Serway, raymond a physics for scientists and engineers (6e) solutions

1
CHAPTER OUTLINE
1.1 Standards of Length, Mass,
and Time
1.2 Matter and Model-Building
1.3 Density and Atomic Mass
1.4 Dimensional Analysis
1.5 Conversion of Units
1.6 Estimates and Order-of-
Magnitude Calculations
1.7 Significant Figures
Physics and Measurement
ANSWERS TO QUESTIONS
Q1.1 Atomic clocks are based on electromagnetic waves which atoms
emit. Also, pulsars are highly regular astronomical clocks.
Q1.2 Density varies with temperature and pressure. It would be
necessary to measure both mass and volume very accurately in
order to use the density of water as a standard.
Q1.3 People have different size hands. Defining the unit precisely
would be cumbersome.
Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms
Q1.5 (b) and (d). You cannot add or subtract quantities of different
dimension.
Q1.6 A dimensionally correct equation need not be true. Example:
1 chimpanzee = 2 chimpanzee is dimensionally correct. If an
equation is not dimensionally correct, it cannot be correct.
Q1.7 If I were a runner, I might walk or run 101
miles per day. Since I am a college professor, I walk about
100
miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on
vacation.
Q1.8 On February 7, 2001, I am 55 years and 39 days old.
55
365 25
1
39 20 128
86 400
1
1 74 10 109 9
yr
d
yr
d d
s
d
s s
.
. ~
F
HG
I
KJ+ =
F
HG I
KJ = × .
Many college students are just approaching 1 Gs.
Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.
Q1.10 The mass of the forty-six chapter textbook is on the order of 100
kg .
Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.
1

2 Physics and Measurement
SOLUTIONS TO PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
No problems in this section
Section 1.2 Matter and Model-Building
P1.1 From the figure, we may see that the spacing between diagonal planes is half the distance between
diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the
Pythagorean theorem, L L Ldiag = +2 2
. Thus, since the atoms are separated by a distance
L = 0 200. nm , the diagonal planes are separated by
1
2
0 1412 2
L L+ = . nm .
Section 1.3 Density and Atomic Mass
*P1.2 Modeling the Earth as a sphere, we find its volume as
4
3
4
3
6 37 10 1 08 103 6 3 21 3
π πr = × = ×. .m me j . Its
density is then ρ = =
×
×
= ×
m
V
5 98 10
1 08 10
5 52 10
24
21 3
3 3.
.
.
kg
m
kg m . This value is intermediate between the
tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to
3 000 3
kg m . The average density of the Earth is significantly higher, so higher-density material
must be down below the surface.
P1.3 With V = base area heighta fb g V r h= π 2
e j and ρ =
m
V
, we have
ρ
π π
ρ
= =
F
HG
I
KJ
= ×
m
r h2 2
9
4 3
1
19 5 39 0
10
1
2 15 10
kg
mm mm
mm
m
kg m
3
3
. .
. .
a f a f
*P1.4 Let V represent the volume of the model, the same in ρ =
m
V
for both. Then ρiron kg= 9 35. V and
ρgold
gold
=
m
V
. Next,
ρ
ρ
gold
iron
gold
kg
=
m
9 35.
and mgold
3 3
3
kg
19.3 10 kg / m
kg / m
kg=
×
×
F
HG
I
KJ =9 35
7 86 10
23 03
.
.
. .
P1.5 V V V r ro i= − = −
4
3
2
3
1
3
πe j
ρ =
m
V
, so m V r r
r r
= =
F
HG I
KJ − =
−
ρ ρ π
π ρ4
3
4
3
2
3
1
3 2
3
1
3
e j e j .

Chapter 1 3
P1.6 For either sphere the volume is V r=
4
3
3
π and the mass is m V r= =ρ ρ π
4
3
3
. We divide this equation
for the larger sphere by the same equation for the smaller:
m
m
r
r
r
rs s s
= = =
ρ π
ρ π
4 3
4 3
5
3
3
3
3
.
Then r rs= = =5 4 50 1 71 7 693
. . .cm cma f .
P1.7 Use 1 u . g= × −
1 66 10 24
.
(a) For He, m0
24
4 00 6 64 10=
×F
HG
I
KJ = × −
. .u
1.66 10 g
1 u
g
-24
.
(b) For Fe, m0
23
55 9 9 29 10=
×F
HG
I
KJ = × −
. .u
1.66 10 g
1 u
g
-24
.
(c) For Pb, m0
24
22
207
1 66 10
3 44 10=
×F
HG
I
KJ = ×
−
−
u
g
1 u
g
.
. .
*P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m0 of one
atom: m Nm= 0. The first assertion is that the mass of one aluminum atom is
m0
27 26
27 0 27 0 1 66 10 1 4 48 10= = × × = ×− −
. . . .u u kg u kg .
Then the mass of 6 02 1023
. × atoms is
m Nm= = × × × = =−
0
23 26
6 02 10 4 48 10 0 027 0 27 0. . . .kg kg g .
Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be
written m Nm= 0.
0 027 0 6 02 1023
0. .kg = × m , so m0 23
260 027
6 02 10
4 48 10=
×
= × −.
.
.
kg
kg ,
in agreement with the first assertion.
(b) The general equation m Nm= 0 applied to one mole of any substance gives M NMg u= ,
where M is the numerical value of the atomic mass. It divides out exactly for all substances,
giving 1 000 000 0 10 1 660 540 2 103 27
. .× = ×− −
kg kgN . With eight-digit data, we can be quite
sure of the result to seven digits. For one mole the number of atoms is
N =
F
HG
I
KJ = ×− +1
1 660 540 2
10 6 022 137 103 27 23
.
. .
(c) The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one
molecule of H O2 is 2 1 008 0 15 999 18 0. . .b g+ =u u. Then the molar mass is 18 0. g .
(d) For CO2 we have 12 011 2 15 999 44 0. . .g g g+ =b g as the mass of one mole.

