Sol Física 10ed Halliday

Prévia do material em texto

1 
 
 
Chapter 1 
 
 
1. THINK In this problem we’re given the radius of Earth, and asked to compute its 
circumference, surface area and volume. 
 
EXPRESS Assuming Earth to be a sphere of radius 
 
  6 3 36.37 10 m 10 km m 6.37 10 km,ER    
 
 
the corresponding circumference, surface area and volume are: 
 
2 342 , 4 ,
3
E E EC R A R V R
   
. 
The geometric formulas are given in Appendix E. 
 
ANALYZE (a) Using the formulas given above, we find the circumference to be 
 
3 42 2 (6.37 10 km) 4.00 10 km.EC R      
 
(b) Similarly, the surface area of Earth is 
 
 
2
2 3 8 24 4 6.37 10 km 5.10 10 kmEA R     , 
(c) and its volume is 
 
3
3 3 12 34 4 6.37 10 km 1.08 10 km .
3 3
EV R
 
    
 
 
LEARN From the formulas given, we see that 
EC R
, 
2
EA R
, and 
3
EV R
. The ratios 
of volume to surface area, and surface area to circumference are 
/ / 3EV A R
 and 
/ 2 EA C R
. 
 
2. The conversion factors are: 
1 gry 1/10 line
, 
1 line 1/12 inch
and 1 point = 1/72 
inch. The factors imply that 
 
1 gry = (1/10)(1/12)(72 points) = 0.60 point. 
 
Thus, 1 gry
2
 = (0.60 point)
2
 = 0.36 point
2
, which means that 
2 20.50 gry = 0.18 point
. 
 
3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside 
front cover of the textbook (see also Table 1–2). 
 
 CHAPTER 1 2 
(a) Since 1 km = 1  103 m and 1 m = 1  106 m, 
 
  3 3 6 91km 10 m 10 m 10 m m 10 m.    
 
The given measurement is 1.0 km (two significant figures), which implies our result 
should be written as 1.0  109 m. 
 
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10
2
 m, 
 
  2 2 6 41cm = 10 m = 10 m 10 m m 10 m.    
 
We conclude that the fraction of one centimeter equal to 1.0 m is 1.0  104. 
 
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m, 
 
  6 51.0yd = 0.91m 10 m m 9.1 10 m.   
 
4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we 
obtain 
 
1 inch 6 picas
0.80 cm = 0.80 cm 1.9 picas.
2.54 cm 1 inch
  
  
  
 
(b) With 12 points = 1 pica, we have 
 
 
1 inch 6 picas 12 points
0.80 cm = 0.80 cm 23 points.
2.54 cm 1 inch 1 pica
   
   
   
 
 
5. THINK This problem deals with conversion of furlongs to rods and chains, all of 
which are units for distance. 
 
EXPRESS Given that 
1 furlong 201.168 m, 1rod 5.0292 m
 and 
1chain 20.117 m
, 
the relevant conversion factors are 
1 rod
1.0 furlong 201.168 m (201.168 m) 40 rods,
5.0292 m
  
 
and 
1 chain
1.0 furlong 201.168 m (201.168 m) 10 chains
20.117 m
  
. 
Note the cancellation of m (meters), the unwanted unit. 
 
ANALYZE Using the above conversion factors, we find 
 
(a) the distance d in rods to be 
 
40 rods
4.0 furlongs 4.0 furlongs 160 rods,
1 furlong
d   
 
 
 
3 
(b) and in chains to be 
 
10 chains
4.0 furlongs 4.0 furlongs 40 chains.
1 furlong
d   
 
 
LEARN Since 4 furlongs is about 800 m, this distance is approximately equal to 160 
rods (
1 rod 5 m
) and 40 chains (
1 chain 20 m
). So our results make sense. 
 
6. We make use of Table 1-6. 
 
(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? 
We note from the already completed part of the table that 1 cahiz equals a dozen fanega. 
Thus, 1 fanega = 
1
12
 cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the 
already completed part) implies that 1 cuartilla = 
1
48
 cahiz, or 2.08  102 cahiz. 
Continuing in this way, the remaining entries in the first column are 6.94  103 and 
33.47 10
. 
 
(b) In the second (“fanega”) column, we find 0.250, 8.33  102, and 4.17  102 for the 
last three entries. 
 
(c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. 
 
(d) Finally, in the fourth (“almude”) column, we get 
1
2
 = 0.500 for the last entry. 
 
(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our 
amount of 7.00 almudes must be equal to 14.0 medios. 
 
(f) Using the value (1 almude = 6.94  103 cahiz) found in part (a), we conclude that 
7.00 almudes is equivalent to 4.86  102 cahiz. 
 
(g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 
m
3
 or 55501 cm
3
. Thus, 7.00 almudes = 
7.00
12
 fanega = 
7.00
12
 (55501 cm
3
) = 3.24  104 cm3. 
 
7. We use the conversion factors found in Appendix D. 
 
 
2 31 acre ft = (43,560 ft ) ft = 43,560 ft 
 
 
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is 
 
 
2 2 2 7 3(26 km )(1/6 ft) (26 km )(3281ft/km) (1/6 ft ) 4.66 10 ft .V    
 
 
Thus, 
V 

 
  
4 66 10
4 3560 10
11 10
7
4
3.
.
.
ft
ft acre ft
acre ft.
3
3
 
 CHAPTER 1 4 
 
8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is 
equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z. 
 
(a) In units of W, we have 
 
258 W
50.0 S 50.0 S 60.8 W
212 S
 
  
 
 
 
(b) In units of Z, we have 
 
156 Z
50.0 S 50.0 S 43.3 Z
180 S
 
  
 
 
 
9. The volume of ice is given by the product of the semicircular surface area and the 
thickness. The area of the semicircle is A = r2/2, where r is the radius. Therefore, the 
volume is 
2
2
V r z


 
 
where z is the ice thickness. Since there are 10
3
 m in 1 km and 10
2
 cm in 1 m, we have 
 
 
3 2
510 m 10 cm2000km 2000 10 cm.
1km 1m
r
   
     
   
 
 
In these units, the thickness becomes 
 
2
210 cm3000m 3000m 3000 10 cm
1m
z
 
    
 
 
 
which yields 
   
2
5 2 22 32000 10 cm 3000 10 cm 1.9 10 cm .
2
V

    
 
 
10. Since a change of longitude equal to 
360
corresponds to a 24 hour change, then one 
expects to change longitude by
360 / 24 15  
 before resetting one's watch by 1.0 h. 
 
11. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio 
of weeks is simply 10/7 or (to 3 significant figures) 1.43. 
 
(b) In a regular day, there are 86400 seconds, but in the French system described in the 
problem, there would be 10
5
 seconds. The ratio is therefore 0.864. 
 
12. A day is equivalent to 86400 seconds and a meter is equivalent to a million 
micrometers, so 
 
 
5 
37 10
14 86400
31
6.
. .
m m m
day s day
m s
b gc h
b gb g


 
 
13. The time on any of these clocks is a straight-line function of that on another, with 
slopes 

 1 and y-intercepts 

 0. From the data in the figure we deduce 
 
2 594 33 662
, .
7 7 40 5
C B B At t t t   
 
 
These are used in obtaining the following results. 
 
(a) We find 
 
33
495 s
40
B B A At t t t    
 
when t'A  tA = 600 s. 
 
(b) We obtain 
      t t t tC C B B
2
7
2
7
495 141b g b g s.
 
 
(c) Clock B readstB = (33/40)(400) (662/5)  198 s when clock A reads tA = 400 s. 
 
(d) From tC = 15 = (2/7)tB + (594/7), we get tB  245 s. 
 
14. The metric prefixes (micro (), pico, nano, …) are given for ready reference on the 
inside front cover of the textbook (also Table 1–2). 
 
(a) 
 6
100 y 365 day 24 h 60 min
1 century 10 century 52.6 min.
1 century 1 y 1 day 1 h
                 
       
 
 
(b) The percent difference is therefore 
 
52.6 min 50 min
4.9%.
52.6 min


 
 
15. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 
seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix, 
this is roughly 1.21  1012 s. 
 
16. We denote the pulsar rotation rate f (for frequency). 
 
3
1 rotation
1.55780644887275 10 s
f



 
 
 CHAPTER 1 6 
(a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if 
we ignore significant figure considerations for a moment), we obtain the number of 
rotations: 
 3
1 rotation
604800 s 388238218.4
1.55780644887275 10 s
N

 
  
 
 
 
which should now be rounded to 3.88  108 rotations since the time-interval was 
specified in the problem to three significant figures. 
 
(b) We note that the problem specifies the exact number of pulsar revolutions (one 
million). In this case, our unknown is t, and an equation similar to the one we set up in 
part (a) takes the form N = ft, or 
 
6
3
1 rotation
1 10
1.55780644887275 10 s
t

 
   
 
 
 
which yields the result t = 1557.80644887275 s (though students who do this calculation 
on their calculator might not obtain those last several digits). 
 
(c) Careful reading of the problem shows that the time-uncertainty per revolution is 
173 10 s 
. We therefore expect that as a result of one million revolutions, the 
uncertainty should be 
17 6 11 ( 3 10 )(1 10 )= 3 10 s     
. 
 
17. THINK In this problem we are asked to rank 5 clocks, based on their performance as 
timekeepers. 
 
EXPRESS We first note that none of the clocks advance by exactly 24 h in a 24-h period 
but this is not the most important criterion for judging their quality for measuring time 
intervals. What is important here is that the clock advance by the same (or nearly the 
same) amount in each 24-h period. The clock reading can then easily be adjusted to give 
the correct interval. 
 
ANALYZE The chart below gives the corrections (in seconds) that must be applied to 
the reading on each clock for each 24-h period. The entries were determined by 
subtracting the clock reading at the end of the interval from the clock reading at the 
beginning. 
 
