh an object towards east. For 
this, we take a graph paper and choose 
a scale of 1 small unit for 1 N force. 
Then, we have 5 small divisions for 
5 N force. That means, the magnitude 
of the force is equal to the length of 
five small units in the graph paper. If 
we change the scale, the length will 
be different. Since, the force is acting 
due east, we put an arrow pointing 
towards positive x-direction which 
represents east. Ultimately we get a 
segmented arrow which represents 
the force of magnitude 5 N and direc­
ted towards east. The length of the
We can represent a vector by an arrow (a directed 
line segment) and label it by a letter capped by a 
symbolic arrow. The orientation of the arrow shows (taJD 
—>
the direction of vector and the label A gives the
name of the vector (like velocity 7, acceleration 7, force F, position 7 etc.); In 
general A = ~riv,F etc. In some books, vectors are represented by bold letters.
Remember that an arrow is not a vector; it simply represents a vector. If we choose a 
bent arrow ), it may mean a sense of rotation of a point Hence, a bent arrow is not a vector. 
When we slide a vector in space keeping its orientation same, the vector remains constant.
---- --------------- I—J---- --------------------->x(E)
Fig. 1.1 5 N force is acting towards oast is 
graphically depicted as an arrow pointing in 
+x-direction and having a length of 5 units. 
Scale : 1 small unit=1 N
arrow gives the magnitude of the force and tip of the arrow shows the direction of the 
force.
Using the above representations we can specify the force as, F. The origin of the segmented 
arrow is the tail, and tip of the arrow is known as head of the vector.
G.R.B. Understanding Physics MECHANICS (Part-A)
A
1.
2.
(iii) Antiparallel vectors: Two vectors A and B are antiparallel if £ 
they are directed opposite to each other. They need not have equal 
magnitudes.
equal vectors are parallel vectors of same magnitude representing 
same kind of physical quantities.
IN force is not equal to 1 rad/s of angular velocity, having same direction.
———----- B_^
|A| * |B| but A and B 
are unidirectional
Fig. 1.2
--------------------- ► B
|A|* |B|and A and B 
point oppositely directed
Fig. 1.4
4 .
§1.4 Types of Vectors
After learning how to represent a vector, let us now categorise the vectors 
based upon their magnitude and direction (orientation in space) as following:
(i) Parallel vectors: Two vectors A and B are parallel if they point in 
same direction. Their magnitudes need not be equal.
—* —> __
(ii) Equal vectors: If two vectors A and B are equal, they must have ---------------► A
same magnitude and point in the same direction. That means if ► b’ 
A = B, we have |A| = |B| and A and B
have same direction 
| A | = | B | and A and B are parallel. This tells us that, p 3
---- ► 0 ~----- R = - R
F, F2 2
Sum of two vectors of equal magnitude 
and opposite direction gives a null vector
Fig. 1.6
(iv) Negative vectors: Two vectors A and B are negative when their • b’
means that if A = - B, then ] A | = | B | and the vectors are antiparallel |A| = |B| and A and B
are oppositely directed
_________________________ Fig. 1.5
Two negative vectors must be antiparallel but two antiparallel vectors need not be negative vectors.
(v) Null (Zero) vectors: By definition, a null (zero) 
vector has magnitude “zero” and has an arbitrary 
direction. This will be clear in the following 
example:
When two persons push an object with equal 
and opposite forces F( and F2 , the resultant 
force is F = F| + F2 = because the object does 
not accelerate. Since, the particle’s acceleration is zero, logically we can say that it is 
impossible to define the direction of acceleration. In the other words we can say that, the 
direction of a null vector is arbitrary (not specific).
I Zero vector possesses no magnitude and its directions undefined. It is represented as a point 
(•) labeled as 0 , but not simply zero (0), because | 0 | = o.
(vi) Unequal vectors: Two vectors A and B are unequal vectors
either magnitude or direction or both magnitude and direction ’ d'fferent in
5Vectors
y
B
A
x
c
z
(x) Orthogonal vectors: If any two or three vectors are perpen­
dicular to each other, we call them orthogonal vectors.
Since, A, B and C act along x, y and z-axes respectively 
they are orthogonal vectors. However, the orthogonal vectors 
need not be parallel to the coordinate axes.
A, B and C are mutually 
perpendicular 
Fig. 1.11
B
The coplanar vectors
A, B and C lie in the 
plane of the paper
Fig. 1.9
(vii) Collinear vectors: If two or more vectors A, B and C etc., pass through the same 
straight line, they are said to be collinearrT n , cos 9 dtJo
r-ty 
J7 
/Xo
173Relative Motion
...(ii)
0, then
In other words, when observed from different inertial frames a particle appears to move with 
same acceleration whereas velocity of a particle measured from different inertial frames need 
not be equal.
§ 4.7 Relative Acceleration
Galilean Transformation of Acceleration : As discussed earlier, ~vAB =v\c~^bc ■ ^en dif­
ferentiating both sides with respect to time we have
d~^AB _ d^AC d^BC 
dt dt dt
If the relative velocity between the reference frames B and C is constant, 
d^AB _ d^AC 
dt di ’
aAB ~aAC~aBC
The above formula states that, acceleration of a points relative to the point B is equal to the 
acceleration of A relative to C minus acceleration of B relative to C.
Similarly, Jo = r = v Jo
Eliminating cos 9 dt from eq. (i) and (ii), we obtain l' = ~-
That means after a long time (/ -> oo) the bodies maintain a constant separation of .
5. If v d\’po 
tormulae. vp0 = —^—=———
i ~* —>—>—♦ —► 1 2 j 2 2 «—~*
we have VpO=uPO+apOt,SpQ=UpOt + -aPOt and vP0 - uP0 =2aP0-spo .
Let us apply the above formulae in the following example:
‘ Ex. 11. Imagine yourself standing in an elevator which is moving with an upward acceleration a. A 
coin is dropped from rest from the roof of the elevator, relative to you. After what time the coin 
will strike the base of the elevator?
Here, we need to apply the formula s* =~ut+-~a t2 for the coin relative to
2 
the elevator which can be written as
If the elevator accelerates down, ~ace=~ac-a^ = (,-gj)-(-aj)=-(g-a)j. Substituting
-> . 1 -> 2 I r-ace= -(g-a)j, uce=0 and jce=0 in sce =ucet + -acet , we have/= I----- .For
2 y g ~ a
freely falling elevator a = g . Then, t = where ~ue =~at = at J after a time t. The other factors remain unchanged.
-* —> i —> 2
5ce ~ uce + ~2 °ce ’
where/ = time of fall of the coin, and ^ce are the initial velocity and the
acceleration of the coin relative to the elevator, respectively. Since the coin 
was attached with the elevator, both coin and the elevator would move with 
equal velocity at the time of release of the coin. Hence ~uce
of the coin relative to the elevator is ~^cv -~ac-~ae . Since the coin falls freely after losing 
contact with the elevator, ac = -gj. As the elevator accelerates up, "o = aj.
Then’ ’ace =(%.-(% = -(g +a) j
As the coin moves down with a displacement of magnitude h, sce = -hj. Now, substituting 
—> —> . —> . 1 2
uce = 0, oCe ~ +fl) J and sce = J *n formula sce = ucet+-acet , we have
, - 1, . 2 ••~hj =~~(g + a)f J
~2h
g + a
175
Ans. yj2(g + a)h 1
Relative Motion
Student Taskf
> In the example-] 1, find the velocity of the coin relative to you when it strikes the base of the 
elevator.
G.R.B. Understanding Physics MECHANICS (Part-A)176
Assignments
4.
5.
6.
7.
8.
9.
13.
14.
15.
2.
3.
Can we see a body accelerating with a>g or aChoice Questions
i.
2^1/2
/i-/2 
I I
In the absence of air flow, the rate of collection of water is R in the rain falling with a velocity u. 
If the air flows with a velocity v, the rate of collection of water is:
(a) 7»2+v2 (b) 4^R (c) -R (d)R
u
When we release a stone from a balloon which ascends with a velocity v and acceleration a, the 
velocity and acceleration of the stone just after release are:
(a) vt, (g + a')l (b) vl, (g-a) 1
(c) vT, gi (d) vt, (a-g)?
A man swims to and fro along the bank of a river with a velocity v relative to water. If the velocity 
of flow is u. the average speed of the man (for to and fro motion) is:
, It + V v2 - u
(a) 0 (b) (c) — (d) ——2 2v
Referring to Q. No. 3, if the man takes time r, and r2 to move to and fro journey respectively, 
the time taken by him to go downstream (while the man does not swim) is: 
7'iM (c) 7^7
178
5.
(a)
6.
7.
8.
(d) tan
9.
(b) 90° (d) tan
Referring to Q. No. 8 and 9, when v v, 0 is:
. -1 u(a) sm -
• -i v (c)sm
u
i u I V 1 U | V(a) sin — or sin — (b) tan" — or tan — 
v u v u
A block B moves with a velocity u relative to the 
wedge A. If the velocity of the wedge is v as 
shown in the figure, for the block to move verti-
n u •callv up, — is:
v
(a) cos
-1 u cos -
G.R.B. Understanding Physics MECHANICS (Part-A)
Referring to Q. No. 4, the time taken by the man in moving to and fro through equal distance 
when there is no flow of water is:
zl+z2 ... ziz2 4ziz2 ,—
~ (C>r,-H2
Let the distance between two moving particles at an instant be a. Let v be their relative velocity 
and V] and v2 be the components of v along and perpendicular to the line of separation of the 
particles. The time when the particles will be closest and the corresponding minimum distance of 
separation between the particles are:
av7 ay, aw av7 av av av av2(d)Tv
Consider a large number of particles each with speed v. The direction of velocity is randomly 
distributed in the collection. The magnitude of relative velocity between a pair of particles aver­
aged over all the pairs is:
(a) 0 (b) > v (c) - (b) sin -
v v
_l U -1 V
(c) tan (d) cos -
A rocket is fired with a velocity v relative to earth; its velocity relative to space just after the firing 
can be: 
(a) >v
_l u _| v (c) cos — or cos — 
v u
13.
if the wind blows along the line AB
(b) = if the wind blows perpendicular to the line AB
14.
,2
,2
15.
(d) V4?
1.
2.
3.
and vw = u.
(a) = — 
v'
R: vab=^a~^b-
A : A boat has to swim in order to attain the right opposite point of its starting point at an angle 
0 = sin"1 relative to the water current (u v
7777777777777777777/7777777777777777"
nuAAnnnnti
'n*>>vw(a) Minimum distance for v, (p) 6 = sin“1p^ 
k vw .
I
I
(d) Minimum time for vm(0 = tan(d)
3.
4.
(d)rf(c) c
1.
(b) 8 km/hr
(d) 5>/3 km/hr
vr=vrm + vm
Similarly, the velocity of the boat relative to ground can be given as
vb=vbw+vw
All the symbols bear usual meaning as described in this chapter. Using the above two equations, 
answer the following questions:
A boat moves relative to water with a velocity = 10 km/hr. It crosses a river of width km 
in minimum time. If the drift of the boat along the shore is equal to 400 m in crossing the river, 
the speed of water is: 
(a) 6 km/hr
(c) 5 km/hr
(c) 7vo + 2gM = tan
(a) 7vo= tan
Relative Motion 18]
Referring to Q. No. 1, if the stone hit the trolley car, the velocity v of striking of the stone relative 
to the trolley car can be given as (4> = angle made by the velocity v with horizontal):
I >/2gA
v0
-I v0
V2g/?
-1 J2g^
i -------------
v0
v0
v0
Referring to Q. No. 1, if we replace the stone by a raindrop falling with a downward velocity v, 
the:
(a) path traced by the raindrop relative to the trolley car is parabolic
(b) raindrop appears to move in a straight line relative to the trolley car as shown in the dashed 
lined
(c) velocity of raindrop at the time of striking is equal to ^v2 + vq relative to the trolley car
hv
(d) raindrop strikes the trolley car at a distance of —- backward (behind the man sitting on the 
v
trolley car)
Referring to Q. No. 3, if the trolley car accelerates, the path followed bythe raindrop relative to 
trolley car is given by the dashed line: 
(a) a (b) b
Passage-2 (Man in a Rain and Boat in a River)
While tackling the problems of man in a rain using Galilean transformation of velocities, the 
velocity of rain relative to ground is given as
3.
