capítulo 2

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CHAPTER 2 
 
 
Motion in One Dimension 
 
 
 
 
1* · What is the approximate average velocity of the race cars during the Indianapolis 500? 
Since the cars go around a closed circuit and return nearly to the starting point, the displacement is nearly zero, 
 and the average velocity is zero. 
 
2 · Does the following statement make sense? “The average velocity of the car at 9 a.m. was 60 km/h.” 
No, it does not. Average velocity must refer to a finite time interval. 
 
3 · Is it possible for the average velocity of an object to be zero during some interval even though its average 
 velocity for the first half of the interval is not zero? Explain. 
Yes, it is. In a round trip, A to B and back to A, the average velocity is zero; the average velocity between A and B 
is not zero. 
 
4 · The diagram in Figure 2-21 tracks the path of an object moving in a straight line. At which point is the object 
 farthest from its starting point? (a) A (b) B (c) C (d) D (e) E 
(b) Starting point is at x = 0; point B is farthest from x = 0. 
 
5* · (a) An electron in a television tube travels the 16-cm distance from the grid to the screen at an average 
speed of 4 ´ 107 m/s. How long does the trip take? (b) An electron in a current-carrying wire travels at an 
 average speed of 4 ´ 10-5 m/s. How long does it take to travel 16 cm? 
(a) From Equ. 2-3, Dt = Ds/(av. speed) 
(b) Repeat as in (a) 
 Dt = (0.16 m)/(4 ´ 107 m/s) = 4 ´ 10-9 s = 4 ns 
Dt = (0.16 m)/(4 ´ 10-5 m/s) = 4 ´ 103 s = 4 ks 
 
6 · A runner runs 2.5 km in 9 min and then takes 30 min to walk back to the starting point. (a) What is the 
runner’s average velocity for the first 9 min? (b) What is the average velocity for the time spent walking? (c) What 
is the average velocity for the whole trip? (d) What is the average speed for the whole trip? 
Take the direction of running as the positive direction. 
(a) Use Equ. 2-2 
(b) Use Equ. 2-2 
(c) Dx = 0 
(d) Total distance = 5.0 km; Dt = 39 min 
 
vav = (2.5 km)/[(9 min)(1 h/60 min)] = 16.7 km/h 
vav = (-2.5 km)/(0.5 h) = -5.0 km/h 
vav = 0 
Av. speed = (5.0 km)/[(39 min)(1 h/60 min)] 
= 7.7 km/h 
 
Chapter 2 Motion in One Dimension 
 
7 · A car travels in a straight line with an average velocity of 80 km/h for 2.5 h and then with an average velocity of 
40 km/h for 1.5 h. (a) What is the total displacement for the 4-h trip? (b) What is the average velocity for the total 
trip? 
(a) 1. Find the displacements of each segment; 
 Use Equ. 2-2 
2. Add the two displacements 
(b) Use Equ. 2-2 
 S1 = (80 km/h)(2.5 h) = 200 km; 
S2 = (40 km/h)(1.5 h) = 60 km 
Stot = 260 km 
vav = (260 km)/(4 h) = 65 km/h 
 
8 · One busy air route across the Atlantic Ocean is about 5500 km. (a) How long does it take for a supersonic jet 
flying at 2.4 times the speed of sound to make the trip? Use 340 m/s for the speed of sound. (b) How long does it 
take a subsonic jet flying at 0.9 times the speed of sound to make the same trip? (c) Allowing 2 h at each end of the 
trip for ground travel, check-in, and baggage handling, what is your average speed door to door when traveling on 
the supersonic jet? (d) What is your average speed taking the subsonic jet? 
(a) Find v, then use Equ. 2-2 
 
(b) Repeat as for part (a) 
 
(c) Find the total time, then use Equ. 2-2 
 
(d) Repeat as for part (c) 
 v = 2 ´ (0.340 km/s)(3600 s/h) = 2448 km/h; 
t = (5500 km)/(2448 km/h) = 2.25 h 
v = 0.9 ´ (0.340 km/s)(3600 s/h) = 1102 km/h; 
t = (5500 km)/(1102 km/h) = 5.0 h 
ttot = (2.25 + 4)h = 6.25 h; 
vav = (5500 km)/(6.25 h) = 880 km/h 
ttot = 9 h; vav = (5500 km)/(9 h) = 611 km/h 
 
9* · As you drive down a desert highway at night, an alien spacecraft passes overhead, causing malfunctions in your 
speedometer, wristwatch, and short-term memory. When you return to your senses, you can’t tell where you are, 
where you are going, or even how fast you are traveling. The passenger sleeping next to you never woke up during 
this incident. Although your pulse is racing, hers is steady at 55 beats per minute. (a) If she has 45 beats between 
the mile markers posted along the road, determine your speed. (b) If you want to travel at 120 km/h, how many 
heartbeats should there be between mile markers? 
(a) Find the time between mile markers 
v = Ds/Dt 
(b) N = (beats/min)(60 min/h)/(v mi/h) 
 Dt = (45 beats/mile)/(55 beats/min) = 0.818 min 
v = 1 mi/0.818 min = 1.22 mi/min = 73.3 mi/h 
N = (55 ´ 60 beats/h)/[(120 km/h)/(1.61 km/mi)] = 44.3 
 
10 · The speed of light, c, is 3 ´ 108 m/s. (a) How long does it take for light to travel from the sun to the earth, a 
distance of 1.5 ´ 1011 m? (b) How long does it take light to travel from the moon to the earth, a distance of 3.84 ´ 
108 m? (c) A light-year is a unit of distance equal to that traveled by light in 1 year. Convert 1 light-year into 
kilometers and miles. 
(a) Use Equ. 2-3 
(b) Use Equ. 2-3 
(c) See Problem 1-54(c) 
Express km in miles 
t = (1.5 ´ 1011 m)/(3 ´ 108 m/s) = 500 s 
t = (3.84 ´ 108 m)/(3 ´ 108 m/s) = 1.28 s 
1 l-y = 9.48 ´ 1015 m = 9.48 ´ 1012 km 
= (9.48 ´ 1012 km)(1 mi/1.61 km) = 5.9 ´ 1012 mi. 
 
Chapter 2 Motion in One Dimension 
 
11 · The nearest star, Proxima Centauri, is 4.1 ´ 1015 km away. From the vicinity of this star, Gregor places an 
order at Tony’s Pizza in Hoboken, New Jersey, communicating via light signals. Tony’s fastest delivery craft travels 
at 10-4c (see Problem 10). (a) How long does it take Gregor’s order to reach Tony’s Pizza? (b) How long does 
Gregor wait between sending the signal and receiving the pizza? If Tony’s has a 1000-years-or-it’s-free delivery 
policy, does Gregor have to pay for the pizza? 
(a) Use Equ. 2-3 
 
(b) Traveling at 10-4c, time will be 104 times (a) 
t = (4.1 ´ 1018 m)/(3 ´ 108 m/s) = 1.37 ´ 1010 s 
= 434 y 
t = 4.34 ´ 106 y; he need not pay for the pizza 
 
12 · A car making a 100-km journey travels 40 km/h for the first 50 km. How fast must it go during the second 
50 km to average 50 km/h? 
1. Find the time for the total journey 
2. Find the time for the first 50 km 
3. Find the speed in the remaining 0.75 h 
ttot = (100 km)/(50 km/h) = 2 h 
t1 = (50 km)/(40 km/h) = 1.25 h 
v = (50 km)/(0.75 h) = 66.7 km/h 
 
13* ·· John can run 6.0 m/s. Marcia can run 15% faster than John. (a) By what distance does Marcia beat John in a 
100-m race? (b) By what time does Marcia beat John in a 100-m race? 
(a) 1. Find the running speed for Marcia 
2. Find the time for Marcia 
3. Find the distance covered by John in 14.5 s 
(b) Find the time required by John 
vM = 1.15(6.0 m/s) = 6.9 m/s 
tM = (100 m)/6.9 m/s) = 14.5 s 
sJ = (6.0 m/s)(14.5 s) = 87 m; distance = 13 m 
tJ = (100 m)/(6 m/s) = 16.7 s; time difference = 2.2 s 
14 ·· Figure 2-22 shows the position of a particle versus time. Find the average velocities for the time intervals a, b, c, 
and d indicated in the figure. 
From the figure: Dsa = 0, so vav = 0; Dsb = 1 m, Dtb = 3 s, so vav = 0.33 m/s; Dsc = -6 m, Dtc = 3 s, so 
 vav = -2 m/s; Dsd = 3 m, Dtd = 3 s, so vav = 1 m/s. 
15 ·· It has been found that galaxies are moving away from the earth at a speed that is proportional to their distance 
from the earth. This discovery is known as Hubble’s law. The speed of a galaxy at distance r from the earth is given 
by v = Hr, where H is the Hubble constant, equal to 1.58 ´ 10-18 s-1. What is the speed of a galaxy 
(a) 5 ´ 1022 m from earth and (b) 2 ´ 1025 m from earth? (c) If each of these galaxies has traveled with constant 
 speed, how long ago were they both located at the same place as the earth? 
(a) Use Hubble’s law 
(b) Use Hubble’s law(c) Use Equ. 2-3 for both galaxies 
va = (5 ´ 1022 m)(1.58 ´ 10-18 s-1) = 7.9 ´ 104 m/s 
vb = (2 ´ 1025 m)(1.58 ´ 10-18 s-1) = 3.16 ´ 107 m/s 
t = r/v = r/rH = 1/H = 6.33 ´ 1017 s = 2 ´ 1010 y 
 
Chapter 2 Motion in One Dimension 
 
16 ·· Cupid fires an arrow that strikes St. Valentine, producing the usual sound of harp music and bird chirping as 
Valentine swoons into a fog of love. If Cupid hears these telltale sounds exactly one second after firing the arrow, 
and the average speed of the arrow was 40 m/s, what was the distance separating them? Take 340 m/s for the 
speed of sound. 
Let t1 be the travel time of the arrow, and let t2 be that of the sound. Both the sound and arrow travel a distance D. 
1. Write expressions for D in terms of t1 and t2 
2. Write t1 in terms of t2 
3. The sum of t1 and t2 equals 1 s 
4. Solve (1) and (2) for t1 and t2 
5. Use (1) to find D 
D = (40 m/s)t1 = (340 m/s)t2 (1) 
t1 = (340/40)t2 = 8.5t2 (2) 
t1 + t2 = 1 s (3) 
9.5t2 = 1 s; t2 = 0.105 s; t1 = 0.895 s (4) 
D = 35.8 m 
 
17* · If the instantaneous velocity does not change, will the average velocities for different intervals differ? 
No, they will not. For constant velocity, the instantaneous and average velocities are equal. 
 