Clocks C and D are both good timekeepers in the sense that each is consistent in its daily 
drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and 
predictable corrections. The correction for clock C is less than the correction for clock D, 
so we judge clock C to be the best and clock D to be the next best. The correction that 
must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range 
from 5 s to +10 s, for clock E it is in the range from 70 s to 2 s. After C and D, A has 
 
 
7 
the smallest range of correction, B has the next smallest range, and E has the greatest 
range. From best to worst, the ranking of the clocks is C, D, A, B, E. 
 
CLOCK Sun. Mon. Tues. Wed. Thurs. Fri. 
-Mon. -Tues. -Wed. -Thurs. -Fri. -Sat. 
A 16 16 15 17 15 15 
B 3 +5 10 +5 +6 7 
C 58 58 58 58 58 58 
D +67 +67 +67 +67 +67 +67 
E +70 +55 +2 +20 +10 +10 
 
LEARN Of the five clocks, the readings in clocks A, B and E jump around from one 24-
h period to another, making it difficult to correct them. 
 
18. The last day of the 20 centuries is longer than the first day by 
 
   20 century 0.001 s century 0.02 s.
 
 
The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. 
Since the increase occurs uniformly, the cumulative effect T is 
 
   
 
average increase in length of a day number of days
0.01 s 365.25 day
2000 y
day y
7305 s
T 
   
    
   

 
 
or roughly two hours. 
 
 
19. When the Sun first disappears while lying 
down, your line of sight to the top of the Sun 
is tangent to the Earth’s surface at point A 
shown in the figure. As you stand, elevating 
your eyes by a height h, the line of sight to the 
Sun is tangent to the Earth’s surface at point 
B. 
 
 
Let d be the distance from point B to your eyes. From the Pythagorean theorem, we have 
 
 
2 2 2 2 2( ) 2d r r h r rh h     
 
 
 CHAPTER 1 8 
or 
2 22 ,d rh h 
where r is the radius of the Earth. Since 
r h
, the second term can be 
dropped, leading to 
2 2d rh
. Now the angle between the two radii to the two tangent 
points A and B is , which is also the angle through which the Sun moves about Earth 
during the time interval t = 11.1 s. The value of  can be obtained by using 
 
 
360 24 h
t


. 
This yields 
 
(360 )(11.1 s)
0.04625 .
(24 h)(60 min/h)(60 s/min)
   
 
 
Using 
tand r 
, we have 
2 2 2tan 2d r rh , or 
 
 
2
2
tan
h
r


 
 
Using the above value for  and h = 1.7 m, we have 
65.2 10 m.r  
 
 
20. (a) We find the volume in cubic centimeters 
 
 
3
3
5 3231 in 2.54 cm193 gal = 193 gal 7.31 10 cm
1gal 1in
   
    
  
 
 
and subtract this from 1  106 cm3 to obtain 2.69  105 cm3. The conversion gal  in3 is 
given in Appendix D (immediately below the table of Volume conversions). 
 
(b) The volume found in part (a) is converted (by dividing by (100 cm/m)
3
) to 0.731 m
3
, 
which corresponds to a mass of 
 
1000 kg m 0.731 m = 731 kg3 2c h c h
 
 
using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be 
filled in 
5731kg 4.06 10 min = 0.77 y
0.0018 kg min
 
 
 
after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h). 
 
21. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the 
number of atoms, then ME = Nm or N = ME/m. We convert mass m to kilograms using 
Appendix D (1 u = 1.661  1027 kg). Thus, 
 
 
 
9 
N
M
m
E 


 

598 10
40 1661 10
9 0 10
24
27
49.
.
. .
kg
u kg ub g c h
 
 
22. The density of gold is 
3
3
19.32 g
19.32 g/cm .
1 cm
m
V
   
 
 
(a) We take the volume of the leaf to be its area A multiplied by its thickness z. With 
density  = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be 
 
V
m
 

1430. .cm3
 
We convert the volume to SI units: 
 
3
3 6 31 m1.430 cm 1.430 10 m .
100 cm
V 
 
   
 
 
 
Since V = Az with z = 1  10-6 m (metric prefixes can be found in Table 1–2), we obtain 
 
A 





1430 10
1 10
1430
6
6
.
. .
m
m
m
3
2
 
 
(b) The volume of a cylinder of length 

 is 
V A 
 where the cross-section area is that of 
a circle: A = r2. Therefore, with r =2.500  106 m and V = 1.430  106 m3, we obtain 
 
4
2
7.284 10 m 72.84 km.
V
r   
 
 
23. THINK This problem consists of two parts: in the first part, we are asked to find the 
mass of water, given its volume and density; the second part deals with the mass flow 
rate of water, which is expressed as kg/s in SI units. 
 
EXPRESS From the definition of density: 
/m V 
, we see that mass can be calculated 
as 
m V
, the product of the volume of water and its density. With 1 g = 1  103 kg 
and 1 cm
3
 = (1  102m)3 = 1  106m3, the density of water in SI units (kg/m3) is 
 
3 3
3 3 3
3 6 3
1 g 10 kg cm
1 g/cm 1 10 kg m .
cm g 10 m
 

    
       
     
 
 
To obtain the flow rate, we simply divide the total mass of the water by the time taken to 
drain it. 
 
ANALYZE (a) Using 
m V
, the mass of a cubic meter of water is 
 CHAPTER 1 10 
 
3 3 3(1 10 kg/m )(1m ) 1000 kg.m V    
 
(b) The total mass of water in the container is 
 
3 3 3 6(1 10 kg m )(5700 m ) 5.70 10 kgM V    , 
 
and the time elapsed is t = (10 h)(3600 s/h) = 3.6  104 s. Thus, the mass flow rate R is 
 
6
4
5.70 10 kg
158 kg s.
3.6 10 s
M
R
t

  

 
 
LEARN In terms of volume, the drain rate can be expressed as 
 
3
3
4
5700 m
0.158 m /s 42 gal/s.
3.6 10 s
V
R
t
    

 
 
The greater the flow rate, the less time required to drain a given amount of water. 
 
24. The metric prefixes (micro (), pico, nano, …) are given for ready reference on the 
inside front cover of the textbook (see also Table 1–2). The surface area A of each grain 
of sand of radius r = 50 m = 50  106 m is given by A = 4(50  106)2 = 3.14  108 
m
2
 (Appendix E contains a variety of geometry formulas). We introduce the notion of 
density, 
/m V 
, so that the mass can be found from m = V, where  = 2600 kg/m3. 
Thus, using V = 4r3/3, the mass of each grain is 
 
 
3
63
9
3
4 50 10 m4 kg
2600 1.36 10 kg.
3 m 3
r
m V
             
  
 
 
We observe that (because a cube has six equal faces) the indicated surface area is 6 m
2
. 
The number of spheres (the grains of sand) N that have a total surface area of 6 m
2
 is 
given by 
2
8
8 2
6 m
1.91 10 .
3.14 10 m
N

  

 
 
Therefore, the total mass M is 
   8 91.91 10 1.36 10 kg 0.260 kg.M Nm     
 
 
25. The volume of the section is (2500 m)(800 m)(2.0 m) = 4.0  106 m3. Letting “d” 
stand for the thickness of the mud after it has (uniformly) distributed in the valley, then 
its volume there would be (400 m)(400 m)d. Requiring these two volumes to be equal, 
we can solve for d. Thus, d = 25 m. The volume of a small part of the mud over a patch 
of area of 4.0 m
2
 is (4.0)d = 100 m
3
. Since each cubic meter corresponds to a mass of 
 
 
11 
1900 kg (stated in the problem), then the mass of that small part of the mud is 
51.9 10 kg
. 
 
26. (a) The volume of the cloud is (3000 m)(1000 m)2 = 9.4  109 m3. Since each cubic 
meter of the cloud contains from 50  106 to 500  106 water drops, then we conclude 
that the entire cloud contains from 4.7  1018 to 4.7  1019 drops. Since the volume of 
each drop is 
4
3
 (10  106 m)3 = 4.2  1015 m3, then the total volume of water in a cloud 
is from 
32 10
 to 
42 10
 m
3
. 
 
(b) Using the fact that 
3 3 3 31 L 1 10 cm 1 10 m   
, the amount of water estimated in 
part (a) would fill from 
62 10
 to 
72 10
bottles. 
 
(c) At 1000 kg for every cubic meter, the mass of water is from 
62 10
 to 
72 10
kg. 
The coincidence in numbers between the results of parts (b) and (c) of this problem is due 
to the fact that each liter has a mass of one kilogram when water is at its normal density 
(under standard conditions). 
 
27. We introduce the notion of density, 
/m V 
, and convert to SI units: 1000 g = 1 kg, 
and 100 cm = 1 m. 
 
(a) The density  of a sample of iron is 
 
 
3
3 31 kg 100 cm7.87 g cm 7870 kg/m .
1000 g 1 m
        
   
 
 
If we ignore the empty spaces between the close-packed spheres, then the density of an 
individual iron atom will be the same as the density of any iron sample. That is, if M is 
the mass and V is the volume of an atom, then 
 
26
29 3
3 3
9.27 10 kg
1.18 10 m .
7.87 10 kg m
M
V

   

 
 
(b) We set V = 4R3/3, where R is the radius of an atom (Appendix E contains several 
geometry formulas). Solving for R, we find 
 
 
1 3
29 31 3
10
3 1.18 10 m3
1.41 10 m.
4 4
V
R


              
 
 
The center-to-center distance between atoms is twice the radius, or 2.82  1010 m. 
 
 CHAPTER 1 12 
28. If we estimate the “typical” large domestic cat mass as 10 kg, and the “typical” atom 
(in the cat) as 10 u  2  1026 kg, then there are roughly (10 kg)/( 2  1026 kg)  5  
10
26
 atoms. This is close to being a factor of a thousand greater than Avogadro’s number. 
Thus this is roughly a kilomole of atoms. 
 