4.
5.
vbw
y
X
vj'b(=jf!
vw=(^P
1.
2.
3.
4.
5.
(b) V3v0
(d) 2>/2v0
182
2.
actual velocity of the rain is:
(a) V2v0
(c) 2v0
True or False
The velocity of a boat in still water is defined as velocity of the boat relative to ground, when 
water moves.
To cross a river in minimum time, the boat should head perpendicular to the bank.
• ' 8
(') | hr
Referring to Q. No. 3, the drift of the boat is:
J3 2 '
(a) —- km (b) — km
2 V3
1 3(c) ~f= km - (d) - km
V2 4
A man moves relative to boat at an angle of 45° with the direction
v0
of flow of water with a speed —t= as shown in the figure. Water ---- ■
v0. V2 --------
flows with a velocity — i and the boat moves perpendicular to h------
2 Z ;. ■ Vn vrr
water current relative to water with a velocity -y. If the rain - 
appears to fall vertically with a speed v0 relative to man, the H * -
G.R.B. Understanding Physics MECHANICS (Part-A)
Referring to Q. No. 1, if the boat tries to minimise the drift, it will cross the river in time:
I 1
(a) hr (b) — hr
(c) hr (d) hr
Referring to Q. No. 1, if the speed of the boat relative to water decreases to 4 km/hr, the time 
required by the boat to cross the river for its minimum drift is:
(a) e hr (b) | hr
(d) 4?3 hr
In order to cross a river with a minimum distance, the boat should head at an angle 0 = sin-1 
vmw when vw = /i+r2jand 
re = (J2 -1) i - / j. Find ^2 at t = 1 s.
Relative Motion
Fill In The Blanks
The magnitude of relative velocity between two objects moving perpendicular to each other with 
constant speed v is
A river of width ^km flowing from west to east at a speed of 5 km/hr. A man on the south bank 
of river capable of swimming at a rate of 10 km/hr in still water, wants to swim across the river 
in the minimum time. He must swim 
is
Four particles A, B, C and D situated at the vertices of a square of length of 
sides / start moving. Always, A aims at B, B aims at C, C aims at D and D 
aims at A, with equal speeds v. The magnitude of average velocity of each 
particle over the time of their motion till they meet is
The boat moves relative to water with a speed of 5 km/hr. If it crosses the
Galilean Transformation of Velocities and Acceleration
Suppose you are riding a bike with a speed of 10 m/s due east relative to a person A who is 
walking on the ground toward east. If your friend B walking on the ground due west measures 
your speed as 15 m/s, using Galilean transformation of velocity, find the relative velocity be­
tween two reference frames A and B.
A dog appears to move with a velocity of 5 m/s due N-E when viewed by a person A who is 
moving due south with a velocity of 3 m/s. Find the velocity of the dog with respect to another 
person B who moves with a speed of 4 m/s in 30° north of west relative to ground.
When you are walking due east, the rain appears to fall vertically with a speed of 2 m/s. When 
you walk with same speed due north, in which direction and with what speed does the rain 
appear to fall?
An aeroplane moves with a speed of 300 km/hr due north. If the air flows due east with a speed 
of 100 km/hr due east, find the velocity of the plane relative to air. At what angle does the plane 
head? z
A fish swims with a velocity ~v = (2/ + 3j - 4A) m/s relative to a 
boat at O which moves with a velocity u = (-2/ + j) m/s relative to 
 
 water. If water moves with u = 3/ m/s, find the velocity of fish.
G.R.B. Understanding Physics MECHANICS (Part-A)184
7.
8.
9.
10.
11.
12.
13.
14. t = 0 t = O
A B
15.
16.
z2/zzz2Z7yz/zzzzzzzz/zzzzzzzzzzzzzzzzzzzz
H-------Xo------- H
A boat which moves relative to water with a velocity vbw starts from one of the bank of a river 
of width d. If it always aims at the right opposite point “0” on the other bank of the river, find the 
equation of path traced by the boat with respect to the point 0. The speed of water is vw.
A body is projected vertically down from the top of a tower of height h (= 40 m) with a velocity 
V| (= 5 m/s). Another body is projected up from the bottom of the tower with vertically upward 
velocity V2 (= 10 m/s). When and where will the bodies meet?
2 
A 1*------- vi V2 2, 2 —> 3 and 3 -> 1 with a constant speed v.
(i) When and where do the persons meet?
(ii) After what time the distance of separation between the persons 
becomes half of the initial distance of separation?
(iii) Find the average speed and average velocity of the persons over the time of their motion.
A bird flies to and fro between two cars which move 
with velocities V] and V2 ■ If the speed of the bird is 
vj and the initial distance of separation between them 
is d, find the total distance covered by the bird till the 
cars meet.
In the example-10, let us replace the missile by a projectile. If we throw the projectile so as to hit 
the rocket. Find the (a) time of hitting (b) minimum initial speed of projection and the direction of 
projection. Take I = h.
A person starts moving with a constant velocity v = 8 
m/s from a point A at a distance xo = 12 m behind the 
car 5; If the car starts from rest and moves with a 
constant acceleration a = 1 m/s2, when and where do 
they (person and car) meet?
17.
18.
I.
a2.
Six particles move in a cyclic manner along the sides of a regular3.
vi
4.
I
5.
A balloon B descends with a velocity u while moving with an upward acceleration 
a as shown in the figure. A man moves relative to a rope hanging from the balloon, 
with a velocity v. Find the time taken by the man to move through a distance I 
relative to the(i) rope (ii) ground.
particle is v, the particles lie always at the vertices of a hexagon, 
(a) When and where do the particles merge? (b) When will the 
size of the hexagon be halved?
A train moves with a speed v inside the station.
(a) In which direction the passenger must head such that he can most conveniently enter into 
the train?
(b) Find the velocity of entry, assuming the speed of the passenger is u.
Relative Motion 185
In the example-l I, find the velocity of the coin relative to ground when it strikes the base of the 
elevator.
Two cars A and B are moving in the same direction with speeds vA and vB respectively. When 
the distance between the cars is d, the car A applies its brakes and moves with a constant 
retardation a. Derive the condition if there will (i) be a collision (ii) not be a collision.
v
IVMjj
)Q
\30°
V2
(ii) If the points P and Q in part (i) are not rigidly connected, find the 
relative velocity between them. p
Assume V| = v2 = 1 m/s.
Problems
A ship is travelling due east with a speed of 60 km/hr relative to water which flows due north 
with a speed of 20 km/hr. A south-eastern wind flows with a speed of 100^2 km/hr relative to 
ground. Find the velocity of wind relative to the ship.
u|0i
(i) A rigid rod moves such that its end points P and Q move with ..gw 
velocities V| and v2 as shown in the figure. Find the relative velocity p (east) and j -> (north)
(a) [-7/ -5 j + i]m/s 
5V37km/hr 4> = sin-11
v of the train.-if v
2(b) >/w2 -V
188
Problems
80(-2f + j)
(v-m)±7(v~«)2 -2a/
a
(ii) (v + »)±V(v + »)2 -2o/
a
_ . 2/ „ , ,(a) —, centre of the hexagon.
v
(b) -v
1 +V3 .-------m/s
2
5/+ 2.5 j
I v V3v' u 
(2+-2 77
(a) No. 
vl » •
velocity and 
acceleration of a particle in circular motion without enquiring the cause of circular motion. Hence, we 
call this chapter “Kinematics of circular motion”. The “cause of circular motion” will be dealt in the 
chapter-9 called “Dynamics of circular motion”.
§ 5.3 Linear Quantities
To express the linear quantities, i.e., position vector, displacement, velocity, let us choose the 
centre of the circle as the origin of the coordinate system.
Position vector : Let us start calculating time of motion of the particle when it just passes
—> —> A
through the point M. That means at t = 0, r0 = OM = Ri. If the particle passes through the point P after
a time / describing an angle 9, its position vector is ~r = OP = R(cos9i + sin By).
Displacement: As the particle moves from M to P, its 
displacement is given as 7 = Ar=7-^. The magnitude of
5 *s ’ I ? I = I 'b I = +ft>2 -2rr0cos6
Substituting r = r0 = R ,
+ R2 -2R ■ RcosQ.
|7| = | A?| = 2/?sin—.
2/? sin
At
|7| = T?
Average velocity: Since, the particle changes its position vector by A r during the time interval
A/(=f) while moving from Mto Pin the circle, its velocity averaged over that time is ~vm
= 0
Kinematics of Circular Motion 191
Sol.
we have
Then,
v
0
Fig. 5.5
Z?9 
t
v 1 Fwhen Fis 
x defined relative to 0
Fig. 5.4 Ar tends to 
be tangential, that is, 
Ari r, when At-*0
^av 
Vav
f
/ y
V 
'*\Tangent to 
11 the circle
. « dQ
= (—2?sin 9/ + J?cos9 j)~^-
. . 9
"•"2
AZ ’
| and V.v=2^sin(y) /10 
v.v=im/s-
The ratio of magnitude of average velocity to the average speed during any time interval is
O'!
— > in uniform circular motion.s /
r need not be perpendicular 
to v in general curvilinear motion
r* I
9
2sin|
-» dr
V = — dt
If we take the dot product of ~r and 7 we will get zero. That means, 7-7 = 0. This tells us that
D
Average speed : Hence, the average speed «av of the particle over the time t is «av =
Instantaneous velocity : Let us, now diminish the time interval, Az to zero to define the 
instantaneous velocity at any time /. When AZ -» 0, we can see that A r tends to be tangential as shown 
in Fig. 5.4. That means 717.
Even though 7 is tangential for all curvilinear motion, it is not always essentially perpendicular 
to the position vector 7. However, in circular motion, the velocity of the particle is always perpendicular 
to its position vector drawn relative to the centre of the circle. We verify this as following:
P 
g 2/?sin-
s = 27?sin— and vav =
s = 2(l/4)sin[y]
J = lm and vav = 4()......
Distance : The length of the path, that is, arc MP is given as D (or /) = RQ.
Ex. 1. Find the magnitude of displacement and average velocity of the tip of the seconds hand of a 
clock of dial radius R = 25 cm during AZ = 10 j.
During Az = 10 s the tip of the seconds hand rotates through 9 = 60°. Then substituting
9 = 60°, /? = 4m, Az = 10 s in
4 t = 10 s
ro
Fig. 5.3
Ans.
Since 7 = T?cos9i + /?sin9j, the velocity ofthe particle is
Fig. 5.7
we have
Sol.
IA v | = v0 Ans.Then,
Fig. 5.8
£ 
where 71 r (but not r').
dr* -» 
~di~ V
| A v | will be more if 0 increases upto 90°
Change in velocity : When the particle describes (revolves) an angle 0, its velocity vector 
rotates through same angle 0. During this time let the velocity change from, = 7 to y£ = v*.
Then the change in velocity is A v = v2 -Vj
Then, | Av | = | v2 -V] | = Jv2 + v2 - 2V]V2 cos0 
Substituting v, = v and v2 = v', we have
| A v* | = 7v2 + v,2-2w'cos0
If the particle moves with uniform speed, v' = v,
Average acceleration : Since, the particle changes its velocity by A 7, 
during time Ar = t its acceleration averaged over this time interval is
Rate of change in position vector r relative to any fixed point is equal to velocity 7 
measured relative to that point. 717 for all times, relative to the centre of the circle. Linear 
velocity of a particle is same with respect to all fixed points of reference, but angular velocity 
may not be same.
Fig. 5.6. Vpo b7P0' , whereas 
topQ * I n
| A v | = 2v0 sin — where I ~-R. 
IK 3
Kinematics of Circular Motion 193
Hence
I °av I =we have
.2
7
Since
Fig. 5.9
we have
->
is directed towards the centre of the circle. Hence
Fig. 5.10
| a | = lim
A/->0 
e->o
-> 
a =-
2vsin- 
I ^av I ~
Consequently, A 7 tends to be perpendicular to v(. Since, v 
radially inward as shown in Fig. 5.9.