18 · If vav = 0 for some time interval Dt, must the instantaneous velocity v be zero at some point in the interval? 
 Support your answer by sketching a possible x-versus-t curve that has Dx = 0 for some interval Dt. 
Yes, it must. 
In the adjoining graph of x versus t, Dx = 0 in the 
interval between t = 0 and t = 4.0 s. Consequently, 
vav = Dx/Dt = 0, although the instantaneous velocity 
is zero only at the point t = 2 s. 
 
 
19 · An object moves along the x axis as shown in Figure 2-23. At which point or points is the magnitude of its 
velocity at a minimum? (a) A and E (b) B, D, and E (c) C only (d) E only (e) None of these is correct. 
(b) At these points the slope of the position-versus-time curve is zero; therefore the velocity is zero. 
 
20 · For each of the four graphs of x versus t in Figure 2-24, answer the following questions. (a) Is the velocity at 
time t2 greater than, less than, or equal to the velocity at time t1? (b) Is the speed at time t2 greater than, less than, 
or equal to the speed at time t1? 
We shall use v to denote velocity and v to denote speed. 
(a) curve a: v(t2) < v(t1); curve b: v(t2) = v(t1); curve c: v(t2) > v(t1); curve d: v(t2) < v(t1). 
(b) curve a: v(t2) < v(t1); curve b: v(t2) = v(t1); curve c: v(t2) < v(t1); curve d: v(t2) > v(t1). 
21* · Using the graph of x versus t in Figure 2-25, (a) find the average velocity between the times t = 0 and t = 2 s. 
(b) Find the instantaneous velocity at t = 2 s by measuring the slope of the tangent line indicated. 
(a) Find Dx from graph; vav = Dx/Dt Dx = 2 m, Dt = 2 s; vav = 1 m/s 
Chapter 2 Motion in One Dimension 
 
(b) From graph, tangent passes through points 
x = 0, t = 1 s; x = 4 m, t = 3 s. 
Slope of tangent line is (4 m)/(2 s) = 2 m/s 
v(t = 2 s) = 2 m/s 
 
22 · Using the graph of x versus t in Figure 2-26, find (a) the average velocity for the time intervals Dt = 
t2 - 0.75 s when t2 is 1.75, 1.5, 1.25, and 1.0 s; (b) the instantaneous velocity at t = 0.75 s; (c) the approximate 
 time when the instantaneous velocity is zero. 
(a) From graph find x(0.75), x(1.75), x(1.5), 
x(1.25), x(1.0) Find vav = Dx/Dt 
 
(b) Draw tangent at t = 0.75 s and find slope 
(c) Find the value of t where slope is zero 
 x(0.75) = 4.0 m, x(1.75) = 5.9 m, x(1.5) = 6 m, 
x(1.25) = 5.6 m, x(1.0) = 5 m 
1.75 s, vav = 1.9 m/s; 1.5 s, vav = 2.67 m/s; 1.25 s, 
vav = 3.2 m/s; 1.0 s, vav = 4 m/s 
Slope = v(0.75) = 4.2 m/s 
t = 1.6 s 
 
23 ·· The position of a certain particle depends on the time according to x = (1 m/s2)t2 - (5 m/s)t + 1 m. (a) Find the 
displacement and average velocity for the interval 3 s £ t £ 4 s. (b) Find the general formula for the displacement 
for the time interval from t to t + Dt. (c) Use the limiting process to obtain the instantaneous velocity for any time t. 
(a) 1. Find x(4) and x(3) 
 
2. Find Dx 
3. Use Equ. 2-2 
(b) 1. Find x(t + Dt) 
2. Find x(t + Dt) - x(t) = Dx 
(c) From (b) find Dx/Dt as Dt ® 0 
x(4) = (16 - 20 + 1) m = -3 m; 
x(3) = (9 - 15 + 1) m = -5 m 
Dx = x(4) - x(3) = 2 m 
vav = Dx/Dt = (2 m)/(1 s) = 2 m/s 
x(t + Dt) = [(t2 + 2tDt + Dt2) - 5(t + Dt) + 1] m 
Dx = [(2t - 5)Dt + Dt2] m 
v = limt®0(Dx/Dt) = (2t - 5) m/s 
24 ·· The height of a certain projectile is related to time by y = -5(t - 5)2 + 125, where y is in meters and t is in 
seconds. (a) Sketch y versus t for 0 £ t £ 10 s. (b) Find the average velocity for each of the 1-s time intervals 
between integral time values from 0 £ t £ 10 s. Sketch vav versus t. (c) Find the instantaneous velocity as a function 
of time. 
(a) The plot of y versus t is shown 
 
(b) Substitute t = 0,1,2,...,10 into the expression for y. The table below lists the values. Then evaluate 
Dy/Dt = Dy/(1 s) to get vav, shown in the table. 
t y, m vav , m/s 
Chapter 2 Motion in One Dimension 
 
 0 0 0 
 1 45 45 
 2 80 35 
 3 105 25 
 4 120 15 
 5 125 5 
 6 120 -5 
 7 105 -15 
 8 80 -25 
 9 45 -35 
 10 0 -45 
 (c) To find the instantaneous velocity, take the derivative of the expression for y(t); 
 v(t) = dy/dt = (50 - 10t) m/s 
 
25* ·· The position of a body oscillating on a spring is given by x = A sin wt, where A and w are constants with values 
 A = 5 cm and w = 0.175 s-1. (a) Sketch x versus t for 0 £ t £ 36 s. (b) Measure the slope of your graph at t = 0 
to find the velocity at this time. (c) Calculate the average velocity for a series of intervals beginning at t = 0 and 
ending at t = 6, 3, 2, 1, 0.5, and 0.25 s. (d) Compute dx/dt and find the velocity at time t = 0. 
(a) The plot of x versus t is shown 
 
 (b) The slope of the dotted line (tangent at t = 0) 
 is 0.875; v(0) = 0.875 cm/s 
(c) t x Dx/Dt 
6 4.34 0.723 
Chapter 2 Motion in One Dimension 
 
3 2.51 0.835 
2 1.71 0.857 
1 0.174 0.871 
0.5 0.437 0.874 
0.25 0.219 0.875 
(d) dx/dt = Aw cos wt; at t= 0 cos wt = 1; 
dx/dt at t = 0 is Aw = 0.875 cm/s 
 
 
26 · To avoid falling too fast during a landing, an airplane must maintain a minimum airspeed (the speed of the plane 
relative to the air). However, the slower the ground speed (speed relative to the ground) during a landing, the safer 
the landing. Is it safer for an airplane to land with the wind or against the wind? 
It is safer to land against the wind. 
 
27 ·· Two cars are traveling along a straight road. Car A maintains a constant speed of 80 km/h; car B maintains a 
constant speed of 110 km/h. At t = 0, car B is 45 km behind car A. How far will car A travel from t = 0 before it is 
overtaken by car B? 
1. Find the velocity of car B relative to car A 
2. Find the time before overtaking 
3. Find the distance traveled by car A in 1.5 h 
vrel = vB - vA = (110 - 80) km/h = 30 km/h 
t = s/vrel = (45 km)/(30 km/h) = 1.5 h 
d = (1.5 h)(80 km/h) = 120 km 
 
28 ·· A car traveling at constant speed of 20 m/s passes an intersection at time t = 0, and 5 s later another car 
traveling 30 m/s passes the same intersection in the same direction. (a) Sketch the position functions x1(t) and x2(t) 
for the two cars. (b) Determine when the second car will overtake the first. (c) How far from the intersection will 
the two cars be when they pull even? 
(a) The plot of x1(t) and x2(t) are shown 
 
(b) From the plot, the secondwill overtake the first at t = 15 s 
(c) The two cars will be 300 m from the intersection 
 
 
 
 
Chapter 2 Motion in One Dimension 
 
29* ·· Margaret has just enough gas in her speedboat to get to the marina, an upstream journey that takes 4.0 hours. 
Finding it closed for the season, she spends the next 8.0 hours floating back downstream to her shack. The entire 
trip took 12.0 h; how long would it have taken if she had bought gas at the marina? 
Let D = distance to marina, vW = velocity of stream, vrel = velocity of boat under power relative to stream. 
1. Express the times of travel with gas in terms of 
D, vW, and vrel 
2. Express the time required to drift distance D 
3. From t1 = 4 h, find vrel 
4. Solve for t2 
5. Add t1 and t2 
t1 = D/(vrel - vW) = 4 h 
t2 = D/(vrel + vW) 
t3 = D/vW = 8 h; vW = D/(8 h) 
4(vrel - D/8) = D; vrel = 1.5D/(4 h) 
t2 = D/[(1.5D/4 h) + (D/8 h) = 2 h 
ttot = t1 + t2 = 6 h 
30 ·· Joe and Sally tend to argue when they travel. Just as they reached the moving sidewalk at the airport, their 
struggle for itinerary-making powers peaked. Though they stepped on the moving belt at the same time, Joe chose 
to stand and ride, while Sally opted to keep walking. Sally reached the end in 1 min, while Joe took 2 min. How long 
would it have taken Sally if she had walked twice as fast? 
Let vB = velocity of belt, vS velocity of Sally when walking normally, and D the length of the belt. 
1. Write D in terms of vB; Joe’s speed on the belt 
2. Write D in terms of vB + vS; Sally’s speed on the 
belt 
3. Solve for vS 
4. Write t2 = time for fast walk 
D = (2 min)(vB); vB = D/(2 min) 
D = (1 min)(vB + vS) = (1 min)[(D/2 min) + vS] 
 
vS = D/(2 min) 
t2 = D/(vB + 2vS) = D/[(D/2 min) + (D/1 min) = 40 s 
31 · Walk across the room in such a way that, after getting started, your velocity is negative but your acceleration is 
positive. (a) Describe how you did it. (b) Sketch a graph of v versus t for your motion. 
 (a) Start walking with velocity in the negative direction. Gradually slow the 
speed of walking, until the other end of the room is reached. 
(b) A sketch of v versus t is shown 
 
 
32 · Give an example of a motion for which both the acceleration and the velocity are negative. 
Let up be the positive direction. A falling object then has a negative velocity and a negative acceleration. 
 
33* · Is it possible for a body to have zero velocity and nonzero acceleration? 
Yes, it is. An object tossed up has a constant downward acceleration; at its maximum height, its instantaneous 
 velocity is zero. 
 