29. The mass in kilograms is 
 
28 9
100 16 10 10 0 3779
.
.
piculs
gin
1picul
tahil
1gin
chee
1tahil
hoon
1 chee
g
1hoon
b g FHG
I
KJ
F
HG
I
KJ
F
HG
I
KJ
F
HG
I
KJ
F
HG
I
KJ
 
 
which yields 1.747  106 g or roughly 1.75 103 kg. 
 
30. To solve the problem, we note that the first derivative of the function with respect to 
time gives the rate. Setting the rate to zero gives the time at which an extreme value of 
the variable mass occurs; here that extreme value is a maximum. 
 
(a) Differentiating 
0.8( ) 5.00 3.00 20.00m t t t  
with respect to t gives 
 
0.24.00 3.00.
dm
t
dt
 
 
 
The water mass is the greatest when 
/ 0,dm dt 
 or at 
1/0.2(4.00 /3.00) 4.21s.t  
 
 
(b) At 
4.21s,t 
 the water mass is 
 
0.8( 4.21s) 5.00(4.21) 3.00(4.21) 20.00 23.2 g.m t     
 
 
(c) The rate of mass change at 
2.00 st 
 is 
 
0.2
2.00 s
2
g 1 kg 60 s
4.00(2.00) 3.00 g/s 0.48 g/s 0.48
s 1000 g 1 min
2.89 10 kg/min.
t
dm
dt



       
 
 
 
(d) Similarly, the rate of mass change at 
5.00 st 
 is 
 
0.2
2.00 s
3
g 1 kg 60 s
4.00(5.00) 3.00 g/s 0.101g/s 0.101
s 1000 g 1 min
6.05 10 kg/min.
t
dm
dt



         
  
 
 
31. The mass density of the candy is 
 
 
 
13 
 
4 3 4 3
3
0.0200 g
4.00 10 g/mm 4.00 10 kg/cm .
50.0 mm
m
V
        
 
If we neglect the volume of the empty spaces between the candies, then the total mass of 
the candies in the container when filled to height h is 
,M Ah
 where 
2(14.0 cm)(17.0 cm) 238 cmA 
 is the base area of the container that remains 
unchanged. Thus, the rate of mass change is given by 
 
4 3 2( ) (4.00 10 kg/cm )(238 cm )(0.250 cm/s)
0.0238 kg/s 1.43 kg/min.
dM d Ah dh
A
dt dt dt
     
 
 
 
32. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m 
and base area A = 20  12 = 240 m2) in addition to a rectangularbox (height h´ = 6.0 m 
and same base). Therefore, 
31 1800 m .
2 2
h
V hA h A h A
 
      
 
 
 
(a) Each dimension is reduced by a factor of 1/12, and we find 
 
Vdoll
3 3m m
F
HG
I
KJ 1800
1
12
10
3
c h . .
 
 
(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. 
Therefore, 
Vminiature
3m 6.0 10 m
F
HG
I
KJ  
1800
1
144
3
4 3c h .
 
 
33. THINK In this problem we are asked to differentiate between three types of tons: 
displacement ton, freight ton and register ton, all of which are units of volume. 
 
EXPRESS The three different tons are defined in terms of barrel bulk, with 
31 barrel bulk 0.1415 m 4.0155 U.S. bushels 
(using 
31 m 28.378 U.S. bushels
). 
Thus, in terms of U.S. bushels, we have 
 
4.0155 U.S. bushels
1 displacement ton (7 barrels bulk) 28.108 U.S. bushels
1 barrel bulk
 
   
 
 
4.0155 U.S. bushels
1 freight ton (8 barrels bulk) 32.124 U.S. bushels
1 barrel bulk
4.0155 U.S. bushels
1 register ton (20 barrels bulk) 80.31 U.S. bushels
1 barrel bulk
 
   
 
 
   
 
 
 
 CHAPTER 1 14 
ANALYZE (a) The difference between 73 “freight” tons and 73 “displacement” tons is 
 
73(freight tons displacement tons) 73(32.124 U.S. bushels 28.108 U.S. bushels)
293.168 U.S. bushels 293 U.S. bushels
V    
 
 
(b) Similarly, the difference between 73 “register” tons and 73 “displacement” tons is 
 
3
73(register tons displacement tons) 73(80.31 U.S. bushels 28.108 U.S. bushels)
3810.746 U.S. bushels 3.81 10 U.S. bushels
V    
  
 
LEARN With 
1 register ton 1 freight ton 1displacement ton, 
 we expect the difference 
found in (b) to be greater than that in (a). This is indeed the case. 
 
34. The customer expects a volume V1 = 20  7056 in
3
 and receives V2 = 20  5826 in.
3
, 
the difference being 
3
1 2 24600 in.V V V   
, or 
 
 
3
3
3
2.54cm 1L
24600 in. 403L
1 inch 1000 cm
V
   
     
  
 
 
where Appendix D has been used. 
 
35. The first two conversions are easy enough that a formal conversion is not especially 
called for, but in the interest of practice makes perfect we go ahead and proceed formally: 
 
(a) 
 
2 peck
11 tuffets = 11 tuffets 22 pecks
1 tuffet
 
 
 
. 
 
(b) 
 
0.50 Imperial bushel
11 tuffets = 11 tuffets 5.5 Imperial bushels
1 tuffet
 
 
 
. 
 
(c) 
 
36.3687 L
11 tuffets = 5.5 Imperial bushel 200 L
1 Imperial bushel
 
 
 
. 
 
36. Table 7 can be completed as follows: 
 
(a) It should be clear that the first column (under “wey”) is the reciprocal of the first 
row – so that 
9
10
 = 0.900, 
3
40
 = 7.50  102, and so forth. Thus, 1 pottle = 1.56  103 wey 
and 1 gill = 8.32  106 wey are the last two entries in the first column. 
 
(b) In the second column (under “chaldron”), clearly we have 1 chaldron = 1 chaldron 
(that is, the entries along the “diagonal” in the table must be 1’s). To find out how many 
 
 
15 
chaldron are equal to one bag, we note that 1 wey = 10/9 chaldron = 40/3 bag so that 
1
12
 
chaldron = 1 bag. Thus, the next entry in that second column is 
1
12
 = 8.33  102. 
Similarly, 1 pottle = 1.74  103 chaldron and 1 gill = 9.24  106 chaldron. 
 
(c) In the third column (under “bag”), we have 1 chaldron = 12.0 bag, 1 bag = 1 bag, 1 
pottle = 2.08  102 bag, and 1 gill = 1.11  104 bag. 
 
(d) In the fourth column (under “pottle”), we find 1 chaldron = 576 pottle, 1 bag = 48 
pottle, 1 pottle = 1 pottle, and 1 gill = 5.32  103 pottle. 
 
(e) In the last column (under “gill”), we obtain 1 chaldron = 1.08  105 gill, 1 bag = 9.02 
 103 gill, 1 pottle = 188 gill, and, of course, 1 gill = 1 gill. 
 
(f) Using the information from part (c), 1.5 chaldron = (1.5)(12.0) = 18.0 bag. And since 
each bag is 0.1091 m
3
 we conclude 1.5 chaldron = (18.0)(0.1091) = 1.96 m
3
. 
 
37. The volume of one unit is 1 cm
3
 = 1  106 m3, so the volume of a mole of them is 
6.02  1023 cm3 = 6.02  1017 m3. The cube root of this number gives the edge length: 
5 38.4 10 m
. This is equivalent to roughly 8  102 km. 
 
38. (a) Using the fact that the area A of a rectangle is (width) length), we find 
 
    
 
  
total
2
2
3.00acre 25.0perch 4.00perch
40 perch 4 perch
3.00 acre 100perch
1acre
580 perch .
A  
 
  
 

 
 
We multiply this by the perch
2
  rood conversion factor (1 rood/40 perch2) to obtain the 
answer: Atotal = 14.5 roods. 
 
(b) We convert our intermediate result in part (a): 
 
 
2
2 5 2
total
16.5ft
580 perch 1.58 10 ft .
1perch
A
 
   
 
 
 
Now, we use the feet  meters conversion given in Appendix D to obtain 
 
2
5 2 4 2
total
1m
1.58 10 ft 1.47 10 m .
3.281ft
A
 
    
 
 
 
 CHAPTER 1 16 
39. THINK This problem compares the U.K. gallon with U.S. gallon, two non-SI units 
for volume. The interpretation of the type of gallons, whether U.K. or U.S., affects the 
amount of gasoline one calculates for traveling a given distance. 
 
EXPRESS If the fuel consumption rate is 
R
 (in miles/gallon), then the amount of 
gasoline (in gallons) needed for a trip of distance d (in miles) would be 
 
 
(miles)
(gallon)
(miles/gallon)
d
V
R

 
 
Since the car was manufactured in U.K., the fuel consumption rate is calibrated based on 
U.K. gallon, and the correct interpretation should be “40 miles per U.K. gallon.” In U.K., 
one would think of gallon as U.K. gallon; however, in the U.S., the word “gallon” would 
naturally be interpreted as U.S. gallon. Note also that since 
1 U.K. gallon 4.5460900 L
 
and 
1 U.S. gallon 3.7854118 L
, the relationship between the two is 
 
1 U.S. gallon
1 U.K. gallon (4.5460900 L) 1.20095 U.S. gallons
3.7854118 L
 
  
 
 
 
ANALYZE (a) The amount of gasoline actually required is 
 
750 miles
18.75 U.K. gallons 18.8 U.K. gallons
40 miles/U.K. gallon
V    
 
 
This means that the driver mistakenly believes that the car should need 18.8 U.S. gallons. 
 
(b) Using the conversion factor found above, this is equivalent to 
 
 
1.20095 U.S. gallons
18.75 U.K. gallons 22.5 U.S. gallons
1 U.K. gallon
V
 
    
 
 
 
LEARN One U.K. gallon is greater than one U.S gallon by roughly a factor of 1.2 in 
volume. Therefore, 40 mi/U.K. gallon is less fuel-efficient than 40 mi/U.S. gallon. 
 
40. Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to 
kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the 
mass of one atom (1.0 u, but converted to kilograms), then the computation reduces to 
simply taking the reciprocal of the number given in Eq. 1-9 and rounding off 
appropriately. Thus, the answer is 6.0  1026. 
 
41. THINK This problem involves converting cord, a non-SI unit for volume, to SI unit. 
 
EXPRESS Using the (exact) conversion 1 in. = 2.54 cm = 0.0254 m for length, we have 
 
 
 
17 
 
0.0254 m
1 ft12 in (12 in.) 0.3048 m
1in
 
    
 
. 
Thus, 
3 3 31 ft (0.3048 m) 0.0283 m 
 for volume (these results also can be found in 
Appendix D). 
 
ANALYZE The volume of a cord of wood is 
3(8 ft) (4 ft) (4 ft) 128 ftV    
. Using 
the conversion factor found above, we obtain 
3
3 3 3
3
0.0283 m
1cord 128 ft (128 ft ) 3.625 m
1 ft
V
 
     
 
 
which implies that 
3 11 m cord 0.276 cord 0.3 cord
3.625
 
   
 
. 
 
LEARN The unwanted units ft
3
 all cancel out, as they should. In conversions, units obey 
the same algebraic rules as variables and numbers. 
 
42. (a) In atomic mass units, the mass of one molecule is (16 + 1 + 1)u = 18 u. Using Eq. 
1-9, we find 
 
27
261.6605402 10 kg18u = 18u 3.0 10 kg.
1u

    
 
 
 
(b) We divide the total mass by the mass of each molecule and obtain the (approximate) 
number of water molecules: 
21
46
26
1.4 10
5 10 .
3.0 10
N


  

 
 
43. A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3  106 mg/week. 
Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds 
are equivalent to one week. Thus, (2.3  106 mg/week)/(604800 s/week) = 3.8 mg/s. 
 
44. The volume of the water that fell is 
       
   
2
2 2
6 2
6 3
1000 m 0.0254 m
26 km 2.0 in. 26 km 2.0 in.
1 km 1 in.
26 10 m 0.0508 m
1.3 10 m .
V
   
     
   
 
 
 
 
We write the mass-per-unit-volume (density) of the water as: 
3 31 10 kg m .
m
V
  
 
The mass of the water that fell is therefore given by m = V: 
 
   3 3 6 3 91 10 kg m 1.3 10 m 1.3 10 kg.m     
 
 CHAPTER 1 18 
 
45. The number of seconds in a year is 3.156  107. This is listed in Appendix D and 
results from the product 
 
(365.25 day/y) (24 h/day) (60 min/h) (60 s/min). 
 
(a) The number of shakes in a second is 10
8
; therefore, there are indeed more shakes per 
second than there are seconds per year. 
 
(b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during 
which humans have existed is given by 
 
10
10
10
6
10
4  u - day,
 
 
which may also be expressed as 
10
86400
1
8 64
F
HG
I
KJ u - day
u - sec
u -day
u - sec.c h .
 
 
46. The volume removed in one year is 
V = (75 10 m ) (26 m) 2 10 m4 2 7 3  
, 
which we convert to cubic kilometers: 
V  
F
HG
I
KJ 2 10
1
0 0207
3
m
km
1000 m
km3 3c h . .
 
 
47. THINK This problem involves expressing the speed of light in astronomical units per 
minute. 
 
EXPRESS We first convert meters to astronomical units (AU), and seconds to minutes, 
using 
81000 m 1 km, 1 AU 1.50 10 km, 60 s 1 min.
 
 
ANALYZE Using the conversion factors above, the speed of light can be rewritten as 
 
 8
8
8
3.0 10 m 1 km AU 60 s
3.0 10 m/s 0.12 AU min.
s 1000 m 1.50 10 km min
c
       
          
      
 
 
LEARN When expressed the speed of light c in AU/min, we readily see that it takes 
about 8.3 (= 1/0.12) minutes for sunlight to reach the Earth (i.e., to travel a distance of 1 
AU). 
 
48. Since one atomic mass unit is 
241u 1.66 10 g 
(see Appendix D), the mass of one 
mole of atoms is about 
24 23(1.66 10 g)(6.02 10 ) 1g.m    
 On the other hand, the mass 
of one mole of atoms in the common Eastern mole is 
 
 
 
19 
 
75 g
10 g
7.5
m  
 
 
Therefore, in atomic mass units, the average mass of one atom in the common Eastern 
mole is 
23
23
10 g
1.66 10 g 10 u.
6.02 10A
m
N
    

 
 
49. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain 
 
1
1
197
1
3 88
2ken
m
m
m
2
2
2
2
 
.
. .
 
(b) Similarly, we find 
1
1
197
1
7 65
3
3ken
m
m
m
3 3
3
 
.
. .
 
 
(c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus, 
 
   
22 33.00 5.50 156 ken .r h   
 
(d) If we multiply this by the result of part (b), we determine the volume in cubic meters: 
(155.5)(7.65) = 1.19  103 m3. 
 
50. According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would 
be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 
39.4205 km. The difference is 5.95 km. 
 
51. (a) For the minimum (43 cm) case, 9 cubits converts as follows: 
 
 
0.43m
9cubits 9cubits 3.9m.
1cubit
 
  
 
 
And for the maximum (53 cm) case we have 
 
0.53m
9cubits 9cubits 4.8m.
1cubit
 
  
 
 
 
(b) Similarly, with 0.43 m  430 mm and 0.53 m  530 mm, we find 3.9  103 mm and 
4.8  103 mm, respectively. 
 
(c) We can convert length and diameter first and then compute the volume, or first 
compute the volume and then convert. We proceed using the latter approach (where d is 
diameter and 

 is length). 
 CHAPTER 1 20 
 
3
2 3 3 3
cylinder, min
0.43m
28 cubit 28 cubit 2.2 m .
4 1 cubit
V d
  
    
 
 
 
Similarly, with 0.43 m replaced by 0.53 m, we obtain Vcylinder, max = 4.2 m
3
. 
 
52. Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the 
ratio 25 wp/11 barn along with appropriate conversion factors: 
 
       
   
2
28 2
36
100 hide 110 acre 4047 m
1 wp 1acre1 hide
1 10 m
1 barn
25 wp
1 10 .
11 barn

 
 
 
53. THINK The objective of this problem is to convert the Earth-Sun distance (1 AU) to 
parsecs and light-years. 
 
EXPRESS To relate parsec (pc) to AU, we note that when  is measured in radians, it is 
equal to the arc length s divided by the radius R. For a very large radius circle and small 
value of , the arc may be approximated as the straight line-segment of length 1 AU. 
Thus, 
  6
1 arcmin 1 2 radian
1 arcsec 1 arcsec 4.85 10 rad
60 arcsec 60 arcmin 360
           
   
. 
Therefore, one parsec is 
 
5
6
1 AU
1 pc 2.06 10 AU
4.85 10
s
    
. 
 
Next, we relate AU to light-year (ly). Since a year is about 3.16  107 s, 
 
   7 121ly 186,000mi s 3.16 10 s 5.9 10 mi   
. 
 
ANALYZE (a) Since 
51 pc 2.06 10 AU 
, inverting the relation gives 
  65
1 pc
1 AU 1 AU 4.9 10 pc.
2.06 10 AU
    
 
 
 
(b) Given that 
61AU 92.9 10 mi
 and 
121 ly 5.9 10 mi 
, the two expressions 
together lead to 
6 6 5
12
1ly
1AU 92.9 10 mi (92.9 10 mi) 1.57 10 ly
5.9 10 mi
       
 
. 
 
 
 
21 
LEARN Our results can be further combined to give 
1pc 3.2 ly.
From the above 
expression, we readily see that it takes 
51.57 10 y
, or about 8.3 min, for Sunlight to 
travel a distance of 1 AU to reach the Earth. 
 
54. (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.
3
, and 1 in.
3
 = 1.639  
10
2
 L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460 
ft
2
/gal becomes 
2
2
2 2460 ft 1 m 1gal460 ft /gal 11.3 m L.
gal 3.28 ft 3.79 L
    
    
    
 
 
(b) Also, since 1 m
3
 is equivalent to 1000 L, our result from part (a) becomes 
 
2
2 4 1
3
11.3 m 1000L
11.3 m /L 1.13 10 m .
L 1 m
      
  
 
 
(c) The inverse of the original quantity is (460 ft
2
/gal)
1
 = 2.17  103 gal/ft2. 
 
(d) The answer in (c) represents the volume of the paint (in gallons) needed to cover a 
square foot of area. From this, we could also figure the paint thickness [it turns out to be 
about a tenth of a millimeter, as one sees by taking the reciprocal of the answer in part 
(b)]. 
 
55. (a) The receptacle is a volume of (40 cm)(40 cm)(30 cm) = 48000 cm
3
 = 48 L = 
(48)(16)/11.356 = 67.63 standard bottles, which is a little more than 3 nebuchadnezzars 
(the largest bottle indicated). The remainder, 7.63 standard bottles, is just a little less 
than 1 methuselah. Thus, the answer to part (a) is 3 nebuchadnezzars and 1 methuselah. 
 
(b) Since 1 methuselah.= 8 standard bottles, then the extra amount is 8  7.63 = 0.37 
standard bottle. 
 
(c) Using the conversion factor 16 standard bottles = 11.356 L, we have 
 
11.356 L
0.37 standard bottle (0.37 standard bottle) 0.26 L.
16 standard bottles
 
  
 
 
 
56. The mass of the pig is 3.108 slugs, or (3.108)(14.59) = 45.346 kg. Referring now to 
the corn, a U.S. bushel is 35.238 liters. Thus, a value of 1 for the corn-hog ratio would 
be equivalent to 35.238/45.346 = 0.7766 in the indicated metric units. Therefore, a value 
of 5.7 for the ratio corresponds to 5.7(0.777)  4.4 in the indicated metric units. 
 