It means a = —
di
-» A 7 , . a
°av = ~Ki~ ’ where IA v I= 2vsin 2 and A/ = 
. 0
"" 2
t
Kv
JfenP
A— v
When At -> 0
Av tends to be 
_rJ perpendicular to v
I Rd RdFor uniform circular motion, t = - = —. Substituting t = —,
Instantaneous accelerations
(i) Radial acceleration : When we gradually decrease the time interval Ar to zero, the angle 0 
will reduce to zero. Then aav tends to be an instantaneous acceleration which can be given 
as
~r where r = unit vector of position vector r
The above acceleration is known as centripetal or normal or radial acceleration. It arises from 
the change in direction of velocity of the particle. As r changes its direction with respect to 
time, centripetal acceleration must change its direction.
In uniform circular motion, position vector, velocity and acceleration (r, v 
and a) of the particle change their directions continuously. At any point
. t -» , -» v2 . , . ~r 7possessing the position vector r, we have ar = ——r, where r = — = —.
R |r| R
lim ^7^ = 1, where ^-7,
HO 9 2
V2 ■
|a| R
The direction of ~a can be found by analysis. When 0 tends to zero, v' rotates towards 7.
is tangential, Av must be
2v2sin
Rd
2v2sm- v2 ail,-.
--------2- lim
Rd R'~
. d sin-
0
2
194
al =
V
is positive. Then o, points
is negative. *acceleration”. If the speed | v | decreases,
•i
R
Sol.
Then,
Fig. 5.11 |v| increases; 
at is parallel to v, a"= a”t + a"r
dv
It
v
d\ v I 
dt
IM 
dt
G.R.B. Understanding Physics MECHANICS (Part-A)
(ii) Tangential acceleration (Non-uniform circular motion) :
When a particle moves in a curve, the magnitude of its velocity, 
that is, speed of the particle may change. As we know, the 
velocity changes its magnitude along the tangent to the path.
Hence, the rate at which the velocity changes along the tangent 
is called "tangential acceleration”.
The tangential acceleration is given as
a
Fig. 5.12 |v| decreases; ajand 
v are antiparallel, a"= aj + aTf
a” \
V| 
dt
in the direction of motion (7). We can call it “tangential
IM
dt
Hence a, is directed opposite to the direction of motion (7),
We can call it “retardation” or “deceleration”.
In addition to this, at any position, the particle must experience a centripetal acceleration due
-» v2
to the change in direction of velocity of the particle which is given as ar = —-r as discussed
R
earlier. Hence, the total instantaneous acceleration a of the particle is equal to the vector 
sum of tangential and radial accelerations. Thus,
The total acceleration, ~a = = a^+ a^,
where tangential acceleration, a, = 
_> v2 
and radial acceleration, ar = ——r
Let us apply the above ideas in the following examples.
Ex. 3. A particle moves in a circle in such a way that, its tangential deceleration is numerically equal 
to its radial acceleration. If the initial velocity of the particle is v0, find the variation of its 
velocity with time.
Let at any instant t, the velocity of the particle be v. As per the questiona, = ar
dw v2
where a, = —-r as the particle slows down and ar = —
vi, where t = -^~ 
M
If the speed |7| increases,
R
Kinematics of Circular Motion 195
This gives
This gives
Then, Ans.
vi
4
Fig. 5.13
function of linear distance.
The total accelerationSol.
Differentiating
we have -r = a
a =
This gives Ans.4 R2a = a
When t = 0, v = v0 and when / -> a>, v -> 0. 
Then, the particle slows down hyperbolicaliy.
R
Since the particle slows down from speed v0 to v during a time t, we have 
rv dv I rt ,
v v0~ R
where v = afs
dv
7
v = -A_
Student Task^7
> Derive an expression for velocity and acceleration of the particle in the above example as the
Ans. v = voe *, = — voe K
Ex. 4. A particle moves in a circular path such that its speed v varies with distance s as v = aTs 
where a is a positive constant. Find the acceleration of the particle after traversing a distance s.
+(y)
v = a Vs with respect to time, 
dv s'"2 ds
— = a—-—t- dt 2 dt
„ . . . r , dv a2
Substituting ^- = avs, we have = 
dv
Now substituting and v in the expression of a, we have
ds
dt
196 G.R.B. Understanding Physics MECHANICS (Part-A)
Student Task
r
9
O o P
Fig. 5.14
A Find the angle between the velocity and acceleration of the particle in the above example.
. a ♦ -iM Ans. = tan I — I
„ d2Q 
a = ~y
co Increases; a 11 co.
Fig. 5.16
§ 5.4 Angular Quantities
Introduction and need of angular quantities : So far we explained the circular motion of a 
particle by using the linear quantities. In practice, when we analyse the rotation of rigid bodies, i.e., 
rotating disc, fan etc; we can see that all the points of the bodies move in circular paths about the axis 
of rotation. Furthermore, all points of a rotating body move with different linear displacements, velocities 
and accelerations. This makes us difficult to explain the rotation of rigid bodies by using linear quantities, 
by particle to particle basis. Here we need to search for some quantities, which are common to all the 
particles of a rotating rigid body. Let us discuss these special quantities.
Angular speed and angular velocity : When the point P 
moves in a circular path, the angle 9 made by radius vector 
r changes with time. The time rate at which 9 changes is called
„ J9
(0 = -r .dt
----
P v
;, a is directed opposite to © .
angular speed denoted as ©. Symbolically,
The unit of o is radian per second. If © is more, the particle 
turns more quickly about 0 and vice-versa.
The angular velocity is associated with a direction, which 
can be given by “right hand thumb rule’’. The magnitude of the 
angular velocity is given as angular speed.
If we wrap the fingers of the right hand in the sense of 
revolution (turning) of the point P, the extended thumb will give 
us the direction of the angular velocity.
Angular acceleration : The angular acceleration of P relative to 0 can be given as,
„ In the foregoing example, find the angular acceleration of the aeroplane when its angular 
speed is zero. Ans. - 6>/2 rad/s2
Ex. 6. Two satellites I and 2 orbiting with the time periods Tj and T2 respectively, 
lie on the same line as shown in the figure. After what minimum time, 
again the satellites will remain on the same line? Assume that the two 
satellites should lie in same side of the centre of their concentric circular 
paths-.
Let the satellites 1 and 2 describe angles 0] and. 9? at the centre of earth 
during a time t. Suppose, 2 is ahead of 1 by an angle 0. Hence we can 
write
n da .a = — = -18/ dt
„ dQ 0,2(0 = — = 2 - 9/ 
dt
Kinematics of Circular Motion 197
Ex. 5. If the angular displacement of an aeroplane moving in a circular path about the centre of the 
circle varies with time as 0 = 2/ -3/3, find the initial angular (/) velocity, (ii) acceleration.
Q = 2t-3t3
r t2 T] 
r=2iZL
A-t2
0 = 02 ~0]
They will be again lie on same line after a minimum time T say. For this 0 = 27t. This gives 
02 -0| = 2rt. Dividing T' in both sides, we have 
02 0| _ 2n 
7'* T'1 ” T' 
02 ^0IY = co2 and = co, 
2n . 2n 2n— = co2 -CO], where co2 =— and CO] = —
G.R.B. Understanding Physics MECHANICS (Part-A)198
Student Taskf
(a) zero (*) coe
ra
P
rw2
0
0
Fig. 5.19
_ 271 |
= V
\s 
M
> In the foregoing example, if coe = angular velocity of spin of earth, the angular velocity of 
a geostationary satellite relative to an observer on earth's surface is :
co.
T
where = v and = co. at at
where = at and = a at at
(d) none of these
Ans. (b) 
Hint : When an observer fixed with earth observes a satellite always lying at same point 
relative to him, we call it geostationary satellite. Hence a geostationary satellite must cover
same angle as covered by the observer about the centre of earth. That means, cof = coe 
where Tt = 24 hr but I se * - ©e; rather it is equal to
| v — ve |
e • Since vs-ve* 0, the geostationary satellite appears to move relative to earth. We 
rs re
will discuss geostationary satellite in detail in the Chapter -18 (Gravitation).
§ 5.5 Relation between Linear and Angular Quantities
Let the point P describe an arc MP whose length s can be given as
|j = r 6]
Differentiating both sides with time, we have
dt~r dt'
Then, |v = real
Again differentiating both sides with time, we have 
dv d® 
dt dt’
Since the satellites execute uniform circular motion, the angular speeds can be given by 
_ Angle 2n (for one revolution)
2^ " Time T (time period)
If we write — = w2I = CD2 ”®1 and term “®21” as angular velocity (or angular speed) of 
2 relative to 1, it is wrong. However, the expressions co[2 = C0]-©2 ®12 -®1 -c02 do not
have any clear meaning. Please make it clear that (©2 - coj) is the subtraction of angular speeds 
2tt(not angular velocities) which may be equal to — when the satellites will again stay on the 
same radial line in one side of a centre. This becomes possible when the satellite 2 will cover 
2n radian more than the angle covered by satellite 1. If one satellite covers n radian relative to 
the other, they will be on the same line but in opposite side of the centre O.
Kinematics of Circular Motion 199
|«/4
> In the above example, if the radius of the circle is r = \ m, find the magnitude of total 
acceleration at t = | J . Ans. a = 4^2 m/s2
£0 = ^ = 4/ 
dt
v2 :ar =—, putting v = rco, we have
2
2. However, simply we can write s = rQ,v = ra, a, = ra and ar = ra .
0 = 2/2,
ra = rco2
ar = ra1
a- da
200 G.R.B. Understanding Physics MECHANICS (Part-A)
Sol.
yo
°av B- A
vwhere Fig. 5.20
Then, ...(i)
Since
T,Then, j =
where
Then, •••(>’)
Ans.
This gives
Then,
This gives
Fig. 5.21 K v; ap at and IF of a 
particle in circular motion
vp + aT + Vo
2
Ex. 8. A particle in a circular path speeds up with a uniform rate between two diametrically opposite 
points of a circle of radius R. If its time of motion between these two points is equal to T, find 
the acceleration of the particle averaged over thetime T.
The average acceleration of the particle when it revolves half of the circle is:
Jv-vpl
AT
Student Task
> In the previous example, if you blindly (by mistake) assume zero initial velocity, what will be 
the average acceleration?
§ 5.6 Use of Polar Coordinates in Circular Motion
The position vector r of the particle at any angular position 0 is
given as
Lx
V = -V j, v0 = v0 j and AT = T (given) 
_v + vo_ 
“av ~ p
Let us express v + v0 in terms of R and T.
—L—(= a), we have dv = adt
[ adt = at 
Jo
v = vq + at
f vdt = [ (vq + at)dt = |
Jo Jo V
s = nR and v0 + aT = v 
2nR 
v + vo=—
Substituting the value of v + v0 from eq. (ii) in eq. (i), we have 
2ttR
aav ='ZF T2 ,
Ans. 2nR/T2
"r = R(cosQi +sin0 j)
Differentiating ~r with time, we have
dz • o* . ^0v = —= J?(-sin0/ +cosG])~r, where — dt J dt ' dt
7 = R(n(-sin 0z +cos9 j)
 A bead kept on the top of a hemispherical surface of radius R is •
projected horizontally with a velocity v = y]2gR. The radius °f f *\ \
- . . - . . . . . /zz/zzzzzzzzzzzzzzzzzz/zzZ/
curvature of the path of the bead at the given instant is : Fig. 5.26
p
(/>) — (c) 2R (d) infinite
Ans. (c)
Fig. 5.24
Resolving the acceleration a perpendicular to v, we have 
an = asin 30’ = 2 • ^ = I m/s^
Ex. 9. A particle moves with a velocity v = 1 m/s and acceleration a = 2 m/s at a given position P. 
Find the radius of curvature of the path at the given point.
Kinematics of Circular Motion 203
Assignments
,2 in circular3.
11.
12.
13.
T
valid in circular motion?and vav =14.
15.
1.
2.
4.
5.
_V|+V2 
2
6.
7.
8.
9.
10.
f v dt 
TAre the expressions vav
Explain.
Can a particle have zero (i) instantaneous acceleration (ii) average acceleration, while moving in 
a circular path?
16. A particle moves with a constant speed along a path comprising of two semi-circles as shown. 
Describe the change in acceleration while travelling from A to C.
-> -> 1
Compare the s = u t + - a t 
motion.
What is the ratio of angular velocities of the hands of a watch?
In circular motion, if we reverse the velocity of a particle, will its radial acceleration reverse its 
direction?
Does acceleration change in a uniform circular motion?
What is centrifugal acceleration?
Is circular motion a periodic motion? Explain the cases.
Is there any difference between average angular speed and average angular velocity?