34 · True or false: (a) If the acceleration is zero, the body cannot be moving. (b) If the acceleration is zero, the 
x-versus-t curve must be straight line. 
(a) False (b) True 
Chapter 2 Motion in One Dimension 
 
35 ·· State whether the acceleration is positive, negative, or zero for each of the functions x(t) in Figure 2-27. 
(a) a = 0, constant velocity. (b) a > 0, v changes from negative to positive. (c) a < 0. (d) a = 0. 
 
36 ·· Answer the following question for each of the graphs in Figure 2-28: (a) At what times are the accelerations of 
the objects positive, negative, and zero? (b) At what times are the accelerations constant? (c) At what times are the 
instantaneous velocities zero? 
Curve 1: v versus t. Hence, a < 0 for t < 3 s and t > 7 s; a = 0 for t = 3 s and 6 s £ t £ 7 s; a > 0 for 3 s £ t £ 6 s. 
(b) a is constant for t < 2.7 s, 3.2 s £ t £ 5.8 s, 6 s £ t £ 7 s, 7.1 s £ t. (c) v = 0 at t = 8.6 s. 
Curve 2: x versus t. Here, a < 0 for 0 £ t £ 3 s and t > 7 s; a = 0 for 3 s £ t £ 5 s; a > 0 for 5 s £ t £ 7 s. 
(b) It is difficult to tell where a is constant from the curve; if the curved segments are parabolas, then a is 
 constant everywhere. (c) v = 0 at t = 2 s, 5.8 s, and 7.8 s. 
 
37* · A BMW M3 sports car can accelerate in third gear from 48.3 km/h (30 mi/h) to 80.5 km/h (50 mi/h) in 3.7 s. 
(a) What is the average acceleration of this car in m/s2? (b) If the car continued at this acceleration for another 
second, how fast would it be moving? 
(a) 1. Use Equ. 2-8 
2. Convert to m/s2 
(b) In 1 s, its speed increases by 8.7 km/h 
aav = [(80.5 - 48.3) km/h]/(3.7 s) = 8.7 km/h.s 
(8.7 ´ 103 m/h×s)(1 h/3600 s) = 2.42 m/s2 
v = (80.5 + 8.7) km/h = 89.2 km/h 
 
38 · At t = 5 s, an object at x = 3 m is traveling at 5 m/s. At t = 8 s, it is at x = 9 m and its velocity is -1 m/s. Find 
the average acceleration for this interval. 
Use Equ. 2-8, aav = Dv/Dt aav = [(-1 m/s) - (5 m/s)]/[(8 s) - (5 s)] = -2 m/s2 
 
39 ·· A particle moves with velocity v = 8t - 7, where v is in meters per second and t is in seconds. (a) Find the 
 average acceleration for the one-second intervals beginning at t = 3 s and t = 4 s. (b) Sketch v versus t. What is 
the instantaneous acceleration at any time? 
 
(a) 1. Determine v at t = 3 s, t = 4 s, t = 5 s 
2. Find aav for the two 1-s intervals 
 (b) v versus t is shown in the adjacent plot 
Use Equ. 2-10: a = dv/dt = 8 m/s2. 
 
v(3) = 17 m/s; v(4) = 25 m/s; v(5) = 33 m/s 
aav(3 - 4) = 8/1 m/s2 = 8 m/s2; aav(4 - 5) = 8 m/s2 
 
 
 
40 ·· The position of an object is related to time by x = At2 - Bt + C, where A = 8 m/s2, B = 6 m/s, and C = 4 m. 
Find the instantaneous velocity and acceleration as functions of time. 
1. Use Equs. 2-5 and 2-10 v = dx/dt = 2At - B; a = dv/dt = 2A 
Chapter 2 Motion in One Dimension 
 
2. Substitute numerical values for A and B v = (16t - 6) m/s; a = 16 m/s2 
 
41* · Identical twin brothers standing on a bridge each throw a rock straight down into the water below. They throw 
rocks at exactly the same time, but one hits the water before the other. How can this occur if the rocks have the 
same acceleration? 
The initial downward velocities of the two rocks are not the same. 
 
42 · A ball is thrown straight up. What is the velocity of the ball at the top of its flight? What is its acceleration at that 
point? 
At the top of its flight, its velocity is instantaneously zero; its acceleration is 9.81 m/s2 downward. 
 
43 · An object thrown straight up falls back to the ground T seconds later. Its maximum height is H meters. Its 
average velocity during those T seconds is (a) H/T, (b) 0, (c) H/2T, (d) 2H/T. 
(b) The displacement is zero. 
 
44 · For an object thrown straight up, which of the following is true while it is in the air? (a) The acceleration is 
always opposite to the velocity. (b) The acceleration is always directed downward. (c) The acceleration is always 
in the direction of motion. (d) The acceleration is zero at the top of the trajectory. 
(b) The acceleration is -g, always directed downward. 
 
45* · An object projected up with initial velocity v attains a height H. Another object projected up with initial velocity 
2v will attain a height of (a) 4H, (b) 3H, (c) 2H, (d) H. 
(a) 4H; from Equ. 2-15, with a = -g and v = 0 at top of trajectory, H = 2g / v20 , v0 is initial velocity. So H µ v02. 
 
46 · A ball is thrown upward. While it is in the air, its acceleration is (a) decreasing, (b) constant, (c) zero, (d) 
 increasing. 
(b) constant; downward with value g. 
 
47 · At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a 
window 10 m below the roof. During their descent to the ground the distance between the two objects (a) is 
proportional to t. (b) is proportional to t2. (c) decreases. (d) remains 10 m throughout. 
(d) Both move with the same acceleration and velocity at all times; the distance between them remains that at 
t = 0. 
 
48 ·· A Porsche accelerates uniformly from 80.5 km/h (50 mi/h) at t = 0 to 113km/h (70 mi/h) at t = 9 s. Which 
graph in Figure 2-29 best describes the motion of the car? 
(c) v(0) > 0 and v(t) increases at a uniform rate. 
 
49* ·· An object is dropped from rest. If the time during which it falls is doubled, the distance it falls will (a) double, 
(b) decrease by one-half, (c) increase by a factor of four, (d) decrease by a factor of four, (e) remain the same. 
(c) increase by a factor of 4; see Equ. (2-14): Dx µt2 if v0 = 0. 
50 ·· A ball is thrown upward with an initial velocity v0. Its velocity halfway to its highest point is (a) 0.5v0, 
(b) 0.25v0, (c) v0, (d) 0.707v0, (e) cannot be determined from the information given. 
(d) 0.707v0; H = v02/2g, v2 = v02 - 2g(H/2) = 1/2v02; v = 0.707v0. 
51 · A car starting at x = 50 m accelerates from rest at a constant rate of 8 m/s2. (a) How fast is it going after 
10 s? (b) How far has it gone after 10 s? (c) What is its average velocity for the interval 0 £ t £ 10 s? 
Chapter 2 Motion in One Dimension 
 
(a) Use Equ. 2-12 
(b) Use Equ. 2-14 
(c) Use Equ. 2-2 
 v = 0 + (8 m/s2)(10 s) = 80 m/s 
Dx = 1/2(8 m/s2)(10 s)2 = 400 m 
vav = (400 m)/(10 s) = 40 m/s 
 
52 · An object with an initial velocity of 5 m/s has a constant acceleration of 2 m/s2. When its speed is 15 m/s, how 
far has it traveled? 
Use Equ. 2-15 Dx = [(152 - 52) m2/s2]/[2(2 m/s2)] = 50 m 
 
53* · An object with constant acceleration has velocity v = 10 m/s when it is at x = 6 m and v = 15 m/s when it is at 
x = 10 m. What is its acceleration? 
Use Equ. 2-15 a = [(152 - 102) m2/s2]/[2(4 m)] = 15.6 m/s2 
 
54 · An object has constant acceleration a = 4 m/s2. At t = 0, its velocity is 1 m/s and it is at x = 7 m. How fast is it 
moving when it is at x = 8 m? What is t at that point? 
1. Use Equ. 2-15 
2. Use Equ. 2-12 
 v2 = 1 m2/s2 + 2(4 m/s2)(1 m) = 9 m2/s2; v = 3 m/s 
t = [(3 - 1) m/s]/(4 m/s2) = 0.5 s 
 
55 · If a rifle fires a bullet straight up with a muzzle speed of 300 m/s, how high will the bullet rise? (Ignore air 
resistance.) 
At top, v = 0; use Equ. 2-15 H = v02/2g = 4.59 ´ 103 m 
 
56 · A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 
98 km/h to zero is 50 m. Find the acceleration, assuming it to be constant, and express your answer as a fraction 
of the free-fall acceleration due to gravity. How long does the car take to stop? 
1. Use Equ. 2-15 
2. Divide a by g = -9.81 m/s2 
3. Use Equ. 2-12 
 a = -(98 km/h)2/(100 m) = -9.6 ´ 107 m/h2 = -7.41 m/s2 
a = 0.755g 
t = (98 km/h)/(9.6 ´ 107 m/h2) = 1.02 ´ 10-3 h = 3.67 s 
 
57* ·· A ball is thrown upward with an initial velocity of 20 m/s. (a) How long is the ball in the air? (b) What is the 
greatest height reached by the ball? (c) When is the ball 15 m above the ground? 
(a) 1. Take upward as positive; use Equ. 2-14 
2. Solve for t 
(b) See Problem 55 
(c) 1. Use Equ. 2-14 
2. Use quadratic formula 
2a
4ac - b b- = t
2±
 
Dx = 0 = (20 m/s)t - 1/2(9.81 m/s2)t2 
t = 0; t = 4.08 s; t = 4.08 s is the proper result 
H = (400 m2/s2)/[2(9.81 m/s2)] = 20.4 m 
15 m = (20 m/s)t - 1/2(9.81 m/s2)t2 
t = 0.991 s, t = 3.09 s; both are acceptable solutions 
 
58 ·· A particle moves with constant acceleration of 3 m/s2. At t = 4 s, it is at x = 100 m; at t = 6 s, it has a velocity 
v = 15 m/s. Find its position at t = 6 s. 
1. Find v(4 s) = v0 using Equ. (2-12) 
2. Use Equ. 2-14; elapsed time = 2 s 
 v0 = (15 - 2 ´ 3) m/s = 9 m/s 
x = [100 m + (9 m/s)(2 s) + 1/2(3 m/s2)(2 s)2] = 124 m 
 