57. Two jalapeño peppers have spiciness = 8000 SHU, and this amount multiplied by 400 
(the number of people) is 3.2 106 SHU, which is roughly ten times the SHU value for a 
 CHAPTER 1 22 
single habanero pepper. More precisely, 10.7 habanero peppers will provide that total 
required SHU value. 
 
58. In the simplest approach, we set up a ratio for the total increase in horizontal depth x 
(where x = 0.05 m is the increase in horizontal depth per step) 
 
 steps
4.57
0.05 m 1.2 m.
0.19
x N x
 
    
 
 
 
However, we can approach this more carefully by noting that if there are N = 4.57/.19  
24 rises then under normal circumstances we would expect N  1 = 23 runs (horizontal 
pieces) in that staircase. This would yield (23)(0.05 m) = 1.15 m, which - to two 
significant figures - agrees with our first result. 
 
59. The volume of the filled container is 24000 cm
3
 = 24 liters, which (using the 
conversion given in the problem) is equivalent to 50.7 pints (U.S). The expected number 
is therefore in the range from 1317 to 1927 Atlantic oysters. Instead, the number received 
is in the range from 406 to 609 Pacific oysters. This represents a shortage in the range of 
roughly 700 to 1500 oysters (the answer to the problem). Note that the minimum value 
in our answer corresponds to the minimum Atlantic minus the maximum Pacific, and the 
maximum value corresponds to the maximum Atlantic minus the minimum Pacific. 
 
60. (a) We reduce the stock amount to British teaspoons: 
 
1
6 2 2 24
breakfastcup = 2 8 2 2 = 64 teaspoons
1 teacup = 8 2 2 = 32 teaspoons
6 tablespoons = teaspoons
1 dessertspoon = 2 teaspoons
  
 
  
 
 
which totals to 122 British teaspoons, or 122 U.S. teaspoons since liquid measure is being 
used. Now with one U.S cup equal to 48 teaspoons, upon dividing 122/48  2.54, we find 
this amount corresponds to 2.5 U.S. cups plus a remainder of precisely 2 teaspoons. In 
other words, 
 
122 U.S. teaspoons = 2.5 U.S. cups + 2 U.S. teaspoons.
 
 
(b) For the nettle tops, one-half quart is still one-half quart. 
 
(c) For the rice, one British tablespoon is 4 British teaspoons which (since dry-goods 
measure is being used) corresponds to 2 U.S. teaspoons. 
 
(d) A British saltspoon is 
1
2
 British teaspoon which corresponds (since dry-goods 
measure is again being used) to 1 U.S. teaspoon. 
 
23 
 
 
Chapter 2 
 
 
1. The speed (assumed constant) is v = (90 km/h)(1000 m/km)  (3600 s/h) = 25 m/s. 
Thus, in 0.50 s, the car travels a distance d = vt = (25 m/s)(0.50 s)  13 m. 
 
2. (a) Using the fact that time = distance/velocity while the velocity is constant, we 
find 
avg 73.2 m 73.2 m
3.05 m1.22 m/s
73.2 m 73.2 m
1.74 m/s.v

 

 
 
(b) Using the fact that distance = vt while the velocity v is constant, we find 
 
vavg
 m / s)(60 s) m / s)(60 s)
 s
 m / s.


( . ( .
.
122 305
120
214
 
 
(c) The graphs are shown below (with meters and seconds understood). The first 
consists of two (solid) line segments, the first having a slope of 1.22 and the second 
having a slope of 3.05. The slope of the dashed line represents the average velocity (in 
both graphs). The second graph also consists of two (solid) line segments, having the 
same slopes as before — the main difference (compared to the first graph) being that 
the stage involving higher-speed motion lasts much longer. 
 
 
 
3. THINK This one-dimensional kinematics problem consists of two parts, and we 
are asked to solve for the average velocity and average speed of the car. 
 
EXPRESS Since the trip consists of two parts, let the displacements during first and 
second parts of the motion be x1 and x2, and the corresponding time intervals be t1 
and t2, respectively. Now, because the problem is one-dimensional and both 
displacements are in the same direction, the total displacement is simply x = x1 + 
x2, and the total time for the trip is t = t1 + t2. Using the definition of average 
velocity given in Eq. 2-2, we have 
1 2
avg
1 2
.
x xx
v
t t t
 
 
  
 
 
CHAPTER 2 
 
24 
 
To find the average speed, we note that during a time t if the velocity remains a 
positive constant, then the speed is equal to the magnitude of velocity, and the 
distance is equal to the magnitude of displacement, with 
| | .d x v t   
 
 
ANALYZE 
(a) During the first part of the motion, the displacement is x1 = 40 km and the time 
taken is 
t1
40
133 
(
.
 km)
(30 km / h)
 h.
 
Similarly, during the second part of the trip the displacement is x2 = 40 km and the 
time interval is 
t2
40
0 67 
(
.
 km)
(60 km / h)
 h.
 
 
The total displacement is x = x1 + x2 = 40 km + 40 km = 80 km, and the total time 
elapsed is t = t1 + t2 = 2.00 h. Consequently, the average velocity is 
 
avg
(80 km)
40 km/h.
(2.0 h)
x
v
t

  

 
 
(b) In this case, the average speed is the same as the magnitude of the average 
velocity: 
avg 40 km/h.s 
 
 
(c) The graph of the entire trip, shown below, consists of two contiguous line 
segments, the first having a slope of 30 km/h and connecting the origin to (t1, x1) = 
(1.33 h, 40 km)
 and the second having a slope of 60 km/h and connecting (t1, x1) 
to (t, x) = (2.00 h, 80 km). 
 
 
 
From the graphical point of view, the slope of the dashed line drawn from the origin 
to (t, x) represents the average velocity. 
 
LEARN The average velocity is a vector quantity that depends only on the net 
displacement (also a vector) between the starting and ending points. 
 
4. Average speed, as opposed to average velocity, relates to the total distance, as 
opposed to the net displacement. The distance D up the hill is, of course, the same as 
the distance down the hill, and since the speed is constant (during each stage of the 
 
 
25 
 
motion) we have speed = D/t. Thus, theaverage speed is 
 
D D
t t
D
D
v
D
v
up down
up down
up down




2 
 
which, after canceling D and plugging in vup = 40 km/h and vdown = 60 km/h, yields 48 
km/h for the average speed. 
 
5. THINK In this one-dimensional kinematics problem, we’re given the position 
function x(t), and asked to calculate the position and velocity of the object at a later 
time. 
 
EXPRESS The position function is given as x(t) = (3 m/s)t – (4 m/s2)t2 + (1 m/s3)t3. 
The position of the object at some instant t0 is simply given by x(t0). For the time 
interval 
1 2t t t 
, the displacement is 
2 1( ) ( )x x t x t  
. Similarly, using Eq. 2-2, 
the average velocity for this time interval is 
 
2 1
avg
2 1
( ) ( )
.
x t x tx
v
t t t

 
 
 
 
ANALYZE (a) Plugging in t = 1 s into x(t) yields 
 
x(1 s) = (3 m/s)(1 s) – (4 m/s2)(1 s)2 + (1 m/s3)(1 s)3 = 0. 
 
(b) With t = 2 s we get x(2 s) = (3 m/s)(2 s) – (4 m/s2) (2 s)2 + (1 m/s3)(2 s)3 = –2 m. 
 
(c) With t = 3 s we have x (3 s) = (3 m/s)(3 s) – (4 m/s2) (3 s)2 + (1 m/s3)(3 s)3 = 0 m. 
 
(d) Similarly, plugging in t = 4 s gives 
 
x(4 s) = (3 m/s)(4 s) – (4 m/s2)(4 s)2 + (1 m/s3) (4 s)3 = 12 m. 
 
(e) The position at t = 0 is x = 0. Thus, the displacement between t = 0 and t = 4 s is 
(4 s) (0) 12 m 0 12 m.x x x     
 
 
(f) The position at t = 2 s is subtracted from the position at t = 4 s to give the 
displacement: 
(4 s) (2 s) 12 m ( 2 m) 14 mx x x      
. Thus, the average velocity 
is 
avg
14 m
7 m/s.
2 s
x
v
t

  

 
 
(g) The position of the object for the interval 0  t  4 is plotted below. The straight 
line drawn from the point at (t, x) = (2 s, –2 m) to (4 s, 12 m) would represent the 
average velocity, answer for part (f). 
 
CHAPTER 2 
 
26 
 
 
 
LEARN Our graphical representation illustrates once again that the average velocity 
for a time interval depends only on the net displacement between the starting and 
ending points. 
 
6. Huber’s speed is 
v0 = (200 m)/(6.509 s) =30.72 m/s = 110.6 km/h, 
 
where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat 
Huber by 19.0 km/h, his speed is v1 = (110.6 km/h + 19.0 km/h) = 129.6 km/h, or 36 
m/s (1 km/h = 0.2778 m/s). Thus, using Eq. 2-2, the time through a distance of 200 m 
for Whittingham is 
1
200 m
5.554 s.
36 m/s
x
t
v

   
 
 
7. Recognizing that the gap between the trains is closing at a constant rate of 60 km/h, 
the total time that elapses before they crash is t = (60 km)/(60 km/h) = 1.0 h. During 
this time, the bird travels a distance of x = vt = (60 km/h)(1.0 h) = 60 km. 
 