Magnitude of angular velocity is defined as angular speed. Furthermore, angular speed is defined 
as 2if, where/is the frequency of revolution; f = y,>7’ = time period of revolution. Hence, we 
can call co as the angular frequency. Keeping this argument, distinguish between the terms 
angular velocity, angular speed, angular frequency in circular motion.
Discussion Type Questions
Can you call “centripetal deceleration” instead of “centripetal acceleration”? Explain.
Is there any difference between centripetal acceleration and radial acceleration? Discuss.
1 2 
for constant accelerated motion with s = ut +—at
Can we write I (b) = 1 (c)case (n) jr-
where, 7j and T2 are the time periods and the others bear usual meaning for uniform circular 
motion.
10. The ratio of centripetal acceleration of a particle at any latitude of angle 0 and a particle on the 
equator is:
(a) cosO (b) cos20 (c) 1 (d) sin2 0
11. A pendulum is projected downwards from the horizontal position 1 as shown in the figure. The
change in speed and magnitude of change in velocity of the pendulum 
between points 1 and 2 are x and y respectively. Then: 
(a)x>y ' (b) x=y
(c)xy
(a)
1.
*A
BA -7
2.
3.
“ax
4.
5.
206
12.
_ Vabl
dt AB '
A : Referring to Q. no. 3, = -cofl/2e = —, where vTE is equal for both trains. 
R
A : The angular velocities of A and B relative to 0 are co^ and
respectively. The angular velocity of B relative to A is given as
__ cogj?g
rb~ra
de
R: ®ab =
a = total acceleration of the particle and
a2
 e/|v| v-——
(C)----- dt
v
207
(a) aABr (P)
(c) \aAB/ | (r)
(s) aAB*rAB(d)
y
2. In oblique projection of a particle:
(a) Radius of curvature at A (P) A.
x
(b) Radius of curvature at B where = tan(q)
(c) ®BA (r)
(s)(d) afl/4
1
(b) As you know,
v°
B
dQ AB 
dt
Kinematics of Circular Motion
Match The Column
i.
Xeo
tan 0O 
2
Va
9ab\7ab
B
VAg
(9) ®ABXVAB
I I 
rAB
1^1-^ y r 
r12 2
which is valid for two particles (points) moving in any path.
©12 =©2i» whereas V]2 =-v2|
where a>BA and aBA
point B relative to the point of projection A.
Comprehension
Passage-1 (Relative Velocity, Relative Angular Velocity and Relative Angular Acceleration)
It is a common mistake to accept that angular velocity of the particle in circular motion is a 
constant quantity, but it is not always true. As you know, the particle in uniform circular motion 
has constant magnitude of velocity relative to any fixed point. On the contrary, the angular 
velocity of the particle has different magnitudes, relative to different fixed reference points. 
However, in a uniform circular motion, linear velocity has constant magnitude, whereas, angular 
velocity has constant magnitude and direction only about the centre of the circle.
(a) The magnitude of angular velocity of point 1 relative to point 
2 is given as f v’2
i /
 -» 7
r12 *
2 g cos 0O sin 20 
v0 sin2 0O 
g2 sin 2 
vosin20Q ’
VQ2 
gcos0o 
Vp COS2 0q
g
are angular velocity and angular acceleration of the particle at the highest
A particle A moves in a circle of radius r. Assuming ideal symbols, match the following terms:
dt
(b) uAB
208
(c)
(d)
1.
(a) /
2.
3.
(C) W12
(d) an
4.
(b)
(c) (d)
2 iv2*11
(because v)2j = v)2 as vi2|l =0)
This is valid for two points, moving relative to the other in circular paths. Keeping the 
above points in view, answer the following questions:
Two particles I and 2 separated through a distance I have velocities v( and v2 as shown in the 
figure. The angular velocity of 1 relative to 2 is :
G.R.B. Understanding Physics MECHANICS (Part-A)
0The average angular speed is co = —, which gives 0 = ©/.
The magnitude of centripetal acceleration of I relative to 2 can be given as |ij)2 \cp = 
r!2
+ v2 
I
(c)zero (d) —
Referring to Q. 1, centripetal acceleration of 1 relative to 2 is :
(a) (b)
v2(c) -y-> (d) zero
Two satellites ^and S2 having circular orbits of radii and R2, are revolving about a fixed 
point 0 with angular velocities coj and co2 respectively, as shown in the figure. The relative 
angular velocity and the relative acceleration between them at the given instant has magnitudes:
(a) (0|2 =| where = and a = —-; 0 = angular
position of the radius vector.
Find the velocity and acceleration of the earth in its motion around the sun. The radius of earth’s 
orbit is 1.49 * IO1’ m and the period of revolution of earth around the sun is T= 1 year = 
3.16 * 107 s.
Find the angular velocity of each of the three hands of a clock of dial radius 2 cm.
3
A ceiling fan has three blades each of length - m. If the rated speed of the fan is 3000 rpm, find 
the (i) frequency, velocity and acceleration of the tip of the blades (ii) average velocity and 
average acceleration of the tip of the blades during one fourth of a complete revolution cycle.
Non-uniform Circular Motion
The angular frequency of a flywheel increases uniformly from co i = 20 rad/s to C02 =30 rad/s in 
Ar = 10 s. Calculate the (i) angular acceleration (ii) total angular distance covered by the flywheel 
during the given time interval.
A peripheral point on a rotating flywheel of radius r = 25 cm is changing its angular position 
relative to the axis (centre) of the flywheel according to the relation 9 = 3/2 + 2r. Assuming 9 = 0 
when co = 0, find the (i) initial angular velocity (ii) angular acceleration (iii) average angular speed 
over first one second (iv) linear speed at t = 1 s (v) angular speed at t = 2 (vi) linear accelerations 
(centripetal, tangential and total) of the point at t = 1 s.
A motorcyclist while negotiating a curve of radius R - 24 m changes its speed. If the rate of 
change in speed of the motorcyclist is 2 m/s2 when its speed is 12 m/s, find the (i) angle between 
the velocity and total acceleration of the motorcyclist (ii) total acceleration of the motorcyclist.
12.
13.
14.
15.
16.
17.
—-v
2v
P
1.
2.
Moving Centre of Circular Motion
A cyclist moves with a velocity of 5 m/s. Find the angular velocity and acceleration of the point
3 
of contact of the tyre relative to the centre of the wheel of radius m.
A bowler swings his hand with an angular velocity co = 3 rad/s while running with a speed of 
v = 5 m/s. If the speed of the ball leaving the hand at its top is v' = 30 m/s. Find the effective 
length of the hand.
A ball placed at the lowest position of the spherical surface 
has a velocity vf = -2v/ when the surface moves with a
Problems
A particle is moving in a circle with constant speed v. If the radius of the circle is R, find
(a) angular velocity of the particle relative to any point P on the 
perimeter of the circle.
(b) angular acceleration a, radial and transverse components 
of acceleration (ar and ae) of the particle relative to P.
(c) Referring to (b), is radial acceleration equal to centripetal 
acceleration of the particle? Is tangential acceleration at equal 
to the transverse acceleration a9?
An aeroplane moves with constant velocity v parallel to x-axis at a height y = h. Find the
(i) angular velocity
Kinematics of Circular Motion 211
An automobile enters a circular path of radius R. If the centers of the wheels of the automobile 
can have maximum acceleration a, find the maximum speed of the automobile. Assume p = 
coefficient of friction.
A space shuttle is moving in a circular orbit with a speed of 7.8 km/s and period of 87 min. In 
order to return to earth, the shuttle fires its retro engines opposite to its direction of motion. The 
engines provide a deceleration of 6 m/s2 which is constant in magnitude and direction. What are 
the radial and tangential acceleration of the shuttle?
An electron in a research apparatus follows a circular path. The initial speed of the electron 
before entering into the apparatus is vq and the radius of its first revolution is Rq. If the electron 
gets a “kick” just after each revolution so that its speed increases by in such a way that its 
centripetal acceleration remains the same, what is the radius of the circular path after the electron 
has received 'n' kicks?
velocity -v‘i and acceleration a =-ai. Find the (i) 
angular velocity of P relative to 0, (ii) angular acceleration 
of P relative to Q, (iii) tangential and radial acceleration of P 
relative to Q.
212
y
v
X
3.
4.
5.
6.
7.
w
8.
v
X0
O
G.R B. Understanding Physics MECHANICS (Part-A)
(ii) angular acceleration of the aeroplane relative to 0 as the function of time t. Assume that at 
t = 0, x = 0 .
0 
•—R
OJ
A particle slides down a parabolic surface having equation y = kx2, with constant speed v. Find 
the tangential and normal acceleration of the particle at P(x}, y,).
y*
I 
11 
1 
0
A point source of light “O” moves in such a way that the tip “P” of 
the shadow of a pole of height h moves towards the base of the 
pole with a constant velocity v. Find the angular velocity of the 
point source relative to P at the given position. h
......... 
"777777777777777777777/777777777777
Shadow
A particle moves in a circular path with increasing speed v = kt. Find the total acceleration of the 
particle when it revolves n times.
A particle moving in a circular path decelerates obeying the relation a = k\[a. If the initial angular 
speed of the particle is coo, find the angular speed averaged over the total time of motion of the 
particle.
A bullet flies with a velocity v horizontally. If it pierces the first v 
rotating disc at t = 0, say, the hole made by the bullet in the disc \ I \ 
1 rotates through an angle 0 relative to the centre of disc 1 just 0 j 0 j 
before the bullet hits the disc 2. Find the distance of separation I / \ /
between the discs. \27
A disc rotates with a constant angular speed co. An insect Q crawls with a Mf 
velocity vr relative to the periphery of the disc as shown in the figure. Find 
the q
(a) acceleration of the insect relative to 0.
(b) acceleration of the insect relative to its point of contact of the disc.
-C/xi.yi) 
zy = kx2
213
9.
v
x0
10.
a a
B
11.
12.
13.
14.
1
C 
(i)
A car moving in a circular path of radius 50 m speeds up from 15 m/s to 27 m/s uniformly during 
3 s. Find the magnitude of the acceleration when the speed of the car is 20 m/s.
A car starts from rest from the position A. Then it speeds up uniformly from A to C with a 
tangential acceleration of magnitude a following a quadrant BC of length I as shown in the figure. 
Find the acceleration at B and C. Draw the variation of acceleration with time.
A B
H
R/2
T
A 
I 
R
A 
—R
•B
Kinematics of Circular Motion
A trolley car moves up with a constant speed v along a sine curve, 
given as y = A sin ax as shown in the figure. Find the minimum 
and maximum acceleration of the trolley car.
Two discs each of radius R are rotating with a constant angular acceleration a as shown in 
the Fig. (i) and (ii). Find the relative velocity, relative angular velocity and relative acceleration 
between the points B and C, after a time t from starting of motion.
v = 0.5e'm.
C
(ii)
A particle moves with a tangential acceleration of constant magnitude along a semi­
circular track of diameter d as shown in the figure. If the particle takes a time T to 
cover the semi-circle, find the:
(a) average speed (b) average velocity
(c) change in velocity (d) average acceleration, over the time T.
A particle moves in a circle of radius R = 1 m such that its speed varies with time as 
If the particle starts from 0 = 0°, find the velocity and acceleration when 0 = 30°.
r
16.
17.
18.
I
vo6-
19.
20. ------21 /—
vo B
I
21.
22. Two particles A and B move in opposite sense with same speed as shown in the figure. If the 
particles start from the diametrically opposite points, find the velocity and acceleration of B
(i)
(ii)
(iii)
H ’V
214
15.
v oc — where s = distance 
s
G.R.B. Understanding Physics MECHANICS (Part-A)
A particle moves with a constant speed v along a curve comprising of a set of semi-circles of 
gradually decreasing radii as shown in the figure. Draw the graph of I a | versus time.
A particle moving in a circle slows down from v0 obeying the relation
covered by the particle. Find a = f (s).
A motorcyclist observes at A that a wall C is situated at a ___
distance of 3/ from him. Hence, he applies his brakes and 
retards uniformly from A to B. Then he rides the motorcycle v
uniformly along the quadrant of a circle to avoid accident 
with the wall at D as shown in the figure. Find the (i) 
average velocity (ii) average acceleration over the time 
of journey of the motorcyclist from A to D.