 
Chapter 2 Motion in One Dimension 
 
59 ·· A bullet traveling at 350 m/s strikes a telephone pole and penetrates a distance of 12 cm before stopping. 
(a) Estimate the average acceleration by assuming it to be constant. (b) How long did it take for the bullet to stop? 
(a) Use Equ. 2-15 
(b) Use Equ. 2-12 
 a = -(350 m/s)2/[2(0.12 m)] = -5.1 ´ 105 m/s2 
t = (350 m/s)/(5.1 ´ 105 m/s2) = 0.686 ms 
 
60 ·· A plane landing on an aircraft carrier has just 70 m to stop. If its initial speed is 60 m/s, (a) what is the 
acceleration of the plane during landing, assuming it to be constant? (b) How long does it take for the plane to stop? 
(a) Use Equ. 2-15 
(b) Use Equ. 2-12 
 a = -(60 m/s)2/[2(70 m)] = -25.7 m/s2 
t = (60 m/s)/(25.7 m/s2) = 2.33 s 
 
61* ·· An automobile accelerates from rest at 2 m/s2 for 20 s. The speed is then held constant for 20 s, after which 
 there is an acceleration of -3 m/s2 until the automobile stops. What is the total distance traveled? 
1. Determine the distance traveled during first 20 s 
 and the speed at the end of first 20 s 
2. Find Dx2 = distance covered 20 s £ t £ 40 s 
3. Find Dx3 = distance during deceleration 
4. Find total distance 
 v(20) = at = (2 m/s2)(20 s) = 40 m/s 
Dx1 = vavt = (20 m/s)(20 s) = 400 m 
Dx2 = (40 m/s)(20 s) = 800 m 
Dx3 = (40 m/s)2/[2(3 m/s2)] = 267 m 
x = Dx1 + Dx2 + Dx3 = 1467 m 
 
62 ·· In the Blackhawk landslide in California, a mass of rock and mud fell 460 m down a mountain and then traveled 
8 km across a level plain on a cushion of compressed air. Assume that the mud dropped with the free-fall 
acceleration due to gravity and then slid horizontally with constant deceleration. (a) How long did the mud take to 
drop the 460 m? (b) How fast was it traveling when it reached the bottom? (c) How long did the mud take to slide 
the 8 km horizontally? 
(a) Use Equ. 2-14 
(b) Use Equ. 2-12 
(c) Use Equ. 2-13 
t2 = 2(460 m)/(9.81 m/s2) = 93.8 s2; t = 9.68 s 
v = (9.81 m/s2)(9.68 s) = 95.0 m/s 
t = 2(8000 m)/(95.0 m/s) = 168 s 
 
63 ·· A load of bricks is being lifted by a crane at a steady velocity of 5 m/s when one brick falls off 6 m above the 
ground. (a) Sketch x(t) to show the motion of the free brick. (b) What is the greatest height the brick reaches 
above the ground? (c) How long does it take to reach the ground? (d) What is its speed just before it hits the 
ground? 
(a) Use Equ. 2-14; x0 = 6 m, v0 = 5 m/s, a = -9.81 m/s2 
(b) Use Equ. 2-15; find Dx and add to x0; 
Dx = (25/2 ´ 9.81) m = 1.27 m; 
x = 7.27 m (also see graph) 
 
(c) Use Equ. 2-14; then solve quadratic equ. for t 
Chapter 2 Motion in One Dimension 
 
0 = 6 + 5t - 1/2 ´ 9.81t2; t = 1.73 s, t = -0.708 s 
Second (negative solution) is non-physical; t = 1.73 s 
 
(d) Use v2 = 2gH, with H = 7.27 m 
v2 = 2 ´ 9.81 ´ 7.27 m2/s2; v = 11.9 m/s 
 
64 ·· An egg with a mass of 50 g rolls off a table at a height of 1.2 m and splatters on the floor. Estimate the 
 average acceleration of the egg while it is in contact with the floor. 
We shall take 5 cm as the diameter of an egg, and consider the time of contact while crashing, during which the 
“egg” travels 5 cm. Just before contact with the floor, v = 2gH » 5 m/s, so the time of contact » (0.05/2.5) s = 
20 ms. In that time the egg decelerates to v = 0. So the magnitude of aav = (5/0.02) m/s2 = 250 m/s2. 
65* ·· To win publicity for her new CD release, Sharika, the punk queen, jumps out of an airplane without a parachute. 
She expects a stack of loose hay to break her fall. If she reaches a speed of 120 km/h prior to impact, and if a 35 g 
deceleration is the greatest deceleration she can withstand, how high must the stack of hay be in order for her to 
survive? Assume uniform acceleration while she is in contact with the hay. 
1. Use Equ. 2-15 
H = m 1.62 = 
)]sm/ x(9.8135x[2 )h 1s/ (3600
)m/h 10x(1.20
22
25
 
 
 
66 ·· A bolt comes loose from underneath an elevator that is moving upward at a speed of 6 m/s. The bolt reaches the 
bottom of the elevator shaft in 3 s. (a) How highup was the elevator when the bolt came loose? (b) What is the 
speed of the bolt when it hits the bottom of the shaft? 
We use up as the positive direction; x = 0 at bottom of shaft. 
(a) Use Equ. 2-14; v0 = 6 m/s, t = 3 s, x(3 s) = 0 
(b) Use Equ. 2-12 
 x0 =[ 0 - (6 ´ 3) - 1/2(-9.81)(9)] m = 26.1 m 
v = [6 - 9.81 ´ 3] m/s = -23.4 m/s; ½v½ = 23.4 m 
 
67 ·· An object is dropped from a height of 120 m. Find the distance it falls during its final second in the air. 
 
1. Find final (impact) speed, vf 
2. Find vf-1, speed 1 s prior to impact 
3. Find the average speed during this 1 s 
4. Find the distance traversed, s = vavDt 
m/s 48.5 = m/s (120) (9.81) 2 = fv 
vf-1 = (48.5 - 9.81) m/s = 38.7 m/s 
vav = 1/2(48.5 + 38.7) m/s = 43.6 m/s 
s = (43.6 ´ 1) m = 43.6 m 
 
68 ·· An object is dropped from a height H. During the final second of its fall, it traverses a distance of 38 m. What 
was H? 
Let vf be the final speed before impact, vf-1 the speed 1 s before impact. 
1. Find the average speed in last second 
2. Express vf-1 in terms of vf and solve for vf 
3. Use vf to determine H 
 vav (1 s) = 38 m = 1/2(vf + vf-1)(1 s); vf + vf-1 = 76 m/s 
vf - vf-1 = 9.81 m/s; vf = 42.9 m/s 
H = vf2/2g = 93.8 m 
 
69* · A stone is thrown vertically from a cliff 200 m tall. During the last half second of its flight the stone travels a 
distance of 45 m. Find the initial velocity of the stone. 
Chapter 2 Motion in One Dimension 
 
We take down as the positive direction. Let v1 be the velocity 1/2 s before impact, vf velocity at impact. 
1. Find vav during last 1/2 s 
2. Write vf in terms of v1 and g 
3. Solve for vf 
4. Use Equ. 2-15 to find v0 
 vav = 1/2(v1 + vf) = (45/0.5) m/s; v1 + vf = 180 m/s 
vf = v1 + gt; vf = v1 + (9.81 ´ 0.5) m/s 
vf = 92.5 m/s 
( ) m/s 68 = m/s2x9.81x200 -92.5 2 ± = v0 ; the 
stone may be thrown either up or down 
 
70 ·· An object in free-fall from a height H traverses 0.4H during the first second of its descent. Determine the 
 average speed of the object during free-fall. 
1. Find Dx for first second and H 
2. Find vf 
3. Determine vav, v0 = 0 
 Dx = 1/2gt2 = 4.9 m = 0.4H; H = 12.3 m 
vf = [2(9.81)(12.3)]1/2 m/s = 15.5 m/s 
vav = 1/2vf = 7.77 m/s 
 
71 ·· A bus accelerates at 1.5 m/s2 from rest for 12 s. It then travels at constant speed for 25 s, after which it slows 
 to a stop with an acceleration of -1.5 m/s2. (a) How far did the bus travel? (b) What was its average velocity? 
(a) 1. Find v(12) and vav(0-12) 
 2. Find x(12) = vav(0-12)Dt 
 3. Find Dx(12-37) = v(12) ´ Dt 
 4.½deceleration½ = acceleration; Dxdec = Dxacc 
 5. Add displacements to find xtot 
(b) vav = xtot/ttot 
 v(12) = (1.5 ´ 12) m/s = 18 m/s; vav(0-12) = 9 m/s 
x(12) = (9 ´ 12) m = 108 m 
Dx(12-37) = (18 ´ 25) m = 450 m 
Dxdec = 108 m and Dtdec = 12 s 
Total distance = 666 m 
vav = (666 m)/(49 s) = 13.6 m/s 
72 ·· A basketball is dropped from a height of 3 m and rebounds from the floor to a height of 2 m. (a) What is the 
velocity of the ball just as it reaches the floor? (b) What is its velocity just as it leaves the floor? (c) Estimate the 
magnitude and direction of its average acceleration during this interval.
(a) gH = v 2 
(b) Use the above expression 
(c) A reasonable estimate for Dt is 0.05 s
v = -(2 ´ 9.81 ´ 3)1/2 m/s = -7.67 m/s 
v = +(2 ´ 9.81 ´ 2)1/2 m/s = 6.26 m/s 
aav = (6.26 + 7.67)/0.05 m/s2 = 279 m/s2 upward
 
73* ·· A rocket is fired vertically with an upward acceleration of 20 m/s2. After 25 s, the engine shuts off and the 
rocket continues as a free particle until it reaches the ground. Calculate (a) the highest point the rocket reaches, 
(b) the total time the rocket is in the air, (c) the speed of the rocket just before it hits the ground. 
Take up as the positive direction. 
(a) 1. Find x1 and v1 at t = 25 s; use Equ. 2-14 
2. Find x2 = distance above x1 when v = 0 
3. Total height = x1 + x2 
(b) 1. Find time, t2, for part (a) 
2. Find time, t3, to drop 19.0 km 
3. Total time, T = 25 s + t2 + t3 
(c) Use Equ. 2-12 
 x1 = 1/2(20)(25)2 m = 6250 m; v1 = 20 ´ 25 m/s = 500 m/s 
x2 = (500)2/(2 ´ 9.81) m = 1.274 ´ 104 m 
H = 1.90 ´ 104 m = 19.0 km 
t2 = x2/vav = (1.274 ´ 104/250) s = 51 s 
t3 = [2(1.90 ´ 104)/9.81]1/2 s = 62.2 s 
T = (25 + 51 + 62.2) s = 138 s = 2 min 18 s 
vf = (9.81)(62.2) m/s = 610 m/s 
 