8. The amount of time it takes for each person to move a distance L with speed 
sv
 is 
/ st L v 
. With each additional person, the depth increases by one body depth d 
 
(a) The rate of increase of the layer of people is 
 
 
(0.25 m)(3.50 m/s)
0.50 m/s
/ 1.75 m
s
s
dvd d
R
t L v L
    

 
(b) The amount of time required to reach a depth of 
5.0 mD 
is 
 
5.0 m
10 s
0.50 m/s
D
t
R
  
 
 
9. Converting to seconds, the running times are t1 = 147.95 s and t2 = 148.15 s, 
respectively. If the runners were equally fast, then 
 
1 2
avg avg1 2
1 2
 .
L L
s s
t t
  
 
From this we obtain 
 
 
27 
 
2
2 1 1 1 1
1
148.15
1 1 0.00135 1.4 m
147.95
t
L L L L L
t
   
         
  
 
 
where we set L1  1000 m in the last step. Thus, if L1 and L2 are no different than 
about 1.4 m, then runner 1 is indeed faster than runner 2. However, if L1 is shorter 
than L2 by more than 1.4 m, then runner 2 would actually be faster. 
 
10. Let 
wv
be the speed of the wind and 
cv
 be the speed of the car. 
 
(a) Suppose during time interval 
1t
, the car moves in the same direction as the wind. 
Then the effective speed of the car is given by
,1eff c wv v v 
, and the distance traveled 
is 
,1 1 1( )eff c wd v t v v t  
. On the other hand, for the return trip during time interval t2, 
the car moves in the opposite direction of the wind and the effective speed would be 
,2eff c wv v v 
. The distance traveled is 
,2 2 2( )eff c wd v t v v t  
. The two expressions 
can be rewritten as 
1 2
andc w c w
d d
v v v v
t t
   
 
Adding the two equations and dividing by two, we obtain 
1 2
1
2
c
d d
v
t t
 
  
 
. Thus, 
method 1 gives the car’s speed
cv
a in windless situation. 
 
(b) If method 2 is used, the result would be 
 
2
2 2
1 2 1 2
2 2
1
( ) / 2
c w w
c c
c c
c w c w
v v vd d d
v v
d dt t t t v v
v v v v
  
        
      
 
. 
The fractional difference is 
2
2 4(0.0240) 5.76 10c c w
c c
v v v
v v
      
 
. 
11. The values used in the problem statement make it easy to see that the first part of 
the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour. 
Expressed in decimal form, the time left is 1.25 hour, and the distance that remains is 
160 km. Thus, a speed v = (160 km)/(1.25 h) = 128 km/h is needed. 
 
12. (a) Let the fast and the slow cars be separated by a distance d at t = 0. If during the 
time interval 
/ (12.0 m) /(5.0 m/s) 2.40 sst L v  
in which the slow car has moved 
a distance of 
12.0 mL 
, the fast car moves a distance of 
vt d L 
 to join the line 
of slow cars, then the shock wave would remain stationary. The condition implies a 
separation of 
 
(25 m/s)(2.4 s) 12.0 m 48.0 m.d vt L    
 
 
(b) Let the initial separation at 
0t 
 be 
96.0 m.d 
 At a later time t, the slow and 
CHAPTER 2 
 
28 
 
the fast cars have traveled 
sx v t
 and the fast car joins the line by moving a distance 
d x
. From 
 
,
s
x d x
t
v v

 
 
we get 
5.00 m/s
(96.0 m) 24.0 m,
25.0 m/s 5.00 m/s
s
s
v
x d
v v
  
 
 
 
which in turn gives 
(24.0 m) /(5.00 m/s) 4.80 s.t  
 Since the rear of the slow-car 
pack has moved a distance of 
24.0 m 12.0 m 12.0 mx x L     
downstream, the 
speed of the rear of the slow-car pack, or equivalently, the speed of the shock wave, is 
 
 
shock
12.0 m
2.50 m/s.
4.80 s
x
v
t

  
 
(c) Since 
x L
, the direction of the shock wave is downstream. 
 
13. (a) Denoting the travel time and distance from San Antonio to Houston as T and D, 
respectively, the average speed is 
 
avg1
(55 km/h)( /2) (90 km/h)( / 2)
72.5 km/h
D T T
s
T T

  
 
 
which should be rounded to 73 km/h. 
 
(b) Using the fact that time = distance/speed while the speed is constant, we find 
 
avg2 / 2 / 2
55 km/h 90 km/h
68.3 km/h
D D
D D
s
T
  

 
 
which should be rounded to 68 km/h. 
 
(c) The total distance traveled (2D) must not be confused with the net displacement 
(zero). We obtain for the two-way trip 
avg
72.5 km/h 68.3 km/h
2
70 km/h.
D D
D
s  

 
 
(d) Since the net displacement vanishes, the average velocity for the trip in its entirety 
is zero. 
 
(e) In asking for a sketch, the problem is allowing the student to arbitrarily set the 
distance D (the intent is not to make the student go to an atlas to look it up); the 
student can just as easily arbitrarily set T instead of D, as will be clear in the following 
discussion. We brieflydescribe the graph (with kilometers-per-hour understood for 
the slopes): two contiguous line segments, the first having a slope of 55 and 
connecting the origin to (t1, x1) = (T/2, 55T/2) and the second having a slope of 90 and 
connecting (t1, x1) to (T, D) where D = (55 + 90)T/2. The average velocity, from the 
 
 
29 
 
graphical point of view, is the slope of a line drawn from the origin to (T, D). The 
graph (not drawn to scale) is depicted below: 
 
 
 
14. Using the general property 
d
dx
bx b bxexp( ) exp( )
, we write 
 
v
dx
dt
d t
dt
e t
de
dt
t
t
 
F
HG
I
KJ   
F
HG
I
KJ

( )
( )
19
19 .
 
 
If a concern develops about the appearance of an argument of the exponential (–t) 
apparently having units, then an explicit factor of 1/T where T = 1 second can be 
inserted and carried through the computation (which does not change our answer). 
The result of this differentiation is 
v t e t  16 1( )
 
 
with t and v in SI units (s and m/s, respectively). We see that this function is zero 
when t = 1 s. Now that we know when it stops, we find out where it stops by 
plugging our result t = 1 into the given function x = 16te
–t
 with x in meters. Therefore, 
we find x = 5.9 m. 
 
15. We use Eq. 2-4 to solve the problem. 
 
(a) The velocity of the particle is 
 
v
dx
dt
d
dt
t t t       ( ) .4 12 3 12 62
 
 
Thus, at t = 1 s, the velocity is v = (–12 + (6)(1)) = –6 m/s. 
 
(b) Since v  0, it is moving in the –x direction at t = 1 s. 
 
(c) At t = 1 s, the speed is |v| = 6 m/s. 
 
(d) For 0  t  2 s, |v| decreases until it vanishes. For 2  t  3 s, |v| increases from 
zero to the value it had in part (c). Then, |v| is larger than that value for t  3 s. 
 
(e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to 
positive (note that as t  + , we have v  + ). One can check that v = 0 when 
2 s.t 
 
CHAPTER 2 
 
30 
 
 
(f) No. In fact, from v = –12 + 6t, we know that v  0 for t  2 s. 
 
16. We use the functional notation x(t), v(t), and a(t) in this solution, where the latter 
two quantities are obtained by differentiation: 
 
v t
dx t
dt
t a t
dv t
dt
b g b g b g b g     12 12and
 
 
with SI units understood. 
 
(a) From v(t) = 0 we find it is (momentarily) at rest at t = 0. 
 
(b) We obtain x(0) = 4.0 m. 
 
(c) and (d) Requiring x(t) = 0 in the expression x(t) = 4.0 – 6.0t2 leads to t = 0.82 s 
for the times when the particle can be found passing through the origin. 
 
(e) We show both the asked-for graph (on the left) as well as the “shifted” graph that 
is relevant to part (f). In both cases, the time axis is given by –3  t  3 (SI units 
understood). 
 
 
(f) We arrived at the graph on the right (shown above) by adding 20t to the x(t) 
expression. 
 
(g) Examining where the slopes of the graphs become zero, it is clear that the shift 
causes the v = 0 point to correspond to a larger value of x (the top of the second curve 
shown in part (e) is higher than that of the first). 
 
17. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and 
work with distances in centimeters and times in seconds. 
 
(a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x2 = 
21.75 cm and x3 = 50.25 cm, respectively. The average velocity during the time 
interval 2.00  t  3.00 s is 
v
x
t
avg 
 cm cm
 s s
 




50 25 2175
300 2 00
. .
. .
 
which yields vavg = 28.5 cm/s. 
 
(b) The instantaneous velocity is 
v tdx
dt
  4 5 2.
, which, at time t = 2.00 s, yields v = 
(4.5)(2.00)
2
 = 18.0 cm/s. 
 
 
31 
 
 
(c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)
2
 = 40.5 cm/s. 
 
(d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)
2
 = 28.1 cm/s. 
 
(e) Let tm stand for the moment when the particle is midway between x2 and x3 (that is, 
when the particle is at xm = (x2 + x3)/2 = 36 cm). Therefore, 
 
x t tm m m   9 75 15 2596
3. . . 
 
 
in seconds. Thus, the instantaneous speed at this time is v = 4.5(2.596)
2
 = 30.3 cm/s. 
 
(f) The answer to part (a) is given by the slope of the straight line between t = 2 and t 
= 3 in this x-vs-t plot. The answers to parts (b), (c), (d), and (e) correspond to the 
slopes of tangent lines (not shown but easily imagined) to the curve at the appropriate 
points. 
 
 
18. (a) Taking derivatives of x(t) = 12t
2
 – 2t3 we obtain the velocity and the 
acceleration functions: 
v(t) = 24t – 6t2 and a(t) = 24 – 12t 
 
with length in meters and time in seconds. Plugging in the value t = 3 yields 
(3) 54 mx 
. 
 
(b) Similarly, plugging in the value t = 3 yields v(3) = 18 m/s. 
 
(c) For t = 3, a(3) = –12 m/s2. 
 
(d) At the maximum x, we must have v = 0; eliminating the t = 0 root, the velocity 
equation reveals t = 24/6 = 4 s for the time of maximum x. Plugging t = 4 into the 
equation for x leads to x = 64 m for the largest x value reached by the particle. 
 