A particle “A” approaches the centre “C” of a rotating disc. If Rw = v, find the velocity and 
acceleration of B relative to v as the function of time before the particle collides with disc.
r/8 
r/2 z~\j(--I—
r/4
The acceleration of a particle in a curve varies with time as a = (16/ + 4t j). Find the trajectory 
of the particle if r = 0 at / = 0. Find the speed and rad: us of curvature of the path at t =1.
Find the maximum acceleration of a particle moving in a parabola y = x2 -9 with a constant 
speed v.
The bob of a pendulum is pushed with velocity v0 > y]2gl as shown in the figure :
Find the acceleration of the bob when the bob swings through an angle Q. 
What is the acceleration of the bob when the string becomes horizontal?
What is the change in (a) velocity (b) speed when the bob moves from initial 
position to horizontal position?
Kinematics of Circular Motion 215
relative to A after a time t and draw the corresponding graphs.
ANSWERS
7. (a,b,c)6. (a,b,c)
4. d 5. a
4. (d)3. (b,c)2. (d)
4. (T) 5. (T)3. (F)
2
3. 4. -1. 0.6 2.
1.
3.
rad/s7.
9.
11.
13.
4. (c)
11. (c)
0)
2
Multiple Choice Questions
1. (a,b,c) 2. (a,b,c,d)
8. (b,c)
5. (c)
12. (a,b,c,d)
n
21600
2sin0
6
dt
3- (b)
9. (a,b,c,d) 10. (a) 
Assertion-Reason Type Questions 
1. e 2. e 3. a
Match the Column
1. a - q; b- r; c- p, s; d - r
2. a - r; b - s; c - p; d - q 
Comprehension 
Passage-1
1. (b) 
True or False
1. (F) 2. (F)
Fill in the Blanks
(ii) sin 9 km/s
(ii) 3* 102 m/s2
Introductory Exercise
(i) km/s
(i) 8x io20 m/s2
5. a2
— rad/s, — ■ rad/s,30 ’ 1800 ’
ar = m/s2 ar,d °/ = 6 m/s2 zol
(i) 1 rad/s2 (ii) 250 rad
(i) at =2 m/s2, ar =6 m/s2 (ii) ^40 m/s2,0 = tan-13
216 G.R.B. Understanding Physics MECHANICS (Part-A)
15.
(«)« = £>A '
17.
Problems
1. (c) No, No
3.
5.
r(a)7. (b)
9.
11. (c)
13.
la |
15.
t
4v2 
r
2v? 
r
nr
v
v2
r
I a I increases by two time after each semi-circle and time taken to over each semi-circle 
decreases by half.
COo
3
I 
I 
I 
I 
I 
I
I 
I 
I 
I 
I 
I 
I 
I 
I 
I
I
2 2
V V
(b) 0, —cos 9,—sin 9
,nd
(d) -y/5 m/s2
v = 5 m/s, a = iy m/s2
7t v2 
Maximum at x = — is —- and minimum at
2a Aq2
(b) yi
V'
2R
v'
2R
ltd
T
Kinematics of Circular Motion 217
17.
19.
VBA=~V21.
♦ ♦♦
v cos—/
R
1 + sin—t / - R J
R ; / v2fl^=’7 I
2v2
CHAPTER
Projectile Motion
t + - 'at1, we have
2
xi +yj = vot cos 0O/ +^v0/ sin 0O ~~g(2
Comparing coefficients of i and J, we have 
x = vot cos Oo 
. _ 1 y = vot sm 0o--gt
G.R.B. Understanding Physics MECHANICS (Part-A)220
-bx2, where
y = x tan 0O -
y
■vo
-VO cosOq
H
°0 -x
0
R =This yields
y
/°
•x
2
v
The above equation can be given as y = ax
0 = R tan 0O -
g
2vq cos2 0o
As the above equation is a parabola, we can confirm that the path of the particle is parabolic.
We can use the equation of trajectory to solve many important problems of projectile motion and 
derive some essential characteristics of the projectile motion as presented below.
§ 6.3 Use of Trajectory Equation (Locus Equation)
Let us recast the trajectory equation
0
The direction of motion of the 
projectile is given by the slope 
m = 4an0o- —--
\ cos2e0/
of its path (parabola)
Fig. 6.3
a = tan 0O and b =
At x = 0, ni = tan 0O; at x = R, m = -tan 0O
If the slope is zero; (m = 0), we have
vq sin 0O cos 0O (_ R
X — I —
g
(c) Maximum height : As the projectile ascends (moves up), the slope of its trajectory de­
creases. At maximum height, the slope becomes zero. That means the projectile moves horizontally at its 
highest position.
gR2 
2vq cos2 0O
gx
Vq cos2 0o
y = x tan 0O - — ^-(1 + tan20o)
2 vo 
g*2 ■
2vq cos2 0o
(a) Horizontal range : When the particle passes through 
the point P on the horizontal (x-axis) as shown in the Fig. 6.2, its y 
coordinate is zero. Then the horizontal distance covered by the 
particle till it comes to the level of projection is termed as horizontal 
range R. Substituting y = 0 and x = R in the locus equation, we 
havevectors. For instance, when several persons pull a 
straight string at its different points, the tensions 7\,7]',T2,T2 etc., in the string are 
collinear.
The vectors A, B and C lie in x-y, y-z 
and z-x plane. Hence they are non-coplanar
Fig. 1.10
All parallel vectors need not be collinear
(viii) Coplanar vectors: If two or more vectors A, B and C , say, lie in 
—> —> —►
the same plane, they are said to be coplanar. Here A,B and C are 
coplanar as they act (lie) in the plane of the paper.
! !/ /-------------- > . >------ ,
ab c /— — — /
A, B, C_act along the same line; the collinear vectors A
A, B, C need not point in same direction
Fig.1.7
L t; ,.t2 t; ,t3 t3 .
All tensions T, ,T|’> T2> T2 etc are collinear 
but not unidirectional 
Fig.1.8
(ix) Non-coplanar vectors: When more than two 
vectors do not lie in the same plane they can be 
termed as non-coplanar vectors.
In the Fig. I.IO, since A, B and C lie in 
x-y, y-z and x-z plane respectively, these 
vectors are non-coplanar.
G.R.B. Understanding Physics MECHANICS (Part-A)6 .
(xi) Axial vectors : Some vectors such as angular velocity y
(1)
1.
+
Fig. 1.14
co, angular momentum L and torque t , etc., are 
directed along the axes of rotation. They are known as 
axial vectors. However, we cannot locate the exact 
position (origin) of the axial vectors along the axes. Hence, 
the axial vectors are known as “Pseudp vectors”^-
E etc., having a fixed origin and 
different orientations in space is known 
as polar vector. For instance, position 
vector 7 is a polar vector. We will 
talk more about position vector in the 
next chapter.
Axial vectors point along 
(or parallel to) the axis of rotation 
Fig. 1.12
into five equal parts each arrow represents a magnitude of one unit. 
Given vector
4--------1
Unit vectors
When we divide a vector (represented by an arrow) 
by its magnitude (length of the arrow) we get a unit vector 
(a)
Signs of polar vectors reverses when the coordinate axes are reversed, where as pseudo 
vectors (axial vectors) remain unchanged (invariant) upon (under) the reversal (inversion) 
of the coordinate axes.
2. Cross product of two polar vectors is a pseudo vector.
3. Cross product of two axial vectors is an axial vector.
4. Cross product of an axial and polar vector is a polar vector. For instance,
cox v =ar, cox r = v etc.
5. Pseudo vectors are not imaginary vectors like pseudo force.
(xiii) Unit vector : When we divide the arrow which represents a vector of magnitude 5 units
A ° lAl a; a shows the direction of A 
(b)
y
or(fp
Reversal (inversion) of 
coordinate axes reverses the sign of a polar vector 
Fig. 1.13
Axial vectors are represented by segmented arrows parallel to (or along) the axes of rotation.
(xii) Polar vectors: Vector such as force
F, velocity 7, electric field intensity
A
A
B n1
(a)
|n|=1
Hence, any vector A can be given as its magnitude times its unit vector, A = | A | a. Hence A |[ a. 
The unit vectors have no dimensions and no units. A unit vector shows the orientation of the 
corresponding vector in space.
A 
XB 
|n| I and n I and | n | 0, B is 
parallel to A and if n = tan
A 
Fig. 1.20
net effect or vector sum of the forces P] and F2. Symbolically, 
F = F| + F2 where | F | = OR, | F^ | = OP and IF2 | = OQ. Let us calcu­
late the magnitude | F| and the direction given as of the net force F 
B —> —> —> —> 
in terms of | F} |,| F2 | and the angle e between the vectors F} and F2 .
RM
OP + PM
Two forces F1 and F2 
act at a point (Box) to 
yield a resultant force F 
Fig. 1.18 
Q
OR = JoP2 + OQ2 + 2(OP)(Og)cos9
Substituting OR = | F |, OP = | A | and OQ = | B |, we have
l^l = W2+l£l2 +2|^|| FJcosG
From &0RM, the orientation (direction) of the resultant force F is given as
= tan
0^
------ ....L — -u-------
Equivalent parallelogram whose 
adjacent sides are represented 
by the vector F1 and F2 
Fig. 1.19
Let us produce OP and drop a perpendicular from R onto the 
produced line at M. Now we have a right angled triangle ORM. In this 
triangle, by Pythagoras theorem, | p | = OR = -foM2 + RM
Then substituting OM = OP + PM, we have
OR = J OP2 + PM2I \P .
R------ >|\~
The projectile has horizontal 
range R and maximum height H
Fig. 6.2
vq sin 20o
g
(b) Direction of motion : The direction of motion (velocity) of 
the particle is given by the angle 0 made by its instantaneous velocity 
~v with horizontal. This can be given as the slope of the path by differ­
entiating the locus equation with respect to x. Hence
c/ySlope = m = — = tan 0O -
dx
Projectile Motion 221
H =
y
v0 cosOq
X
O
and H =2. Since, R = we have
, substituting g/vjj = in the locus equation we have,3. Since, R =
Sol.
y = x tan 0O -
we have,
and ...(ii)
Vp sin2 e0 
2g
/f9o
— R/2
tan 0O = a
Vq sin 20o
g
— = tan 0O
x
Find the (a) speed of projection
(i) horizontal range
(c) maximum height reached by the projectile.
(a) By comparing the given locus equation with the equation of trajectory 
gx2
2vq cos2 0O
Substituting the slope m = — = tan 0O - = 0, we obtain x =
ntan 0O = — 0 R
Substituting the slope m = — = tan 0O - &—
dx vq cos2 0O
substituting this value ofx in the locus equation, we have
vq sin2 0O 
2g
From the above discussion we note the following points.
4
H 
I \P
•I* R/2—ri'
Fig. 6.4
b 
2vq cos 0O
sin 20o
■ * I u dy . n i x j ---, whereas — = tan 0O 1-------RJ_______ dx 2/?J
4. The shape of the curve above x-axis is symmetrical about a vertical line passing through 
the highest point.
Let us use the equation of trajectory to solve the following examples.
Ex. 1. A particle is projected in gravity so that its x andy coordinates are related as
y-ax-bx2
I. The projectile moves horizontally with a constant 
velocity vx = vocos0o as there is no horizontal 
acceleration ax = 0. Due to the effect of gravity its 
vertical velocity decreases from v0 sin 0O to zero at its 
highest position. Hence, the projectile moves horizontally 
with a velocity v0 cos 0O at the highest point of the 
trajectory.
vq sin 20o 
g
2g 2
x
R
G.R.B. Understanding Physics MECHANICS (Part-A)222
Substituting cos 0O = a
Fig. 6.5 Ans.v0 =
Substituting sin 0O = and Vq =
,2,2
Ans.
we will get
in the trajectory equationSol.
vo
P(x. y)y = x tan 0O -
x0
(R-x)tan 0O =We have
y1
Ry 
x(R-x) R-x x
= tan 02 -
a 
\ll + a'
, a we have R = - .
b
Ans.
- bx2 we will obtain the
1
x = R in the equation y = ax
R 
2sin0o cos 0O
from eq. (i) in eq. (ii), we have
— R------
 x--------
Fig. 6.6
By substituting — = 
S
g(a2+l) 
2b
g*2
2 vq cos2 0o
vq sin2 0O
the above answer. 2g
Ex. 2. A particle is projected from a point 0 such that it passes through a point P and again passes 
through a point Q at same level of projection. If /.POQ = a and Z.PQO = p, find the angle 
of projection.
x = — . Substituting 
2b
Alternately, substituting y = 0 for 
same answer.