Chapter 2 Motion in One Dimension 
 
74 ·· A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally 
holding a stopwatch, notices that it takes 0.2 s for the pot to fall past his window, which is 4 m high. How far 
above the top of the window is the ledge from which the pot fell? 
Take down as positive, x = 0 at ledge. Let t = time when pot is at top of window, t + Dt when pot is at bottom of 
window. 
1. Express Dx = 4 m in terms of t, Dt, and g 
2. Solve for t, with Dt = 0.2 s, g = 9.81 m/s2 
3. Find distance pot falls from rest in 1.94 s 
4 m = 1/2g[(t+Dt)2 - t2] = 1/2g(Dt)2 + gtDt 
t = 1.94 s 
H = 1/2(9.81)(1.94)2 = 18.4 m 
 
75 ·· Sharika arrives home late from a gig, only to find herself locked out. Her roommate and bass player Chico is 
practicing so loudly that he can’t hear Sharika’s pounding on the door downstairs. One of the band’s props is a 
small trampoline, which Sharika places under Chico’s window. She bounces progressively higher trying to get 
Chico’s attention. Propelling herself furiously upward, she miscalculates on the last bounce and flies past the 
window and out of sight. Chico sees her face for 0.2 s as she moves a distance of 2.4 m from the bottom to the top 
of the window. (a) How long until she reappears? (b) What is her greatest height above the top of the window? 
(Treat Sharika as a point-particle punk.) 
Take up as positive; x = 0 at bottom of window, v0 = upward velocity at top of window. 
(a) 1. Use Equ. 2-14 and solve for v0 
2. Find the time to reach v = 0 
3. Time when her face reappears is 2t 
(b) Find H using Equ. 2-15 
2.4 m = v0(0.2 s) + 1/2(9.81)(0.2)2 m; v0 = 11.0 m/s 
t = (11.0/9.81) s = 1.12 s 
Time till her face reappears is 2.24 s 
H = v02/2g = 6.17 m 
 
76 ·· In a classroom demonstration, a glider moves along an inclined air track with constant acceleration a. It is 
 projected from the start of the track (x = 0) with an initial velocity v0. At time t = 8 s, it is at x = 100 cm and is 
 moving along the track at velocity v = -15 cm/s. Find the initial speed v0 and the acceleration a. 
1. Use Equ. (2-13) and solve for v0 
2. Use the definition of a = Dv/Dt 
 v0 = 2x/t - v = [(200/8) - (-15)] m/s = 40 cm/s 
a = (-55/8) cm/s2 = -6.88 cm/s2 
 
77* ·· A rock dropped from a cliff falls one-third of its total distance to the ground in the last second of its fall. How 
high is the cliff? 
1. Write the final speed in terms of H and g 
2. Write the average speed in the last second 
3. Set the distance in last second = H/3; solve for vf 
4. Set vf2 = 2gH and solve for H 
 vf2 = 2gH 
vav =1/2[vf + (vf - g)] =(vf - g/2) 
(vf - g/2) = H/3; vf = H/3 + g/2 
2gH = H2/9 + gH/3 + g2/4; H = 14.85g = 145.7 m 
 
78 ··· A typical automobile has a maximum deceleration of about 7 m/s2; the typical reaction time to engage the brakes 
is 0.50 s. A school board sets the speed limit in a school zone to meet the condition that all cars should be able to 
stop in a distance of 4 m. (a) What maximum speed should be allowed for a typical automobile? (b) What fraction 
of the 4 m is due to the reaction time? 
(a) 1. Write the total distance as the sum of distance Dx1 = 0.50v0 ; Dx2 = -v02/2a (a = -7 m/s2) 
Chapter 2 Motion in One Dimensiontraveled in 0.50 s plus that during deceleration 
 2. Solve quadratic equ. keeping positive result 
(b) Find reaction time distance 
4 m = Dx1 + Dx2 = (0.5v0 + v02/14) m 
v0 = 4.76 m/s = 4.76 m/s = 10.6 mi/h 
Dx1 = 2.38m = (2.38/4)? xtot = 59.5% of )? xtot 
 
 
 
79 ·· Two trains face each other on adjacent tracks. They are initially at rest 40 m apart. The train on the left accelerates 
rightward at 1.4 m/s2. The train on the right accelerates leftward at 2.2 m/s2. How far does the train 
 on the left travel before the two trains pass? 
 Take x = 0 as position of train on left at t = 0. 
 1. Write positions of each train as a function of time 
 2. Set xL = xR and solve for t 
 3. Find distance traveled, i.e., xL 
xL = 0.7t2 m; xR = 40 - 1.1t2 
0.7t2 = 40 - 1.1t2; t = 4.71 s 
xL = 15.6 m 
 
 
 80 ·· Two stones are dropped from the edge of a 60-m cliff, the second stone 1.6 s after the first. How far below the 
cliff is the second stone when the separation between the two stones is 36 m? 
 1. Write expressions for x1 and x2 
 2. Set x1 - x2 = 36 m 
 3. Solve for t, then find x2 
x1 = 1/2gt2; x2 = 1/2g(t - 1.6 s)2 
36 m = 1/2g[(3.2 s)t - 2.56 s2] 
t = 3.09 s; x2 = 10.9 m 
 
 
81* ·· A motorcycle policeman hidden at an intersection observes a car that ignores a stop sign, crosses the intersection, and 
continues on at constant speed. The policeman takes off in pursuit 2.0 s after the car has passed the stop sign, accelerates 
at 6.2 m/s2 until his speed is 110 km/h, and then continues at this speed until he catches the car. At that instant, the car is 
1.4 km from the intersection. How fast was the car traveling? 
1. Find the time of travel of policeman: t1 is the time 
 of acceleration, t2 the time of travel at 110 km/h; 
 and d1 and d2 the corresponding distances 
2. The time of travel of car is 2.0 s + t1 + t2 
3. Find the speed of the car 
t1 = (1.10 ´ 103 m/h)/[(3600 s/1 h)(6.2 m/s2)] = 4.93 s 
d1 = 1/2(30.5 m/s)(4.93 s) = 75.3 m; d2 = (1400 - 75.3) m 
= 1325 m; t2 = d2/(30.6 m/s) = 43.3 s 
tC = (2.0 + 43.3 + 4.93) s = 50.2 s 
vC = (1400/50.2) m/s = 27.9 m/s = 100.4 km/h 
 
 
82 ·· At t = 0, a stone is dropped from a cliff above a lake; 1.6 seconds later another stone is thrown downward from the 
same point with an initial speed of 32 m/s. Both stones hit the water at the same instant. Find the height of the cliff. 
 1. Write the distances dropped and equate 
 2. Solve for t 
 3. Find H; Use Equ. 2-14 
1/2gt2 = (32 m/s )(t - 1.6 s) + 1/2g(t - 1.6 s)2 
t = 2.37 s 
H = 1/2gt2 = 27.6 m 
 
 
83 ·· A passenger train is traveling at 29 m/s when the engineer sees a freight train 360 m ahead traveling on the same track 
in the same direction. The freight train is moving at a speed of 6 m/s. If the reaction time of the engineer is 0.4 s, what 
must be the deceleration of the passenger train if a collision is to be avoided? If your answer is the maximum deceleration 
of the passenger train but the engineer’s reaction time is 0.8 s, what is the relative speed of the two trains at the instant of 
Chapter 2 Motion in One Dimension 
 
collision and how far will the passenger train have traveled in the time between the sighting of the freight train and the 
collision? 
 Let x = 0 be the location of the passenger train engine at moment of sighting of freight train’s end; let t = 0 be the 
 instant the passenger train decelerates. 
1. Write expressions for xP and xF, positions of the 
 passenger train engine and the freight train’s end 
2. Set xF = xP and obtain an equation for t 
3. If equation has a real solution collision occurs 
4. Solve for a so there is no real solution for t 
5. Repeat, with a = 0.754 m/s2 and 0.8 s reaction 
 time 
6. Solve for t, keeping only smaller value 
7. Find xP at that instant 
8. Find the speeds of the two trains 
 
xP = [(29 m/s)(t + 0.4 s) - 1/2at2]; xP = 11.6 + 29t -4.91t2 
xF = (360 m) + (6 m/s)(t + 0.4 s); xF = 362.4 + 6t 
1/2at2 - (23 m/s)t + 351 m = 0 
If (232 - 702a) > 0, collision 
a ³ 0.754 m/s2 
quadratic equation is 0.377t2 - 23t + 342 = 0 
 
t = 25.5 s; (t = 35.5 s: trains have already collided) 
xP = (29 m/s)(26.3 s) - ( 0.377 m/s2)(25.5 s)2 = 518 m 
vP = (29 m/s) - (0.754 m/s2)(25.5 s) = 9.77 m/s; 
vF = 6 m/s; vrel = 3.77 m/s 
 
 The adjacent figure shows the location of the trains. The solid straight line is for the freight train; the solid and dashed 
curved lines are for the passenger train, with reaction times of 0.4 s and 0.8 s, respectively. 
 
84 ·· After being forced out of farming, Lou has given up on trying to find work locally and is about to “ride the rails” to 
look for a job. Running at his maximum speed of 8 m/s, he is a distance d from the train when it begins to accelerate from 
rest at 1.0 m/s2. (a) If d = 30 m and Lou keeps running, will he be able to jump into the train? (b) Sketch the position 
function x(t) for the train, with x = 0 at t = 0. On the same graph, sketch x(t) for Lou for various distances d, including d 
= 30 m and the critical separation distance dc, the distance at which he just catches the train. (c) For the situation d = dc, 
what is the speed of the train when Lou catches it? What is the train’s average speed for the time interval between t = 0 
and the moment Lou catches the train? What is the exact value of dc? 
 (a), (b) and (c) For the train, xT(t) = 1/2(1.0 m/s2)t2; for Lou, xL(t) = (8 m/s)t - d 
(a) Yes, he can jump on the train. (see graph) 
(c) When Lou just manages to catch the train, his speed and that of the train are equal. The speed of the train is vT(t) = 8 
Chapter 2 Motion in One Dimension 
 
m/s = at = t m/s. 
Hence the critical time is tC = 8 s. At that time, Lou has run 64 m and the train has traveled 32 m. Consequently, dC = (64 - 
32) m = 32 m. vav = (32/8) m/s = 4 m/s. 
The three straight lines on the graph correspond to d = 30 m, d = dC = 32 m, and d = 40 m. 
 