(e) From (d), we see that the x reaches its maximum at t = 4.0 s. 
 
(f) A maximum v requires a = 0, which occurs when t = 24/12 = 2.0 s. This, inserted 
into the velocity equation, gives vmax = 24 m/s. 
 
(g) From (f), we see that the maximum of v occurs at t = 24/12 = 2.0 s. 
 
(h) In part (e), the particle was (momentarily) motionless at t = 4 s. The acceleration at 
that time is readily found to be 24 – 12(4) = –24 m/s2. 
CHAPTER 2 
 
32 
 
 
(i) The average velocity is defined by Eq. 2-2, so we see that the values of x at t = 0 
and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part (a)). 
Thus, 
vavg = 54 0
3 0


 = 18 m/s. 
 
19. THINK In this one-dimensional kinematics problem, we’re given the speed of a 
particle at two instants and asked to calculate its average acceleration. 
 
EXPRESS We represent the initial direction of motion as the +x direction. The 
average acceleration over a time interval 
1 2t t t 
is given by Eq. 2-7: 
 
2 1
avg
2 1
( ) ( )
.
v t v tv
a
t t t

 
 
 
 
ANALYZE Let v1 = +18 m/s at 
1 0t 
and v2 = –30 m/s at t2 = 2.4 s. Using Eq. 2-7 
we find 
22 1
avg
2 1
( ) ( ) ( 30 m/s) ( 1m/s)
20 m/s
2.4 s 0
v t v t
a
t t
   
   
 
. 
 
LEARN The average acceleration has magnitude 20 m/s
2
 and is in the opposite 
direction to the particle’s initial velocity. This makes sense because the velocity of the 
particle is decreasing over the time interval. With 
1 0t 
, the velocity of the particle 
as a function of time can be written as 
 
2
0 (18 m/s) (20 m/s )v v at t   
. 
 
20. We use the functional notation x(t), v(t) and a(t) and find the latter two quantities 
by differentiating: 
v t
dx t
t
t a t
dv t
dt
tb g b g b g b g      15 20 302 and
 
 
with SI units understood. These expressions are used in the parts that follow. 
 
(a) From 
0 15 202  t
, we see that the only positive value of t for which the 
particle is (momentarily) stopped is 
t  20 15 12/ . s
. 
 
(b) From 0 = – 30t, we find a(0) = 0 (that is, it vanishes at t = 0). 
 
(c) It is clear that a(t) = – 30t is negative for t > 0. 
 
(d) The acceleration a(t) = – 30t is positive for t < 0. 
 
(e) The graphs are shown below. SI units are understood. 
 
 
 
 
33 
 
 
 
21. We use Eq. 2-2 (average velocity) andEq. 2-7 (average acceleration). Regarding 
our coordinate choices, the initial position of the man is taken as the origin and his 
direction of motion during 5 min  t  10 min is taken to be the positive x direction. 
We also use the fact that 
 x v t '
 when the velocity is constant during a time 
interval 
t '
. 
 
(a) The entire interval considered is t = 8 – 2 = 6 min, which is equivalent to 360 s, 
whereas the sub-interval in which he is moving is only 
8 5 3min 180 s.t'    
 
His position at t = 2 min is x = 0 and his position at t = 8 min is 
x v t  
 
(2.2)(180) 396 m
. Therefore, 
vavg
 m
 s
 m / s


396 0
360
110. .
 
 
(b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, 
keeping the answer to 3 significant figures, 
 
aavg
2 m / s
 s
 m / s


2 2 0
360
0 00611
.
. .
 
 
(c) Now, the entire interval considered is t = 9 – 3 = 6 min (360 s again), whereas the 
sub-interval in which he is moving is 
9 5 4min 240 st    
). His position at 
3 mint 
is x = 0 and his position at t = 9 min is
(2.2)(240) 528 mx v t   
. 
Therefore, 
vavg
 m
 s
m / s.


528 0
360
147.
 
 
(d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. 
Consequently, aavg = 2.2/360 = 0.00611 m/s
2
 just as in part (b). 
 
(e) The horizontal line near the bottom of this 
x-vs-t graph represents the man standing at x = 0 
for 0  t < 300 s and the linearly rising line for 
300  t  600 s represents his constant-velocity 
motion. The lines represent the answers to part (a) 
and (c) in the sense that their slopes yield those 
results. 
 
The graph of v-vs-t is not shown here, but would 
consist of two horizontal “steps” (one at v = 0 for 
0  t < 300 s and the next at v = 2.2 m/s for 300  
CHAPTER 2 
 
34 
 
t  600 s). The indications of the average accelerations found in parts (b) and (d) 
would be dotted lines connecting the “steps” at the appropriate t values (the slopes of 
the dotted lines representing the values of aavg). 
 
22. In this solution, we make use of the notation x(t) for the value of x at a particular t. 
The notations v(t) and a(t) have similar meanings. 
 
(a) Since the unit of ct
2
 is that of length, the unit of c must be that of length/time
2
, or 
m/s
2
 in the SI system. 
 
(b) Since bt
3
 has a unit of length, b must have a unit of length/time
3
, or m/s
3
. 
 
(c) When the particle reaches its maximum (or its minimum) coordinate its velocity is 
zero. Since the velocity is given by v = dx/dt = 2ct – 3bt2, v = 0 occurs for t = 0 and 
for 
t
c
b
  
2
3
2 30
3 2 0
10
( . )
( . )
.
 m / s
 m / s
 s .
2
3
 
 
For t = 0, x = x0 = 0 and for t = 1.0 s, x = 1.0 m > x0. Since we seek the maximum, we 
reject the first root (t = 0) and accept the second (t = 1s). 
 
(d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and 
goes back to 
x( ( . )( . ( . )( .4 30 4 0 2 0 4 0 802 s) m / s s) m / s s) m .2 3 3   
 
 
The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m. 
 
(e) Its displacement is x = x2 – x1, where x1 = 0 and x2 = –80 m. Thus, 80 mx  . 
 
The velocity is given by v = 2ct – 3bt2 = (6.0 m/s2)t – (6.0 m/s3)t2. 
 
(f) Plugging in t = 1 s, we obtain 
 
2 3 2(1 s) (6.0 m/s )(1.0 s) (6.0 m/s )(1.0 s) 0.v   
 
 
(g) Similarly, 
2 3 2(2 s) (6.0 m/s )(2.0 s) (6.0 m/s )(2.0 s) 12m/s .v    
 
 
(h) 
2 3 2(3 s) (6.0 m/s )(3.0 s) (6.0 m/s )(3.0 s) 36 m/s .v    
 
 
(i) 
2 3 2(4 s) (6.0 m/s )(4.0 s) (6.0 m/s )(4.0 s) 72 m/s .v    
 
 
The acceleration is given by a = dv/dt = 2c – 6b = 6.0 m/s2 – (12.0 m/s3)t. 
 
(j) Plugging in t = 1 s, we obtain 
2 3 2(1 s) 6.0 m/s (12.0 m/s )(1.0 s) 6.0 m/s .a    
 
 
(k) 
2 3 2(2 s) 6.0 m/s (12.0 m/s )(2.0 s) 18 m/s .a    
 
 
 
 
35 
 
(l) 
2 3 2(3 s) 6.0 m/s (12.0 m/s )(3.0 s) 30 m/s .a    
 
 
(m) 
2 3 2(4 s) 6.0 m/s (12.0 m/s )(4.0 s) 42 m/s .a    
 
 
23. THINK The electron undergoes a constant acceleration. Given the final speed of 
the electron and the distance it has traveled, we can calculate its acceleration. 
 
EXPRESS Since the problem involves constant acceleration, the motion of the 
electron can be readily analyzed using the equations given in Table 2-1: 
 
0
2
0 0
2 2
0 0
(2 11)
1
(2 15)
2
2 ( ) (2 16)
v v at
x x v t at
v v a x x
  
   
   
 
 
The acceleration can be found by solving Eq. 2-16. 
 
ANALYZE With 
5
0 1.50 10 m/sv  
, 
65.70 10 m/sv  
, x0 = 0 and x = 0.010 m, we 
find the average acceleration to be 
 
 2 2 6 2 5 2
15 20 (5.7 10 m/s) (1.5 10 m/s) 1.62 10 m/s .
2 2(0.010 m)
v v
a
x
   
   
 
 
LEARN It is always a good idea to apply other equations in Table 2-1 not used for 
solving the problem as a consistency check. For example, since we now know the 
value of the acceleration, using Eq. 2-11, the time it takes for the electron to reach its 
final speed would be 
6 5
90
15 2
5.70 10 m/s 1.5 10 m/s
3.426 10 s
1.62 10 m/s
v v
t
a
      

 
 
Substituting the value of t into Eq. 2-15, the distance the electron travels is 
 
2 5 9 15 2 9 2
0 0
1 1
0 (1.5 10 m/s)(3.426 10 s) (1.62 10 m/s )(3.426 10 s)
2 2
0.01 m
x x v t at           

 
This is what was given in the problem statement. So we know the problem has been 
solved correctly. 
 
24. In this problem we are given the initial and final speeds, and the displacement, and 
are asked to find the acceleration. We use the constant-acceleration equation given in 
Eq. 2-16, v
2
 = v
2
0 + 2a(x – x0). 
 
(a) Given that 
0 0v 
, 
1.6 m/s,v 
 and 
5.0 m,x  
 the acceleration of the spores 
during the launch is 
CHAPTER 2 
 
36 
 
2 2 2
5 2 40
6
(1.6 m/s)
2.56 10 m/s 2.6 10
2 2(5.0 10 m)
v v
a g
x 

     

 
 
(b) During the speed-reduction stage, the acceleration is 
 
2 2 2
3 2 20
3
0 (1.6 m/s)
1.28 10 m/s 1.3 10
2 2(1.0 10 m)
v v
a g
x 
 
       

 
 
The negative sign means that the spores are decelerating. 
 