(c)y is maximum when — = 0. Hence - = a-2bx = 0. This gives 
dx dr
x = in the given locus equation, we have H = ~.
Alternately, substituting the obtained value of v0 and sin 0O in H =
1 
7777 
|g(o2 + l) 
V 2b
(b) The horizontal range R = ^v° S'n C0S 
g
, cos 0O =
Then substituting — = tan a and —= tan p we have
x R-x
tan 0O = tan a + tan p Ans.
(d) Finding the point of collision between two projected bodies: When two bodies projected 
from same point collide in air, the point of collision of the bodies can be found by solving two trajectory 
equations (parabolas). If (x, y) are the coordinates of the point of collision, we can write,
— = tan 0|------------- = tan 02----- —
x 2v| cos 0, 2v2 cos2 02
where, vt and v2 are the velocities of projection and 0!, 02 are the angles of projection respectively. 
By solving the above equation first of all we can find x. Then substituting x in either locus equation we 
can find}'. Let us look at the following example.
Sol.
y
>>l = X| tan 0| -we have
y} = X] tan 02 -and —(ii) x
O
Fig. 6.7
...(iii)
respectivelyand t2 =
Ar = | z, -t21 = X] —(iv)
Ans.Az =
1.
2.
•v
g cos 9
x
at any
Fig. 6.8 lfe1 =e2 
no collision occurs
v2 COS 02 - V| COS 0] 
Vj COS 0] • v2 cos 02 
Now substituting X| from eq. (iii) in eq. (iv), we have
2V|V2 sin (0, -02) 
(Vj cos 0| + v2 cos 02 )g
Fig. 6.9 The tangential and radial acceleration of 
the projectile are given as a, = g sinO and 
(aj s gcosO where 8 is the angle made by the 
velocity v of the projecti'e with horizontal
For collision, Az = Zj -t2 > 0 Hence we have v2 cos02 > V| cosO| and 
0| > 02. Similarly, if Az(= Z2 — Zt) > 0, we have V] cos0] > v2 cos02 
and 02 > Op
Solving eq. (i) and (ii), we have
 2v\v2 cos 0| cos 02 sin(0| - 02) 
g(v2 cos2 02 -V|2 cos2 0|)
The times of motion of the particles till they collide are 
xi . x.Zi =-------— and Z2 =----- L—
Vj cos 0] v2 cos 02
Then the required time interval is
g*|2 
2v2 cos2 0]
ffll2 
2v2 cos2 02
Projectile Motion 223
Ex. 3. Two particles are projectedfrom the same spot with speeds V| v2 and angles of projection 0j 
and 02 respectively. Find the time intervals of their firing if .hey collide in air.
Let the particles meet at /’(x),^l). Then these coordinates must satisfy the trajectory equa­
tions of both particles at the point of their collision. Substituting xh in the corresponding 
trajectory equations of the particles,
If 0| - 02, no collision takes place.
(e) Tangential and radial acceleration : Since 
velocity ~v of the projectile at any angular elevation 
0 is tangential to its path, we resolve the 
acceleration g* of the projectile parallel to the line of 
motion (along the tangent to the path). Then we have 
a, = gsinO . Similarly resolving g perpendicular to 
the line of motion (along the normal) we have 
| ar | = g cos 0 where 0 can be calculated from the 
expression tan0 = = tan 90---- ■—g*-—
. . , .. , dx ?v2 co where
 2vq sin 90 cos 90
A —
g
projectile Motion 225
. The magnitude of
Fig. 6.12
Sol.
v0 . Since tan 90 = b/a, cos 90 =radius of curvature at P is
we have pp = (a2 + b2)il2lga .
Ans.
y
p,'
Fig. 6.13
Ans.Then.
Student Task
vq cos2 9 
g 
(given)
a
Ja2'+b2'
a
Look at the following example :
Ex. 5. A particle is projected with a velocity v =ai +bj. Find the radius of curvature of the trajec­
tory of the particle at the (/) point of projection (ii) highest point.
(i) Let the angle of projection be 9. At the point of projection P, a„ = g cos 90 . Hence the 
v2 v2 p VP = vo
p an g cos 90
v2
p = —, where an = g cos 9 and v =
I X I9 can be calculated by the expression tan 9 =+ 2(OP)(PM) + RM2
In kRPM, RM2 + PM2 = PR2
Hence OR = yjoP2 + PR2 + 2(OP)(PAf)
Since in APRM, PM = PF cos 9 and substituting PR = OQ in 
O OPRQ, we have
| F2 | sin 9
l + |F2 |cos9
In general, the resultant R of any two vectors A and B acting simultaneously at an angle 9 
can be given as R = A+ B. If A,B are represented by the adjacent sides 
of a parallelogram, its diagonal passing through the point of application of 
the vectors A and B represents the resultant R . The magnitude of R
is | R | = J A2 + B2 +2ABcosQ . The direction of R is given as
_] Zfsin9 -»
$ = tan + gcosg w*to A • *s known as “ parallelogram law of 
vector addition”. This is an “experimental fact”.
RM
OM
Substituting PM = PR cosQ = \F2\ cos9 
and RM = P/?sin9 =| F2 |sin9, we have
Sol.
0^
Fi
0
Fig. 1.21
Ans.
2^3
Sol.
f2
F2
F. 5 ± VlO ' 
Substituting 0 = 150° and solving the quadratic equation, we have — =---- j=—
F2 2V3
a-> 2— = —, we have,
«1 1
• When we reverse the vector F}, the magnitude of the resultant force increases (doubles) 
which eventually doubles the acceleration. We have two values of F|/F2 and hence two differ­
ent arrangements are possible. One we have shown above and the other can be shown by 
interchanging the names of the vectors in the above Fig. 1.21.
Ex. 2. The resultant of p and Q is R. If Q is doubled, R is doubled; when Q is reversed, R is 
again doubled, find P.Q-.R.
—> —>
Let 9 be the angle between p and Q . Then
R2=\P+Q\2=P2+Q2+2PQCQsS ...(i)
If Q is doubled, R is doubled. That means, the magnitude of resultant of 2Q and P is 2R 
(2K)2 = P2 + (2Q)2 +2P(2Q)cos0
This yields, 4P2 = P2 + 4Q2 + 4P0cos0
When Q is reversed, R is doubled. Hence, the magnitude of resultant of P and (-0) is 2R.
F'
- z * F -
10 G.R.B. Understanding Physics MECHANICS (Part-A)
Ex. 1. Two forces of unequal magnitudes simultaneously act on a particle making an angle 0(= 150°) 
with each other. If one of them is reversed, the acceleration of the particle is doubled. Calculate 
the ratio of the magnitude of the forces. 1
Let the two forces be forces F| and F2 •
The resultant of these forces, is Fj' = Fj+ F2. Then | Fj | = | Fj+ F2 |. Using parallelogram law 
of vector addition, we have | Ff | = J^2+F22+2F]F2cos0
If the direction of F| is reversed, new resultant force F2 = (-F1) + F2. Using parallelogram 
law of vectors, the magnitude of new resultant is
| F2 I = 7F12 + F2 ~ cos0
Since force is directly proportional to acceleration,
|f2 Lq2
l^'l 01
Substituting | F/ |,| F2 | and
7f12 + F22-2F|F2cos0 _
^F12 + F22+2F1F2cos0
II
...(Hi)
P£?cos8 =Eq. (i)-Eq. (iii) yields ...(iv)
Eq. (i) + Eq. (iii) yields -(v)
Eq. (ii) + Eq. (iv) yields -.(vi)
Ans.
B
Fig. 1.23
B.
y
Ba' AaC
P
y’P'
Lami’s theorem 
—> —> —>
When three vectors A, B and C , say, acting at a point yield zero resultant, we can represent 
them by a triangle as discussed in previous section.
A
A, B and C represent the 
sides^ofjhe triangle;
A + B + C=0 
Flg.1.22
Then, A+B+C = 0. It means that the sum of three vectors A,B 
and C represented by the sides of a triangle considered in cyclic 
manner (same sense/order) must be zero; A+ B+C = 0.
Vectors
Then,
This yields
_ r— c
When A, B and C represent the side of a triangle taken in same sense, A + B ♦ C ■ 0
Fig. 1.24
I
(2R)2 =P2 +Q2+ 2P0cos(I 80° - 0).
47?2 =P2 + Q2 -2^cos0
-3R2
4
p2+q2 = 5±
2
P2 + 402 = 7/?2
= ^R and p = rSolving Eq. (v) and Eq. (vi) we obtain Q
Hence P.Q-.R = 41 : VJ :>/2
Triangle law of vectors
The vectors A, B and R in the parallelogram law of vectors can 
also be represented by the sides of any triangle of the parallelogram made 
by its diagonal that represents R.
Since, R = A+ B, we conclude that if the vectors A and B are 
represented by two sides of a triangle measured in the same sense, the 
third side taken in the reverse order (sense) represents the resultant R. of 
the vectors A and B. This is what we call as “triangle law of vectors”.
—>
By taking the resultant R in reverse order we have another 
vector C(=-R).
12
Polygon law of vectors
A —> —►
/!„_] respectively, in same cyclic order (sense), the closing side, that is, nA side of the 
polygon taken in the reverse order (sense), represents the resultant R of all the above vectors. This is 
known as “polygon law of vectors".
Ml l?l |C| 
sina sinp siny’
Rn-3
\a3
a2L
+ A-l
+ An-\
A, An_\ are represented by a polygon / 
of n sides, following the logic of triangle ' 
law of vectors, we have
A
When the vectors Av A2... A^ represent (n - 1) 
sides of a polygon, the closing side represented 
by the vector R will be the resultant of all the vectors
Fig. 1.25
If n number of vectors At, a^_,
Vectors 13
A,
Fig. 1.26
Sol.
F = -2(i + j) N
2N
c
3N O
f2
♦xA BFl 4N
You should not be tempted to write the net force as zero because the forces are passing through 
the sides of the square but are not represented by these sides. Had the forces been represented 
by the sides of the square, their magnitudes would have been equal and hence the resultant of 
the forces would have been zero. 
Fig. 1.27
Hence the magnitude of F, that is, | F | = 2^2 is directed in 3rd quadrant, as shown in the
figure 1.27. Ans.
y
♦
D —
\a3
/a2
F*3
-2?
2(2/450;
/t| + zij +.... + An — 0 
That means, if n vectors are represented by 
the sides of a polygon of n sides in cyclic £ 
manner, the resultant of these vectors is a 
zero vector. Triangle and rectangle, etc., are * 
the polygons of sides three and four re- 
spectively. The non-coplanar vectors can An 
be represented by a polygon of sides which 
cannot lie in a single plane. Sometimes, for 
coplanar vectors represented by a polygon, 
the resultant (closing side of the polygon) 
may cross the component vectors (other sides of the polygon) as shown in the above figure.
Let us, now apply the above laws of vector addition.
Ex. 3. Four forces act along the sides of a smooth square frame ABCD in the order A -> B, B -> C, 
CD and D -> A. If the magnitude of the forces are F\,F2,F3 and F4 respectively, find 
the resultant force acting on the frame. Assume F^ = \ N, F2 = 2 N, F3 =-3 N and F4 = 4 N. 
Let us consider x-y coordinate system. The resultant of all the forces is
F = Fl+F2+F3+F4;
After bringing the tails of all the vectors to a point 0 and substituting F} = 1/ N, F2 = 2j N, 
F3=-3i N and F4=-4/N, 
we have
14
Sol.
B,
O' Q
...(H)
QS = | A | =
and Ans.
This yields
Fig. 1.28
OQ = OS cos P + S0cos a
|R|
Fig. 1.29
and | B | =
$P
A
l*l
—>
I /?|sing 
sin(a + P)
|A| 
/l80°-(a+p\ 
P a
OgsinP J?sinP 
sin(a + p) sin(a + P)
G.R.B. Understanding Physics MECHANICS (Part-A)
Ex. 4. (Oblique Components) Find the components of a vector R along two straight lines situated at 
both sides of the vector ft making angles a and p with it.