85* ·· A train pulls away from a station with a constant acceleration of 0.4 m/s2. A passenger arrives at the track 
 6.0 s after the end of the train has passed the very same point. What is the slowest constant speed at which she 
 can run and catch the train? Sketch curves for the motion of passenger and the train as functions of time. 
 1. As in Problem 84, the critical conditions are 
 vT = vP and xT = xP 
 2. Solve for t 
 3. Find vT at t = 12 s, which is also vPC 
vT = 0.4t m/s; xT = 0.2t2 m; xP = vP(t - 6 s) 
0.2t2 = 0.4t(t - 6) 
0.2t2 = 2.4t; t = 12 s 
vPC = 4.8 m/s 
 
 The positions of the train and passenger as functions of time are shown in the adjoining figure. 
 
 
 
86 ··· Lou applies for a job as a perfume salesman. He tries to convince the boss to try his daring, aggressive promotional 
gimmick: dousing prospective customers as they wait at bus stops. A hard ball is to be thrown straight upward with an 
initial speed of 24 m/s. A thin-skinned ball filled with perfume is then thrown straight upward along the same path with a 
speed of 14 m/s. The balls are to collide when the perfume ball is at the high point of its trajectory, so that it breaks open 
Chapter 2 Motion in One Dimension 
 
and everyone gets a free sample. If t = 0 when the first ball is thrown, find the time when the perfume ball should be 
thrown. 
1. Find Dt for the perfume ball to reach the top 
2. Find height at Dt = 1.427 s 
3. Find the time for the hard ball to be at H on its 
 downward motion 
4. Time delay is t - Dt 
v = 0 = (14 - 9.81Dt) m/s; Dt = 1.427 s 
H = (14 m/s)2/2g = 9.99 m 
9.99 = 24t - 1/2(9.81)t2; solve for t 
t = 4.43 s (t = 0.46 s corresponds to upward motion) 
Throw the perfume ball at t = (4.43- 1.43) s = 3 s 
 
 
87 ··· Ball A is dropped from the top of a building at the same instant that ball B is thrown vertically upward from the 
ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. 
At what fraction of the height of the building does the collision occur? 
 Take x = 0 at ground, upward positive. 
 1. Describe the conditions at collision 
 2. Write expressions for xA and xB 
 3. Write expressions for vA and vB 
 4. Set vA = -2vB and solve for t 
 5. Use this result in part 2 with xA = xB 
 6. Write xA (see part 2) in terms of v0 
xA = xB; vA = -2vB (Note: A is moving down) 
xA = H - 1/2gt2; xB = v0t - 1/2gt2 
vA = -gt; vB = v0 - gt 
t = 2v0/3g 
H = 2v02/3g 
xA = 4v02/9g = 2H/3 
 
 
88 ··· Solve Problem 87 if the collision occurs when the balls are moving in the same direction and the speed of A is 4 
times that of B. 
 Proceed as in Problem 87 except that now the condition on velocities is vA = 4vB. One obtains for time 
 t = 4v0/3g, and for velocities vA = -(4/3)v0 and vB = -(1/3)v0. Now H = 4v02/3g, giving xA = H/3. 
89* ··· The Sprint missile, designed to destroy incoming ballistic missiles, can accelerate at 100g. If an ICBM is 
detected at an altitude of 100 km moving straight down at a constant speed of 3 ´ 104 km/h and the Sprint missile 
is launched to intercept it, at what time and altitude will the interception take place? (Note: You can neglect the 
acceleration due to gravity in this problem. Why?) 
 1. Neglect g; see below. Find xICBM, xS 
 2. Express v in m/s 
 3. Set xICBM = xS with H = 105 m and find t 
 4. Find xICBM 
 
xS = 1/2at2; xICBM = H - vt 
v = (3 ´ 107 m/h)(1 h/3600 s) = 8.33 ´ 103 m/s 
1/2 ´ 981t2 + 8.33 ´ 103t - 105 = 0; t = 8.12 s 
xICBM = [981 ´ 8.122/2] m = 3.24 ´ 104 m = 32.4 km 
 Note: In 8 s, the change in velocity Dv of the ICBM due to g is less than 80 m/s , i.e., less than 1% of v; also, g = 
1% of aS. So the result is good to about 1%. Also, if aS = 100g is taken to be the possible horizontal acceleration, 
then both objects suffer the same downward acceleration due to gravity, and this contribution cancels. 
 
90 ··· When a car traveling at speed v1 rounds a corner, the driver sees another car traveling at a slower speed v2 a 
distance d ahead. (a) If the maximum acceleration the driver’s brakes can provide is a, show that the distance d 
must be greater than (v1 - v2)2/(2a) if a collision is to be avoided. (b) Evaluate this distance for v1 = 90 km/h, 
 v2 = 45 km/h, and a = 6 m/s2. (c) Estimate or measure your reaction time and calculate the effect it would have 
Chapter 2 Motion in One Dimension 
 
 on the distance found in part (b). 
 Note: This problem is similar to Problem 83. Again, the critical conditions are x1 = x2 and v1 = v2. 
 (a) 1. Write expressions for x1, x2, v1, and v2 
 2. Set v1(t) = v2 
 3. Set x1 = x2 and solve for d 
(b) Convert km/h to m/s and substitute to find d 
(c) Assume reaction time of 0.8 s. In that time, 
 the distance between cars diminishes by 10 m. 
x1 = v1t - 1/2at2; x2 = d + v2t; v1(t) = v1 - at; v1 = v2 
t = (v1 - v2)/a 
d = (v1 - v2)2/(2a) 
d = [(25 m/s) - (12.5 m/s)]2/[2(6 m/s2)] = 13.0 m 
d¢ = (13.0 + 10) m = 23 m 
 
 
 
91 · The velocity of a particle is given by v = 6t + 3, where t is in seconds and v is in meters per second. (a) 
Sketch v(t) versus t, and find the area under the curve for the interval t = 0 to t = 5 s. (b) Find the position function 
x(t). Use it to calculate the displacement during the interval t = 0 to t = 5 s. 
(a) The graph is shown. The area under the straight line is 90 m. 
 
 (b) x(t) = ò t0 (6t + 3)dt = 3t2 + 3t; Dx = x(5) - x(0) = 90 m. 
 
92 · Figure 2-30 shows the velocity of a particle versus time. (a) What is the magnitude in meters of the area of 
the rectangle indicated? (b) Find the approximate displacement of the particle for the one-second intervals beginning 
at t = 1 s and t = 2 s. (c) What is the approximate average velocity for the interval 1 s £ t £ 3 s? 
(a) Find the area of the rectangle 
(b) 1. Find approx. area under curve 
2. Find approx. area under curve 
(c) vav = Dx/Dt 
 A = (1 m/s)(1 s) = 1 m 
Dx(1 - 2) » 1 m 
Dx(2 - 3) » 3 m 
vav » (4/2) m/s = 2 m/s 
 
93* ·· The velocity of a particle is given by v = 7t2 - 5, where t is in seconds and v is in meters per second. Find 
the general position function x(t). 
C + t - t)( = dt ) - t( = x(t) 32 57/357ò . 
 
94 ·· The equation of the curve shown in Figure 2-30 is v = 0.5t2 m/s. Find the displacement of the particle for the 
Chapter 2 Motion in One Dimension 
 
interval 1 s £ t £ 3 s by integration, and compare this answer with your answer for Problem 92. Is the average 
velocity equal to the mean of the initial and final velocities for this case? 
1. Dx = ò
3
1
0.5t2 dt = (1/6) t3 ú
ú
û
ù
1
3
 = 4.33 m. 
2. a = dv/dt = 1.0t m/s2; a is not constant, therefore vav ¹ vmean. 
 
 
95 ··· Figure 2-31 shows the acceleration of a particle versus time. (a) What is the magnitude of the area of the 
rectangle indicated? (b) The particle starts from rest at t = 0. Find the velocity at t = 1 s, 2 s, and 3 s by counting 
the rectangles under the curve. (c) Sketch the curve v(t) versus t from your results for part (b), and estimate how 
far the particle travels in the interval t = 0 to t = 3 s. 
(a) Find the area of the rectangle 
(b) v = 0 at t = 0; count squares and multiply by 
0.25 
(c) The curve of v(t) versus t is shown. By counting 
squares under the curve, we find the distance 
 traveled is approximately 6.5 m 
 
A = (0.5 m/s2)(0.5 s) = 0.25 m/s 
v(1) = 0.9 m/s, v(2) = 3 m/s, v(3) = 6 m/s 
 
 
 
 
96 ·· Figure 2-32 is a graph of v versus t for a particle moving along a straight line. The position of the particle at 
time t = 0 is x0 = 5 m. (a) Find x for various times t by counting squares, and sketch x versus t. (b) Sketch the 
acceleration a versus t. 
(a) Count squares to t = 1 s, 2 s, 3 s, 4 s, 5 s, 6 s, 7 s, 8 s, 9 s, 10 s; add x(0) = 5 m 
(b) a = dv/dt = slope of v versus t curve. Take slopes at various times. 
Chapter 2 Motion in One Dimension 
 
 
 
 
97* ··· Figure 2-33 shows a plot of x versus t for a body moving along a straight line. Sketch rough graphs of v 
versus t and a versus t for this motion. 
Note: The curve of x versus t appears to be a sine curve; hence v(t) = dx/dt is a cosine curve and a(t) = dv/dt is a 
negative sine curve. 
 
98 · True or false: (a) The equation Dx = v0t + 2
1 at2 is valid for all particle motion in one dimension. (b) If the 
velocity at a given instant is zero, the acceleration at that instant must also be zero. (c) The equation Dx = vavDt 
holds for all motion in one dimension. 
(a) False; valid only for constant a. (b) False (c) True; by definition. 
 
99 · If an object is moving at constant acceleration in a straight line, its instantaneous velocity halfway through 
any time interval is (a) greater than its average velocity. (b) less than its average velocity. (c) equal to its average 
velocity. (d) half its average velocity. (e) twice its average velocity. 
(c) is correct, by definition of vav for constant acceleration. 
100 · On a graph showing position on the vertical axis and time on the horizontal axis, a straight line with a 
negative slope represents (a) a constant positive acceleration. (b) a constant negative acceleration. (c) zero 
velocity. (d) a constant positive velocity. (e) a constant negative velocity. 
(e) The slope represents the velocity; negative slope corresponds to a negative velocity 
 
Chapter2 Motion in One Dimension 
 
101* ·· On a graph showing position on the vertical axis and time on the horizontal axis, a parabola that opens 
upward represents (a) a positive acceleration. (b) a negative acceleration. (c) no acceleration. (d) a positive 
followed by a negative acceleration. (e) a negative followed by a positive acceleration. 
(a) it represents a positive acceleration; the slope—velocity—is increasing. 
 