25. We separate the motion into two parts, and take the direction of motion to be 
positive. In part 1, the vehicle accelerates from rest to its highest speed; we are given 
v0 = 0; v = 20 m/s and a = 2.0 m/s
2
. In part 2, the vehicle decelerates from its highest 
speed to a halt; we are given v0 = 20 m/s; v = 0 and a = –1.0 m/s
2
 (negative because 
the acceleration vector points opposite to the direction of motion). 
 
(a) From Table 2-1, we find t1 (the duration of part 1) from v = v0 + at. In this way, 
120 0 2.0t 
yields t1 = 10 s. We obtain the duration t2 of part 2 from the same 
equation. Thus, 0 = 20 + (–1.0)t2 leads to t2 = 20 s, and the total is t = t1 + t2 = 30 s. 
 
(b) For part 1, taking x0 = 0, we use the equation v
2
 = v
2
0 + 2a(x – x0) from Table 2-1 
and find 
 2 2 2 2
0
2
(20 m/s) (0)
100 m
2 2(2.0 m/s )
v v
x
a
 
  
. 
 
This position is then the initial position for part 2, so that when the same equation is 
used in part 2 we obtain 
2 2 2 2
0
2
(0) (20 m/s)
100 m
2 2( 1.0 m/s )
v v
x
a
 
  

. 
 
Thus, the final position is x = 300 m. That this is also the total distance traveled 
shouldbe evident (the vehicle did not "backtrack" or reverse its direction of motion). 
 
26. The constant-acceleration condition permits the use of Table 2-1. 
 
(a) Setting v = 0 and x0 = 0 in 
2 2
0 02 ( )v v a x x  
, we find 
 
2 6 2
0
14
1 1 (5.00 10 )
0.100 m .
2 2 1.25 10
v
x
a

    
 
 
 
Since the muon is slowing, the initial velocity and the acceleration must have opposite 
signs. 
 
(b) Below are the time plots of the position x and velocity v of the muon from the 
moment it enters the field to the time it stops. The computation in part (a) made no 
reference to t, so that other equations from Table 2-1 (such as 
v v at 0 and
 
 
 
37 
 
x v t at 0
1
2
2
) are used in making these plots. 
 
 
 
27. We use v = v0 + at, with t = 0 as the instant when the velocity equals +9.6 m/s. 
 
(a) Since we wish to calculate the velocity for a time before t = 0, we set t = –2.5 s. 
Thus, Eq. 2-11 gives 
v    ( . . ( . .9 6 32 2 5 16 m/ s) m/ s s) m/ s.2c h
 
 
(b) Now, t = +2.5 s and we find
v   ( . . ( .9 6 32 2 5 18 m/ s) m/ s s) m/ s.2c h
 
 
28. We take +x in the direction of motion, so v0 = +24.6 m/s and a = – 4.92 m/s
2
. We 
also take x0 = 0. 
 
(a) The time to come to a halt is found using Eq. 2-11: 
 
0 2
24.6 m/s
0 5.00 s
4.92 m/s
.v at t

    
 
 
(b) Although several of the equations in Table 2-1 will yield the result, we choose Eq. 
2-16 (since it does not depend on our answer to part (a)). 
 
 
2
2
0 2
(24.6 m/s)
0 2 61.5 m
2 4.92 m/s
.v ax x

     
 
 
(c) Using these results, we plot 
21
0 2
v t at
 (the x graph, shown next, on the left) and 
v0 + at (the v graph, on the right) over 0  t  5 s, with SI units understood. 
 
 
 
CHAPTER 2 
 
38 
 
29. We assume the periods of acceleration (duration t1) and deceleration (duration t2) 
are periods of constant a so that Table 2-1 can be used. Taking the direction of motion 
to be +x then a1 = +1.22 m/s
2
 and a2 = –1.22 m/s
2
. We use SI units so the velocity at t 
= t1 is v = 305/60 = 5.08 m/s. 
 
(a) We denote x as the distance moved during t1, and use Eq. 2-16: 
 
2
2 2
0 1 2
(5.08 m/s)
2 
2(1.22 m/s )
v v a x x     
10.59 m 10.6 m. 
 
(b) Using Eq. 2-11, we have 
0
1 2
1
5.08 m/s
4.17 s.
1.22 m/s
v v
t
a

  
 
 
The deceleration time t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances 
traveled during t1 and t2 are the same so that they total to 2(10.59 m) = 21.18 m. This 
implies that for a distance of 190 m – 21.18 m = 168.82 m, the elevator is traveling at 
constant velocity. This time of constant velocity motion is 
 
t3
16882
508
3321 
.
.
.
 m
 m / s
 s.
 
 
Therefore, the total time is 8.33 s + 33.21 s  41.5 s. 
 
30. We choose the positive direction to be that of the initial velocity of the car 
(implying that a < 0 since it is slowing down). We assume the acceleration is constant 
and use Table 2-1. 
 
(a) Substituting v0 = 137 km/h = 38.1 m/s, v = 90 km/h = 25 m/s, and a = –5.2 m/s
2
 
into v = v0 + at, we obtain 
 
t 



25 38
52
2 5
2
m / s m / s
m / s
s
.
. .
 
 
(b) We take the car to be at x = 0 when the brakes 
are applied (at time t = 0). Thus, the coordinate of 
the car as a function of time is given by 
 
   2 2
1
38 m/s 5.2 m/s
2
x t t  
 
 
in SI units. This function is plotted from t = 0 to t 
= 2.5 s on the graph to the right. We have not 
shown the v-vs-t graph here; it is a descending 
straight line from v0 to v. 
 
31. THINK The rocket ship undergoes a constant acceleration from rest, and we want 
to know the time elapsed and the distance traveled when the rocket reaches a certain 
speed. 
 
 
39 
 
 
EXPRESS Since the problem involves constant acceleration, the motion of the rocket 
can be readily analyzed using the equations in Table 2-1: 
0
2
0 0
2 2
0 0
(2 11)
1
(2 15)
2
2 ( ) (2 16)
v v at
x x v t at
v v a x x
  
   
   
 
 
ANALYZE (a) Given that
29.8 m/sa 
, 
0 0v 
 and 
70.1 3.0 10 m/sv c  
, we can 
solve 
0v v at 
 for the time: 
7
60
2
3.0 10 m/s 0
3.1 10 s
9.8 m/s
v v
t
a
  
   
 
 
which is about 1.2 months. So it takes 1.2 months for the rocket to reach a speed of 
0.1c starting from rest with a constant acceleration of 9.8 m/s
2
. 
 
(b) To calculate the distance traveled during this time interval, we evaluate 
x x v t at  0 0
1
2
2 ,
 with x0 = 0 and 
0 0v
. The result is 
 2 6 2 13
1
9.8 m/s (3.1 10 s) 4.6 10 m.
2
x    
 
 
LEARN In solving parts (a) and (b), we did not use Eq. (2-16):
2 2
0 02 ( )v v a x x  
. 
This equation can be used to check our answers. The final velocity based on this 
equation is 
 
2 2 13 7
0 02 ( ) 0 2(9.8 m/s )(4.6 10 m 0) 3.0 10 m/sv v a x x        
, 
 
which is what was given in the problem statement. So we know the problems have 
been solved correctly. 
 
32. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7). 
 
a
v
t
 
F
HG
I
KJ



1020
1000
3600
14
202 4 2
km / h
m / km
s / h
s
m / s
b g
.
. . 
 
In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s
2
 
as follows: 
2
2
202.4 m/s
21 .
9.8 m/s
a g g
 
  
 
 
 
33. THINK The car undergoes a constant negative acceleration to avoid impacting a 
barrier. Given its initial speed, we want to know the distance it has traveled and the 
time elapsed prior to the impact. 
CHAPTER 2 
 
40 
 
 
EXPRESS Since the problem involves constant acceleration, the motion of the car 
can be readily analyzed using the equations in Table 2-1: 
 
0
2
0 0
2 2
0 0
(2 11)
1
(2 15)
2
2 ( ) (2 16)
v v at
x x v t at
v v a x x
  
   
   
 
 
We take x0 = 0 and v0 = 56.0 km/h = 15.55 m/s to be the initial position and speed of 
the car. Solving Eq. 2-15 with t = 2.00 s gives the acceleration a. Once a is known, the 
speed of the car upon impact can be found by using Eq. 2-11. 
 
ANALYZE (a) Using Eq. 2-15, we find the acceleration to be 
 
  20
2 2
2 (24.0 m) (15.55 m/s)(2.00 s)2( )
3.56m/s ,
(2.00 s)
x v t
a
t

   
 
 
or 
2| | 3.56 m/sa 
. The negative sign indicates that the acceleration is opposite to the 
direction of motion of the car; the car is slowing down. 
 
(b) The speed of the car at the instant of impact is 
 
2
0 15.55 m/s ( 3.56 m/s )(2.00 s) 8.43 m/sv v at     
 
 
which can also be converted to
30.3 km/h.
 
 
LEARN In solving parts (a) and (b), we did not use Eq. 1-16. This equation can be 
used as a consistency check. The final velocity based on this equation is 
 
2 2 2
0 02 ( ) (15.55 m/s) 2( 3.56 m/s )(24 m 0) 8.43 m/sv v a x x       
, 
 
which is what was calculated in (b). This indicates that the problems have been solved 
correctly. 
 
34. Let d be the 220 m distance between the cars at t = 0, and v1 be the 20 km/h = 50/9 
m/s speed (corresponding to a passing point of x1 = 44.5 m) and v2 be the 40 km/h 
=100/9 m/s speed (corresponding to a passing point of x2 = 76.6 m) of the red car. 
We have two equations (based on Eq. 2-17): 
 
d – x1 = vo t1 + 
1
2
 a t1
2