' —> —► —►
Since the components A and B of the given vector R are chosen in the given directions, R 
—► —> —> —> —>
is the resultant of A and B ; then, R = A+ B. Hence R is represented by the diagonal OQ of 
the where the component vectors A and B are represented by the adjacent sides OP 
and OS respectively according to the parallelogram law of vectors. Then we convert the 
parallelogram OSQP to the vector triangle OSQ which represents the vectors R,A and B as 
shown in the Fig. 1.28 (i). By converting to a scalar triangle OSQ as shown in the Fig. 1.28 (ii i), 
we can write..
and SR = (?Ssinp = S£)sina
Solvingthe Eqs. (i) and (ii), we have
Alternate solution
Using Lami’s theorem, we have
Ul_ |?| = 
sin P sin a sin{!80°-(a + P)}
sin(a + P)
os = |?l=-^ 
sin(a + P)
Vectors 15
y
XFig. 1.30
y
R
I
i
I
I
/a
A
Unless it is stated, hereafter we will use the term components of a vector for rectangular 
coordinate system. However, if it is asked to find the components of a vector along two given 
lines, we should not immediately assume it as rectangular components. When the given lines are 
perpendicular to each other, we can call the components “rectangular components”.
The parallelogram and triangle law of vector addition are valid for two vectors only. At a time we 
cannot add more than two vectors using these laws. For more than two vectors acting at a point, we 
will use another method to add them. This is known as component method.
Component method
Suppose four vectors At,A2,A3 and 4t simultaneously act at the origin of the coordinate 
system in four different quadrants, as shown in the Fig. 1.32. How do we find the resultant of these 
vectors?
First of all we have to express each vector in terms of its rectangular components as described 
above. Considering the acute angles 0|,02,03 and 04 made by the vectors 4, ^2,^3 and respectively 
with + x and -x directions, we have
I
a
A
If a + p*-^, the coordinate system (x-y) is known as oblique 
—> —>
coordinate system. Hence A and B are termed as oblique 
—>
components of R.
Rectangular components
If a + P = n/2, the coordinate system (x-y) is known as rectan­
gular coordinate (because the parallelogram containing the vectors 
—► —>
becomes a rectangle). Hence, A and B are termed as rectangular 
components of R ; Then R = A + B, where A =| A | i and B =| B | i. —
—> —> —► —► B
Substituting | A | = | R|cosa and | B | = | R|sina, finally we have, 
R = | R | cos a i +1 R | sina j where a is the acute angle made by R 
with positive (or negative) x-direction. When the components of R lie 
in negative x and y-directions, we must use -/ and -j respectively.
ol
A vector R hasjrectangular 
components A an 
where E AX/ 
i=l
G.R.B. Understanding Physics MECHANICS (Part-A)
A} = A} cos0]/ + At sinG]/, 
—> A A
A2 = A2 cos02(-/) + A2 sin 02j 
/13 = A3 cos03(-/) + A3 sin03(-j) 
and A4 = A4 cos04z" + 44 sin04(-j)
Adding all the vectors we have the resultant, 
we have A = Ai+ A2+ A3+ A4
Substituting the above values of the vectors and adding 
x-components and ^-components separately we have 
—1> A
A = (4] cos 0] - A2 cos 02 - A3 cos 03 + A4 cos 04 )i 
+(4! sin 0] + A2 sin 02 - 43 sin 03 - 44 sin 04 ) j 
In general, the resultant A of n vectors acting at a 
>int is given by adding their rectangular components which 
in be stated as
Four vectors Av A2, A3 and A4 acting at 
the origin at angles 0V 02, 03 and 04 with 
+ x and -x-axes, are vectorically added by 
algebraically adding their rectangular 
components in x and y-axes 
Fig. 1.32
i=n -» 
and E ^y,
An'
Ax
Ax and A are the rectapgular 
components of A 
Fig. 1.33
A,
o2
A3x (
i=n -» —> i=n —» —>
Assuming E A, = 4X and £ Ay = Ay we have 4 = Ax+ A . 
i=\ i=l
If we assume the components in +x and +y-directions, the magnitude
of 4 is
-I I 4 ■ I = tan” ——, with respect to the +x-axis.
_______________Hxl___________________
While writing +ve or -ve, we note the following points:
4X and Ay are the components of the vector 4 , known as vector components.
^x, = |4x|(+z) and Ay, = | 4^ |(±y)-;+ve sign is used when the vectors Ax and 
A,, are directed along +ve axes and -ve sign is used when the vectors point along 
-ve axes.
are the vector additions of the x and y-components of the vectors
-> i=n -> z=n ->
A = £ A*, + £ Ay. • 
(=i i=i
A2y ..
!-
17Vectors
3.
4.
Ex. 5. y
■3N
1
.30°
.3Sol. 45°
y
3N
1N x
2N
X = -
2
,2
1N 2
4 I
and directed at an angle 4> = tan 
figure.
with -ve x-direction, as shown in the
Ans.
2
*
Fig. 1.35
\2 J 
3-2>/2 
.373+2^2-2.
F4=7x2+/ =
| AX/ | = A, cos0, and \ Ay/1 = A, sin 0,, where 0( = acute angle between the vectors A, 
—> 
and +x or -x-directions. These are known as scalar components of A • —>
Alternately, first of all see in which quadrant the vector A, is oriented. A^ is positive 
for 1st and 4th quadrant and -ve for 2nd and 3rd quadrant. Similarly is +ve for 1st 
and 2nd, At is -ve for 3rd and 4th quadrants. Finally we will add all x andy-components 
of the vectors algebraically.
Four ants 1, 2, 3 and 4 are pulling a grain with forces of 
—>
magnitudes 3 N, 1 N, 2 N and | F | N as shown in the Fig. 1.34.
—>
Find the force F if the grain remains in equilibrium under the 
action of the above forces.
The forces exerted by the ants are given as,
F, = 3(cos30°f +sin30° j)
F2 = -\i
F2 = 2 cos 45° i + 2sin45° (-j)
and F4 (= F) = x i + y j, say.
Since the grain is in equilibrium, it experiences a net force
F = F] + F2 + F3 + F4 = 0
Substituting the values of the above forces, we have
i j + ^2 + x'l + f— - 41 + .y'b’ = 0
I 2 ) k2 J
Comparing the coefficients of i and j, we have
- + 72-1'1 and =
.2 J \2 J
Hence the force exerted by the 4th ant has the magnitude
. 2 J
F '2N
Fig. 1.34
zw
3X«°
G.R.B. Understanding Physics MECHANICS (Part-A)18
A
■x
and directed in
Sol.
is given by
y
Sp 
aJ
A
The vector A has three rectangular components 
Ax, Ay and Az in cartesian coordinate system 
along x, y and z-axes respectively;
A = Axi + Ayji + AIk 
= Ax + Ay + Az
Fig. 1.36
—» A —> —> A
= | Ay | j and A, = }AZ | k, we have,
Rectangular components in three dimensions (Optional)
Let us express a vector A oriented in space making angles a, 0 and y with x, y and z-axes 
respectively, in terms of its components along the axes. For this we drop a perpendicular PM from the 
tip of the vector A onto the x-z plane and PT onto the j'-axis, as shown in the Fig. 1.36.
Then we drop perpendiculars MS and MR from 
M onto x and j^-axes. In this way, now we have three 
segments OS, OT and OR. These are termed as the 
projections or components of OP along x, y and z-axes 
—> —>
respectively. Since, OP = | A | we can write OS = | Ax |, 
OT =| Ay | and OR =| A. | ■ It means, Ax, Ay and 
A- are the vector components of A along the 
coordinate axes (x, y and z) respectively. In other words, 
Ax, Av and A. are the rectangular (vector) compo- 
nents of A in cartesian coordinate system comprising 
three mutually perpendicular axes x, y and z. Then, we 
can write
A — Ax+ Av+ A„
* J ~
Since A* = | 119 Ay :
A = | Ax | i +1 Ay | j +1 A. | k. The magnitude of A is | A | = ^A2 + A2 + A2
space with cosa = —— = /, cosp = ^^- = m and cosy = —— = n; I1 + m2 + n2 = 1; we can call 
Ml Ml Ml
them (/, m and n) as direction cosines because the cosines of the angles are used to measure the 
direction of A.
Let us use the above ideas in the following example.
Ex. 6. A bird flies due east through a distance of\ 00 m, then heading due north by a distance of 50 m, 
it flies vertically up through a distance o/20 m. Find the position of the bird relative to its 
initial position.
The final position r of the bird is followed by three successive displacements S], s2 and S3 
as shown in the Fig. 1.37, where = 100/ ,s2 = 50 j and s3 = 20k , as shown in the Fig. 1.38 
Hence the position vector
r = Sj+ s2+ S3 = 100/ + 50J + 20k.
Then, |7| = >/1002 +502 + 202 and the direction of 7
; as
M
,2
7
Vectors 19
zz
-y(N)
y(N)
Fig. 1.38
/ =
and Ans.
Ans. 134 m,a = cos'
Since
2 n= .—
7129
Already we have acquired the basic knowledge of addition of vectors. Using the ideas of vector 
addition and components of vectors, let us try to establish a fact that any vector remains 
invariant when the coordinate axes are rotated.
Any vector A (say) remains unchanged 
on rotation of coordinate axes 
Fig. 1.39
A
20 k
x 
(E)
OP = OF+ FR+ RE+ EP
RE = FM
(E)X
S2
Fig. 1.37
s3 ► A
'2 50 j
Rotation of coordinateaxes
Here, we have taken the coordinate axes x -y and placed a vector A. Then rotating the coordinate 
—>
axes through certain angle 0, say, let us measure the vector A with respect to new coordinate axes 
x'-y'.
In the coordinate axes x'-y\ A = x'i'+ y'j' where
| AI = OP, x' = OR and y = OS\ A = OP = OR+OS
Substituting OR = OF+ FR and OS-= RP = RE+ EP
we have,
I00 IO
107l29 ”7129
5 
m - ■ ____
7129
Student Task^
> What is the maximum length of a rod that can be kept inside a closed box of dimensions 
10 m*5 m*3 ml Find the angle of orientation of the rod.
, B = cos-i-4= and y = cos"’-J= where a,B and Y 
7134 7134 7134
are the angle made by the rod with the sides of lengths of 10 m, 5m, 3m respectively.
G.R.B. Understanding Physics MECHANICS (Part-A)20
and
B
B-A
0
A
A-B
A
Fig. 1.41
§1.7 Subtraction of Vectors
As discussed in previous section, many vectors 
can be added at a time. Here we will talk about the 
subtraction of vectors. When we use the word “subtraction 
of vectors” it implies there are two vectors; one is 
subtracted from the other. Let us subtract a vector B 
from another vector A. We can write it as A- B = C •
-A *■
A-BBy
VAB ~ VA ~ VB
Hence, relative velocity is a subtraction of two velocities.
It means that if we take the negative of vector B 
and then add it with vector A following the law of vector 
addition, we can find A+(-B) = A - B. The magnitude 
of A-B is, H-8| = a/|1|2 +!?|2 -2| ?|| S|cos9, 
where 0 = angle between the vectors A and B .
The subtraction of two vectors is an addition of one vector with the negative of the other 
—> —► —> —> e ,
vector; A - B = A + (-B). Hence subtraction of two vectors is a vector quantity.
We will use the idea of vector subtraction while studying the concept of relative velocity. For 
instance, velocity of A relative to B is given as
The components of a vector changes on the rotation of coordinate axes. However, any vector 
remains invariant under the change in coordinate system (translation and rotation of coordinate 
axes)
That means A — B is represented by the diagonal of the parallelogram passing 
through the tips of the vectors A and B directed from the tip of B to that 
—'> —> —+ —> —>
of A . Since A-B =-(B - /I), vector subtraction is anti-commutative.
-B ?----------------V
A when added with negative of 
B (-B), it gives A-B; 
A-B = -(B-A) 
Fig. 1.40
FR+ EP = ME+ EP = MP, we have
OP = (OF+ FM) + MP = 0M+MP\ finally substituting
OM-xi, MP = yj, we have A - x'i'+ y' j' = xi + yj
* win define it in the next chapter aa 7 . A7. where 1“ctageTXt ^toTlVus
Fig. 1.42
hy
 = tan = tan x0
This yields 5
oAns.Hence
Fig. 1.43
1.
2.
magnitude of the displacement is | A 71 = \]r2 + r2 - 2rtr2 cos 6
Since the insect moves in a circular path, r} = r2 = R
e
2
cot— = 
2J
Ex. 7.