102 ·· On a graph showing velocity on the vertical axis and time on the horizontal axis, zero acceleration is 
represented by (a) a straight line with positive slope. (b) a straight line with negative slope. (c) a straight line 
with zero slope. (d) either (a), (b), or (c). (e) none of the above. 
(c) Zero acceleration means constant velocity. 
 
103 ·· On a graph showing velocity on the vertical axis and time on the horizontal axis, constant acceleration is 
represented by (a) a straight line with positive slope. (b) a straight line with negative slope. (c) a straight line 
with zero slope. (d) either (a), (b), or (c). (e) none of the above. 
(d) any line with constant slope, including zero slope. 
 
104 ·· Which graph of v versus t in Figure 2-34 best describes the motion of a particle with positive velocity and 
negative acceleration? 
(e) v > 0 and the slope of v(t) is negative. 
 
105* ·· Which graph of v versus t in Figure 2-34 best describes the motion of a particle with negative velocity and 
negative acceleration? 
(d) v £ 0 and the slope of v(t) is negative. 
 
106 ·· A graph of the motion of an object is plotted with the velocity on the vertical axis and time on the horizontal 
axis. The graph is a straight line. Whic h of these quantities cannot be determined from this graph? (a) The 
displacement from time t = 0 (b) The initial velocity at t = 0 (c) The acceleration of the object (d) The average 
velocity of the object (e) None of the above. 
(e) All of the quantities can be determined. 
 
107 ·· Figure 2-35 shows the position of a car plotted as a function of time. At which times t0 to t7 is the velocity 
(a) negative? (b) positive? (c) zero? At which times is the acceleration (a) negative? (b) positive? (c) zero? 
Velocity: (a) Negative at t0 and t1. (b) Positive at t3, t6, and t7. (c) Zero at t2, t4, and t5. 
Acceleration: (a) Negatvie at t4. (b) Positive at t2 and t6. (c) Zero at t0, t1, t3, t5, and t7 
108 ·· Sketch v-versus-t curves for each of the following conditions: (a) Acceleration is zero and constant while 
velocity is not zero. (b) Acceleration is constant but not zero. (c) Velocity and acceleration are both positive. 
(d) Velocity and acceleration are both negative. (e) Velocity is positive and acceleration is negative. (f) Velocity 
is negative and acceleration is positive. (g) Velocity is zero at a point but the acceleration is not zero. 
 (a)
 
(b) (c) (d) 
Chapter 2 Motion in One Dimension 
 
 (e) (f) (g) 
 
 
109* ·· Figure 2-36 shows nine graphs of position, velocity, and acceleration for objects in linear motion. Indicate 
the graphs that meet the following conditions: (a) Velocity is constant (b) Velocity has reversed its direction 
(c) Acceleration is constant (d) Acceleration is not constant. Which graphs of velocity and acceleration are 
mutually consistent? 
(a) a, f , i; (b) c, d; (c) a, d, e, f , h, i; (d) b, c, g. The graphs d and h, and f and i are mutually consistent. 
 
110 · Two cars are being driven at the same speed v, one behind the other, with a distance d between them. The 
first driver jams on her brakes and decelerates at a rate a = 6 m/s2. The second driver sees her brake lights and 
reacts, decelerating at the same rate starting 0.5 s later. (a) What is the minimum distance d such that the two cars 
do not collide? (b) Express this answer in meters for v = 100 km/h (62 mi/h). 
(a) Since v1 = v2 = v, d = 0.5v. 
(b) Converting 100 km/ to m/s we obtain v = 27.8 m/s. Hence d = 13.9 m. 
 
111 · The velocity of a particle in meters per second is given by v = 7 - 4t, where t is in seconds. (a) Sketch v(t) 
versus t, and find the area between the curve and the t axis from t = 2 s to t = 6 s. (b) Find the position x(t) by 
integration, and use it to find the displacement during the interval t = 2 s to t = 6 s. (c) What is the average 
velocity for this interval? 
 (a) The sketch is shown here. The area 
between the curve and the t axis between t = 2 
s and t = 6 s is about 36 m. 
 (b) m;2747 C) + t - t( = dt t) - ( = x(t) 2ò 
 Dx = x(6) - x(2) = -36 m 
(c) Note that this is an instance of constant acceleration. Hence, 
 vav = [v(6) + v(2)]/2 = -9 m/s; note: vavDt = Dx. 
112 ·· Estimate how high a ball or small rock can be thrown if it is thrown straight up. 
 1. Estimate the initial velocity 
 2. Use H = v02/2g to find H 
 v0 » 100 km/h » 28 m/s 
H » [(28)2/19.6] m = 40 m 
 
Chapter 2 Motion in One Dimension 
 
113* ·· The cheetah can run as fast as v1 = 100 km/h, the falcon can fly as fast as v2 = 250 km/h, and the sailfish 
can swim as fast as v3 = 120 km/h. The three of them run a relay with each covering a distance L at maximum 
speed. What is the average speed v of this triathlon team? 
1. Express the time required for each animal 
2. Write the total time, Dt 
3. Find v = Dx/Dt = 3L/Dt; use values for v1 etc. 
t1 = L/v1; t2 = L/v2; t3 = L/v3 
Dt = L(1/v1 + 1/v2 + 1/v3) 
v = [3/(0.01 + 0.004 + 0.00833)] m/s = 134 m/s 
 
114 ·· In 1997, the men’s world record for the 50-m freestyle was held by Tom Jager of the United States, who 
covered d = 50 m in t = 21.81 s. Suppose Jager started from rest at constant acceleration a, and reached his 
maximum speed in 2.00 s, which he then kept constant until the finish line. Find Jager’s acceleration a. 
1. Write d in terms of a, v0, and t 
2. Solve for a 
 50 m = [( 1/2a ´ 22) + (2a ´ 19.81)] m 
a = 1.20 m/s2 
 
 
115 ·· The click beetle can project itself vertically with an acceleration of about a = 400 g (an order of magnitude 
more than a human could stand). The beetle jumps by “unfolding” its legs, which are about d = 0.6 cm long. 
How high can the click beetle jump? How long is the beetle in the air? (Assume constant acceleration while in 
contact with the ground, and neglect air resistance.) 
1. Find the time of contact with the ground 
2. Find the velocity at take-off 
3. Find height H 
4. Find the time to return to x = 0 
 1/2(400g)t12 = 6 ´ 10-3 m; t1 = 1.75 ´ 10-3 s 
v = (400g)(1.75 ´ 10-3 s) = 6.86 m/s 
H = v2/2g = 2.4 m 
T = 2v/g = 1.4 s 
 
116 ·· The one-dimensional motion of a particle is plotted in Figure 2-37. (a) What is the acceleration in the 
intervals AB, BC, and CE? (b) How far is the particle from its starting point after 10 s? (c) Sketch the displacement 
of the particle as a function of time; label the instants A, B, C, D, and E on your figure. (d) At what time is the 
particle traveling most slowly? 
(a) a = Dv/Dt; determine a for the three intervals 
 
(b) Find the area under v versus t 
 
(c) The displacement, x, versus t is shown 
in the figure 
 
 
(d) At D, t = 8 s, v = 0 
AB: a = (10/3) m/s2 = 3.33 m/s2; BC: a = 0; 
CD: a = (-30/4) m/s2 = -7.5 m/s2 
Dx = [5 ´ 3 + 1/2(10 ´ 3) + 15 ´ 3] m = 75 m 
 
Chapter 2 Motion in One Dimension 
 
 
 
117* ·· Consider the velocity graph in Figure 2-38. Assuming x = 0 at t = 0, write correct algebraic expressions for 
x(t), v(t), and a(t) with appropriate numerical values inserted for all constants. 
1. Write v(t); note that a is constant and < 0 
2. x(t) = òv(t)dt 
 v(t) = (50 - 10t) m/s; a = -10 m/s2 
x(t) = 50t - 5t2118 ·· Starting at one station, a subway train accelerates from rest at a constant rate of 1.0 m/s2 for half the 
distance to the next station, then slows down at the same rate for the second half of the journey. The total 
distance between stations is 900 m. (a) Sketch a graph of the velocity v as a function of time over the full 
journey. (b) Sketch a graph of the distance covered as a function of time over the full journey. Place 
appropriate numerical values on both axes. 
 
 
 
(a) The graph for v(t) is shown below (b) The graph for x(t) is shown below 
 
 
 
119 ·· The acceleration of a certain rocket is given by a = Ct, where C is a constant. (a) Find the general position 
function x(t). (b) Find the position and velocity at t = 5 s if x = 0 and v = 0 at t = 0 and C = 3 m/s2. 
(a) 1. Integrate a(t) to find v(t) v(t) = òa(t) dt = Còt dt = 1/2Ct2 + A; A is a constant 
Chapter 2 Motion in One Dimension 
 
2. Integrate v(t) to find x(t) 
(b) Find constants A, B from boundary conditions 
Evaluate v(5) and x(5) with A = B = 0 
x(t) = òv(t) dt = ò(1/2Ct2 + A) dt = Ct3/6 + At + B 
v(0) = 0: A = 0; x(0) = 0: B = 0 
v(5) = 37.5 m/s; x(5) = 62.5 m 
 
120 ·· A physics professor demonstrates his new “anti-gravity parachute” by exiting from a helicopter at an 
altitude of 1500 m with zero initial velocity. For 8 s he falls freely. Then he switches on the “parachute” and 
falls with constant upward acceleration of 15 m/s2 until his downward speed reaches 5 m/s, whereupon he 
adjusts his controls to maintain that speed until he reaches the ground. (a) On a single graph, sketch his 
acceleration and velocity as functions of time. (Take upward to be positive.) (b) What is his speed at the end of 
the first 8 s? (c) For how long does he maintain the constant upward acceleration of 15 m/s2? (d) How far 
does he travel during the upward acceleration in part (c)? (e) How many seconds are required for the entire trip 
from the helicopter to the ground? (f) What is his average velocity for the entire trip? 
Note: we shall do part (a) last. 
 
(b) Use Equ. 2-12 
(c) Use Equ. 2-12 
(d) Use Equ. 2-13 
(e) 1. Find the remaining distance 
 2. Find the total time 
 (a) The sketch of a(t) and v(t) is shown. 
v(8) =[-(9.81)(8)] m/s = -78.5 m/s 
t2 = Dv/a = (73.5/15) s = 4.9 s 
Dx2 = 1/2[(-78.5 - 5)(4.9)] m = 205 m 
Dx3 = [1500 - 1/2(78.5)(8) - 205] m = 981 m 
ttot = [(981/5) + 4.9 + 8] s = 209 s 
 
 
121* ·· Without telling Sally, Joe made travel arrangements that include a stopover in Toronto to visit Joe’s old 
buddy. Sally doesn’t like Joe’s buddy and wants to change their tickets. She hops on a courtesy motor 
scooter and begins accelerating at 0.9 m/s2 toward the ticket counter to make arrangements. As she begins 
moving, Joe is 40 m behind her, running at constant speed of 9 m/s. (a) How long does it take for Joe to 
catch up with her? (b) What is the time interval during which Joe remains ahead of Sally? 
(a) 1. Write expressions for xS and xJ 
2. Set xS = xJ to obtain equation for t 
3. Solve for t; keep smallest answer 
(b) Sally will catch Joe at 13.33 s 
 xS = 0.45t2 m; xJ = (-40 + 9t) m 
0.45t2 -9t + 40 = 0 
t = 6.67 s, t = 13.33 s; t = 6.67 s 
Dt = 6.67 s 
 
122 ·· A speeder races past at 125 km/h. A patrol car pursues from rest with constant acceleration of 8 m/h×s 
 until it reaches its maximum speed of 190 km/h, which it maintains until it catches up with the speeder. (a) 
 How long until the patrol car catches the speeder if it starts moving just as the speeder passes? (b) How far 
does each car travel? (c) Sketch x(t) for each car. 
 (a) 1. Write expressions for xS and xP; convert all 
 quantities to m/s and m/s2 
 
 2. Set xS = xP and solve for t 
 (b) The distance traveled is xS 
 (c) The graph of xS and xP is shown; the straight 
 represents xS, the parabola represents xP. 
Chapter 2 Motion in One Dimension 
 
xS = 34.7t m; tacc = 52.8/2.22 s = 23.8 s; 
xacc =[ 1/2(52.8)(23.8)] m; 
xP = [628 + 52.8 ´ (t - 23.8)] m 
18.1t m = 628 m, t = 34.7 s 
xS = 1204 m 
 
 
 
123 ·· When the patrol car in Problem 122 (traveling at 190 km/h), pulls within 100 m behind the speeder 
(traveling at 125 km/h), the speeder sees the police car and slams on his brakes, locking the wheels. (a) 
Assuming that each car can brake at 6 m/s2 and that the driver of the police car brakes instantly as she sees the 
brake lights of the speeder (reaction time = 0 s), show that the cars collide. (b) At what time after the speeder 
applies his brakes do the two cars collide? (c) Discuss how reaction time affects this problem. 
(a) Dx = v2/2a (Equ. 2-15) 
Note that DxS + 100 m < DxP 
(b) 1. Write xS and xP as functions of t; set t = 
0 at brake time, x = 0 at patrol car at t = 0 
2. Set xS = xP and solve for t 
 DxS = (34.72/12) m=100.3 m; DxP = (52.82/12) m 
= 232.3 m. The cars collide. 
xS = (100 + 34.7t - 3t2) m; xP = (52.8t - 3t2) m 
 
t = 5.52 s 
(c) If you take reaction time into account, the collision will occur sooner and be more severe. 
124 ·· The speed of a good base runner is 9.5 m/s. The distance between bases is 26 m, and the pitcher is about 
18.5 m from home plate. If a runner on first base edges 2 m off the base and takes off for second the instant the 
ball leaves the pitcher’s hand, what is the likelihood that the runner will steal second base safely? 
We will neglect the time of acceleration of the runner; we will assume that the speed of a fast ball is 28 m/s, and use 
that also for the speed of the ball thrown by the catcher. We will take 0.6 s as the reaction time of the catcher. The 
time taken by the runner is 24/9.5 = 2.53 s. The time of flight of the ball is 3 ´ 18.5/28 = 1.98 s; add 0.6 s as the 
reaction time to get 2.58 s. It will be a very close call! 
 
125* ·· Repeat Problem 124, but with the runner attempting to steal third base, starting from second base with a 
lead of 3 m. 
We use the same speeds and reaction times as before. The distance the runner travels is now 23 m. The 
distance the ball travels is now (18.5 + 26) = 44.5 m so the running time is 23/9.5 = 2.42 s. The time of flight of the 
ball is 44.5/28 = 1.59 s. Add to this a reaction time of 0.6 s to get 2.19 s. A good umpire will call him out! 
 
126 ·· Urgently needing the cash prize, Lou enters the Rest-to-Rest auto competition, in which each contestant’s 
car begins and ends at rest, covering a distance L in as short a time as possible. The intention is to demonstrate 
mechanical and driving skills, and to consume the largest amount of fossil fuels in the shortest time possible. The 
course is designed so that maximum speeds of the cars are never reached. If Lou’s car has a maximum acceleration 
Chapter 2 Motion in One Dimension 
 
of a, and a maximum deceleration of 2a, then at what fraction of L should Lou move his foot from the gas pedal to 
the brake? What fraction of the time for the trip has elapsed at that point? 
Let t1 be the time when the brake is applied, L1 the distance traveled from t = 0 to t = t1. Do (b) first. 
 
(b) 1. Write the expressions for x, v for 0 £ t £ t1 
2. Write expressions for x, v for t1 £ t £ tfin 
3. At t = tfin v = 0; find t1 in terms of tfin from 2 
(a) 1. Set v = 0 at x = L to find t = tfin 
2. Find x(t1) = L1 
v = at; x = 1/2at2 
v = at1 - 2a(t -t1); x = 1/2at2 + at1(t - t1) -1/2(2a)(t - t1)2 
t1 = (2/3)tfin 
tfin = 3L/a 
L1 = 1/2at12 = 1/2a(4/9)(3L/a) = (2/3)L 
127 ··· The acceleration of a badminton birdie falling under the influence of gravity and a resistive force, such 
as air resistance,is given by a = dv/dt = g - bv, where g is the free-fall acceleration due to gravity and b is 
a constant that depends on the mass and shape of the birdie and on the properties of the medium. Suppose 
the birdie begins with zero velocity at time t = 0. (a) Discuss qualitatively how the speed v varies with time 
from your knowledge of the rate of change dv/dt given by this equation. What is the velocity when the 
acceleration is zero? This is called the terminal velocity. (b) Sketch the solution v(t) versus t without 
solving the equation. This can be done as follows: at t = 0, v is zero and the slope is g. Sketch a straight-line 
segment, neglecting any change in slope for a short time interval. At the end of the interval, the velocity is 
not zero, so the slope is less than g. Sketch another straight-line segment with a smaller slope. Continue 
until the slope is zero and the velocity equals the terminal velocity. 
(a) Initially v = 0 and increases as gt, but as v becomes finite, the acceleration diminishes to g - bv. 
Ultimately, the acceleration approaches zero and v remains constant. To find this terminal velocity set dv/dt 
= 0 and solve for v = vterm = g/b. 
 
(c) A curve showing the general behavior of v as a function of time is shown below. 
 
 
 
128 ··· Suppose acceleration is a function of x, where a(x) = 2x m/s2. (a) If the velocity at x = 1 m is zero, 
what is the speed at x = 3 m? (b) How long does it take to travel from x = 1 m to x = 3 m? 
(a) Since a = dv/dt we shall have to integrate to find v. Write dv/dt as (dv/dx)(dx/dt) = v(dv/dx). Hence, 
the problem statement can be rewritten v(dv/dx) = 2x, or v dv = 2x dx. Integrate: òv dv = òx dx which gives 
the relationship v2 - v02 = 2x2 - 2x02. Now set v0 = 0, x0 = 1 and find v; 1) - x2( = v(x) 2 and v(3) = 4 
m/s. (b) We have the expression for v(x), given the initial conditions. So we can now integrate the 
Chapter 2 Motion in One Dimension 
 
differential equation dx/dt = v(x) or òdt = òdx/v(x), with limits t = 0 to t, and x = 1 to x = 3. From standard 
integral tables we obtain 
 ÷
÷
ø
ö
ç
ç
è
æ
1 - x + x
1 - x + x = t - t
2
00
2
0 ln
2
1
 
 setting t0 = 0, x0 = 1, and x = 3 we find t(3) = 1.25 s. 
129* ··· Suppose that a particle moves in a straight line such that, at any time t, its position, velocity, and 
acceleration all have the same numerical value. Give the position x as a function of time. 
We are given that v = 1 ´ x = 1 ´ a. So dx/dt = x, and integrating we obtain t - t0 = ln(x/x0) or 
e x = x(t) t - t0 0 . The velocity and acceleration are obtained from v = dx/dt and a = dv/dt: 
a(t) = e x = v(t) t - t0 0 . 
 
130 ··· An object moving in a straight line doubles its velocity each second for the first 10 s. Let the initial 
speed be 2 m/s. (a) Sketch a smooth function v(t) that gives the velocity. (b) What is the average 
velocity over the first 10 s? 
 (a) According to the statement v(t) = 2t v0. 
 A plot of this function is shown. 
 
(b) The average velocity is given by vav = Dx/Dt. We need to find Dx. We do so by integrating v(t)= dx/dt 
from t = 0 to t = 10 s. 
; Dx = (v0/ln 2)( 210 - 1); 
 
Setting v0 = 2 m/s, Dx = 2952 m and vav = 295.2 m/s 
131 ··· In a dream, you find that you can run at superhuman speeds, but there is a resistant force that reduces your 
speed by one-half for each second that passes. Assume that the laws of physics still hold in your dreamworld, and 
that your initial speed is 1000 m/s. (a) Sketch a smooth function v(t) that gives your velocity. (b) What is your 
average velocity over the first 10 s? 
Note: This problem is the same as Problem 130 except 
that now v(t) = 2-t v0. We can follow the procedure of 
the preceding problem. 
(a) A sketch of v(t) is shown 
dt v 2 = dx 0t
10
0
x
0 òò
D
Chapter 2 Motion in One Dimension 
 
 
(b) We now find Dx = (v0/ln 2)(1 - 2-10) = 1441 m, 
and vav = 144.1 m/s