Sol.
7?sin 0
J?(l-cos0)
Aliter
We can solve the above example by component method. Since Ar = r2-, substituting q = Ri
------A A ------
and r2 = 7?cos0 i + 7?sin0 j, we have A r =
ar 
\P
a7x
A A —> —>
-/?(! -cos0)z + /?sin0 j = &rx+ &ry, say.
y
X 
(e ri = —
2
Vectors 21
apply this concept blindly for the sake of understanding the idea of vector subtraction in the following 
example. 
An insect moves in a circular path of radius R through an angle 0. Find its displacement. 
When the insect moves from position I to position 2, the displace- 2
ment ~s = change in position vector A r. Since A7 = r2-r], the ~ / \\
-> 9 ->■Hence IA r I = 27? sin — and the direction A r is given by 
2
-> 0 ->Hence | A r | = 27?sin— The direction of A r is
given as 0 = — + —, as shown in the figure.
Ans.
 , n 0P = n- = - + -
|A^|
----— = tan
|a£|
n 0
2~2
| Ar | = 27?sin— * 0 but A | r | = | r2 | — | /] | = 0. Hence, |Ar | may not be equal to 
A|7|. That means |r2-r(| is different from ||r2| —)/, ||. |r2-r( | 90°
22 G.R.B. Understanding Physics MECHANICS (Part-A)
§ 1.8 Scalar Product of Two Vectors
In physics we define some scalar quantities like work, power, flux, 
etc., as the product of two vectors. For example, work done IF by a force—>
F is defined as the product of the displacement of a point at which the 
force is acting and the component Fx of the force along the displacement 
(direction of motion of the point). Symbolically, = || 7|» where
|Fx| = |?|cose.
—> —>
This gives W = | F | | s | cos 0. Since the scalar quantity W is defined
as a product of two vectors F and s , we can call this product “scalar product”. In short, substituting 
cos0 by a dot (•) we write W = F- s . Hence the scalar product can also be termed as a “dot product”.
There are many instances in physics where we need to express some scalar quantities like flux, 
pressure, power as products of two vectors. For this, we need to develop the idea of scalar (dot) 
product of two vectors.
If, in general, a scalar quantity C is defined as the scalar (or dot) product of any two vectors A 
and B, C is given as C = | A || B | cos0 = A- B, where 0 is the smaller angle between the 
vectors A and B.
—X
Vectors 23
Sol.
A
a
and
Sol.
Ans.
4. The magnitude of scalar product 
is minimum, that is, zero, when the 
vectors are orthogonal.
a ■ a = i ■ i = j • j = k ■ k = 1 
i ■ j = j -k -k-i =0
(iv) o-6 = afecos0 = cos0 (•.•o = Z> = l) where -1) A-(-A) = -A-A=-A2
(iii) A-B = AB cos90° = 0
Ans.
Let us use the scalar product to prove some geometrical theorems. 1 am presenting the proof of 
one of them by using dot product.
Ex. 9. Find A-B, if
Z 0=0° 
A
| C Imin = -4 • c°s 0 = 0 when 0 = 90°
(i) /4 J = /L4cosO° = /l2
G.R.B. Understanding Physics MECHANICS (Part-A)
Ans. 0 = cos
is given as
B
A,
AxB
A-
Right hand thumb rule: If you want to find the 
direction of A x B, first of all, curl or wrap the fingers of 
your right hand by swinging (turning) the fingers from A 
to B following the shorter (smaller) angle 0 between these 
vectors. Then the extended thumb will show the direction 
of A x B. This is what we call right hand thumb rule. I 
present hereby another way to use the above rule. 
—> —>
When you write A x B, first of all, bring the tails of 
these two vectors to a point. Then measure the smaller
24
Student Task^
Ifthe curling of fingers of the right hand 
represents the sense of measuring the angle 
9 from A to B, the extended thump 
represents the direction of A x B 
Fig. 1.52
> In the foregoing example, find the angle between A and B.
Ax Bx + Ay By + A. B.
^(A2 + A2 + A2)(B2 + B2 + B2)
§1.9 Vector Product of two vectors
In the last section, we developed the idea of scalar product to define some scalar quantities as the 
scalar (dot) product of two vectors. On the other hand, we have some vector quantities like torque, 
angular momentum which are defined as the product of two vectors. For this, we will develop the idea 
of “vector product”. Let us take an example lifted from our daily experience.
There are many other instances in physics where we define a vector as the product of two 
suitable vectors. Hence we need to develop a general formula (expression) for the vector product.
If a vector C is to be defined (or expressed) as the product of two vectors A and B, C is 
given as C = AxB so as to satisfy the right hand thumb or right hand screw rule. The magnitude of C
|C|(=|JxB|) = |^||B|sine)
where 0 = smaller angle between A and B . Substituting | A | sin 0 = p} Z ip2
-> -+->-» Z p\;
or | B | sin 0 = p2 we can write | C | = px | B | = p2 \A |, where p] and p2 are / r
the lengths of the perpendicular dropped from the tips of the vectors A and |A * B| = |A|p2 = |B|p1 
-» -> -> where p1 and p2 are the
B onto the vectors B and A respectively as shown in the Fig. 1.51. The perpendiculars dropped
from tip of A onto B, tip of 
direction of C is given by the “right hand screw rule” or “right hand thumb B onto A respectively 
rule” as described below. F'9,1,51
Vectors 25
A
e
CD
I
wood
Since, Ax B = -B x A, cross (vector) product is anti-commutative.
Sol.
Sense of measuring the angle© for A*B 
is anticlockwise and for B * A is clockwise.
Hence A » B is directed vertically up if 
the plane of the paper is horizontal 
Fig. 1.53
Fig. 1.55
0 = 180°
The sense of measuring 0 from A to B* 
is represented by the rotation of the 
right hand screw. The direction 
of motion of the tip of the screw^ 
represents the direction of A x B 
Fig. 1.54
angle 0 between the vectors from A to B . If A and 
—>
B are coplanar with the paper, the sense of measuring
—> “■>
the angle 0 is anticlockwise for Ax B. When we curl or 
wrap the fingers of the right hand in the sense of measur- 
ing the angle between the vectors A and B, that is, in 
anticlockwise sense, the extended thumb will point up- 
ward (outward of the paper). Hence the direction of A x B 
is outward which is associated with the anticlockwise 
sense of rotation of the angle 0.
Following the above rule, if we reverse the order of the cross product, its direction of vector (or 
—> —> —■> —>
cross) product reverses. Hence, Ax B = -B x A.
However, the magnitude of the vector product remains 
unchanged upon reversal of the order of the product. This tells 
us that vector product is anti-commutative.
Right hand screw rule: If the above methods are not 
very comfortable for some of you, 1 present another rule to find 
the direction of a vector product. This is known as right hand 
screw rule.
First of all we bring a right handed wood screw (not left 
handed) and put it vertically on a wooden block. When we rotate 
the screw in the sense of measuring the smaller angle between 
A and B, the direction of motjpn of the screw inside the wood 
gives the direction of Ax B. When we reverse the order of the 
vector product, the sense of measuring the angle is reversed.
Eventually the vector product will be oppositely directed as discussed little ago.
B
0 = 0°
-> B
Let us discuss some fundamental vector products through the following examples.
—Ex. 10. Find the cross product of two vectors A and B which are
(i) parallel and equal (if) antiparallel and negative (Hi) orthogonal.
If A is parallel (or equal) to B , 0 = 0°. ------------- >
Hence | A x B | = ^Bsin0° = 0
If A is antiparallel (or negative) to B . 0 = 180°. Hence
| Ax 51 = /IBsin 180° = 0 
A B
Fig. 1.56
G.R.B. Understanding Physics MECHANICS (Part-A)26
B
A x E
,B
AFig. 1.58
Fig. 1.59
Sol.
C ~7
e
B
B
A 
j
c
Since cross product of two parallel vector is a zero vector 
(but not zero) we have, 
± i x i = ± j x j = ± k x k = 6;
i x j’ = k, j*k = i and k*i = j.
Reversing the orders j*i = -k, k* j = -i and i xk = -j.
A = B*C Ans-
The idea of vector product tells us that the magnitude of the area formed by a 
quadrilateral whose adjacent sides are represented by vectors B and C is
| A | =| B || C | sin0, where 0 is the angle between the sides and the direction of the 
area is perpendicular to the plane of the quadrilateral.
—> -> —►
If we choose to write A = C x B, the area vector is directed downwards. The 
choice is yours.
X
Fig. 1.61
C Fig. 1.57 
—If A is perpendicular to B, 0 = 90°.
Then | C | = | A x B | = AB sin 90° = AB and C(= A x B) is directed outward obeying right hand 
thumb rule. A x B must be perpendicular to the plane containing A and B . Ans.
Let us note the following points.
Ex. 11. (Area as a vector) How to find the area of a parallelogram whose adjacent sides have lengths 
B and C making an angle 0 with each other. Mention the direction of the area vector.
Let us take a parallelogram whose adjacent sides a 
have magnitudes B and C respectively.
Area of the parallelogram is A = base x height,
where base = B and height =Csin0
This yields A = BCsin 0
Considering that the sides of the parallelogram
represent position vectors B and C, we have | A | = | B|1C | sin 0. Hence the area A can be
—> —>
represented as the vector product of B and C as
' z&4Csin9/
------ B------ *Z
Fig. 1.60
27Vectors
n
vector of the area can be given as n =
Ex. 12. Find A x B if
A = Axi + Ayj + A_k and B = Bxi + Byj + B.k.
Sol.
C = A x B , we have
A x B = Ans..
B.
i
A.
8
Fig. 1.62
k
A.
' J
Bx By
Substituting
“A A * ~A M *
A = Axi + Ayj + A.k and B = Bxi + Byj + B.k in the formula
C = (Axi + Ayj + A.k) x (Bxi + Byj + B.k)
= AxBxi x i + AxByi x j + AxB.i xk +AyBxj x i + AyByj x J + AyB.jx k
+A.BX kxi + A.Bykx j + A.B.kxk
Since / x j = +k, jxk = i, kxi = j and / x/ = jx j = ic xk = 0, 
we have c = AxByk - AxB.j - AyBxk + AyBJ + A.Bxj - A.Byi 
Rearranging the terms, we have
C = {AyB. - A.By)i-(AXB. - A.BX)J+ (AxBy - AyBx)k
Now we can put it in a compact form which is easier to remember as the follow- k * i = j 
ing determinant. Fig. 1.63
If two vectors B and C are represented by the adjacent sides of 
—> —> —>
a parallelogram, its area is given by A = B x C. Dividing the area 
vector A by its magnitude, we will obtain a unit vector of A 
which is generally normal to the plane of the area. Hence, the unit
— = The choice A 
A \BxC\
of the direction of area vector, that is n is arbitrary. It depends 
upon how you define area (either as Ax B or Bx A )■
G.R.B. Understanding Physics MECHANICS (Part-A)28
by 0 = sin-1 B
0
Fig. 1.64
and p2(= 2?sin0) =Pi(= A sin 0) =
AxB =
B
A
§1.10 Some Fundamental Laws of Vector Algebra
Commutative law
(i) Addition : Since A + B = B + A , addition of two vectors is commutative.
-A
A
J
A+B=B+A
Fig. 1.65
(ii) Subtraction : Since A - B = -{B - A), subtraction of two vectors is anti-commutative.
1. The angle between the vectors A and B can be found
|7xfl|
AB
——>
2. If the angle 0 between the vectors A and B is given, 
the length of the perpendiculars are given as
\ A x B | D-m M x BI--------' and p2(= 5sm0) =-------- .
B A
3. If A = Axi + Ayj + A.k and B = Bxi + Byj + B.k
i j k
Ax Ay 
Bx By B.
p,= Bsin0 ‘-.’(p
I____
---------------------- - A
(A — B) = — (B — A) 
Fig. 1.66
A
y
X
z z‘
C
B B
■*X 
-> 
A
To 
+ 
Tm
Associative law
(A + B} + C = A + (B + C), when we add A 
with B to get (A + B), then add (A + B) with C 
to finally obtain (A + B) + C .
—> —> —> —>
Again we add B and C to get (5+C),then 
add A with (5 + C) to obtain /l+(2? + C). By
B
Vectors 29
(iii) Scalar