P1.9 Mass of gold abraded: ∆m = − = =
F
HG
I
KJ = × −
3 80 3 35 0 45 0 45
1
4 5 10 4
. . . . .g g g g
kg
10 g
kg3b g .
Each atom has mass m0
27
25
197 197
1 66 10
1
3 27 10= =
×F
HG
I
KJ = ×
−
−
u u
kg
u
kg
.
. .
Now, ∆ ∆m N m= 0 , and the number of atoms missing is
∆
∆
N
m
m
= =
×
×
= ×
−
−
0
4
25
214 5 10
3 27 10
1 38 10
.
.
.
kg
kg
atoms.
The rate of loss is
∆
∆
∆
∆
N
t
N
t
=
× F
HG I
KJF
HG I
KJF
HG I
KJF
HG I
KJ
= ×
1 38 10
50
1 1 1 1
8 72 10
21
11
.
. .
atoms
yr
yr
365.25 d
d
24 h
h
60 min
min
60 s
atoms s
P1.10 (a) m L= = × = × = ×− − −
ρ 3 3 6 3 16 19
7 86 5 00 10 9 83 10 9 83 10. g cm cm g kge je j. . .
(b) N
m
m
= =
×
×
= ×
−
−
0
19
27
79 83 10
55 9 1 66 10
1 06 10
.
. .
.
kg
u kg 1 u
atoms
e j
P1.11 (a) The cross-sectional area is
A = +
= × −
2 0 150 0 010 0 340 0 010
6 40 10 3
. . . .
. .
m m m m
m2
a fa f a fa f.
The volume of the beam is
V AL= = × = ×− −
6 40 10 1 50 9 60 103 3
. . .m m m2 3
e ja f .
Thus, its mass is
m V= = × × =−
ρ 7 56 10 9 60 10 72 63 3
. . .kg / m m kg3 3
e je j .
FIG. P1.11
(b) The mass of one typical atom is m0
27
26
55 9
1 66 10
1
9 28 10=
×F
HG
I
KJ = ×
−
−
.
.
.u
kg
u
kga f . Now
m Nm= 0 and the number of atoms is N
m
m
= =
×
= ×−
0
26
2672 6
9 28 10
7 82 10
.
.
.
kg
kg
atoms .

Chapter 1 5
P1.12 (a) The mass of one molecule is m0
27
26
18 0
1 66 10
2 99 10=
×F
HG
I
KJ = ×
−
−
.
.
.u
kg
1 u
kg . The number of
molecules in the pail is
N
m
m
pail
kg
2.99 kg
molecules= =
×
= ×−
0
26
251 20
10
4 02 10
.
. .
(b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere.
N N
m
M
both pail
pail
total
25
(4.02 10 molecules)
kg
kg
=
F
HG
I
KJ = ×
×
F
HG
I
KJ
1 20
1 32 1021
.
.
,
or
Nboth molecules= ×3 65 104
. .
Section 1.4 Dimensional Analysis
P1.13 The term x has dimensions of L, a has dimensions of LT−2
, and t has dimensions of T. Therefore, the
equation x ka tm n
= has dimensions of
L LT T= −2
e j a fm n
or L T L T1 0 2
= −m n m
.
The powers of L and T must be the same on each side of the equation. Therefore,
L L1
= m
and m = 1 .
Likewise, equating terms in T, we see that n m− 2 must equal 0. Thus, n = 2 . The value of k, a
dimensionless constant, cannot be obtained by dimensional analysis .
*P1.14 (a) Circumference has dimensions of L.
(b) Volume has dimensions of L3
.
(c) Area has dimensions of L2
.
Expression (i) has dimension L L L2 1 2 2
e j /
= , so this must be area (c).
Expression (ii) has dimension L, so it is (a).
Expression (iii) has dimension L L L2 3
e j= , so it is (b). Thus, (a) ii; (b) iii, (c) i= = = .

P1.15 (a) This is incorrect since the units of ax are m s2 2
, while the units of v are m s.
(b) This is correct since the units of y are m, and cos kxa f is dimensionless if k is in m−1
.
*P1.16 (a) a
F
m
∝
∑ or a k
F
m
=
∑ represents the proportionality of acceleration to resultant force and
the inverse proportionality of acceleration to mass. If k has no dimensions, we have
a k
F
m
= ,
L
T
1
F
M2
= , F
M L
T2
=
⋅
.
(b) In units,
M L
T
kg m
s2 2
⋅
=
⋅
, so 1 1newton kg m s2
= ⋅ .
P1.17 Inserting the proper units for everything except G,
kg m
s
kg
m
2
L
NM O
QP=
G
2
2
.
Multiply both sides by m
2
and divide by kg
2
; the units of G are
m
kg s
3
2
⋅
.
Section 1.5 Conversion of Units
*P1.18 Each of the four walls has area 8 00 12 0 96 0. . .ft ft ft2
a fa f= . Together, they have area
4 96 0
1
3 28
35 72
2
. .ft
m
. ft
m2
e jF
HG I
KJ = .
P1.19 Apply the following conversion factors:
1 2 54in cm= . , 1 86 400d s= , 100 1cm m= , and 10 19
nm m=
1
32
2 54 10 10
9 19
2 9
in day
cm in m cm nm m
86 400 s day
nm s
F
HG I
KJ =
−
.
.
b ge je j .
This means the proteins are assembled at a rate of many layers of atoms each second!
*P1.20 8 50 8 50
0 025 4
1 39 10
3
4
. .
.
.in in
m
1 in
m3 3 3
=
F
HG I
KJ = × −

Chapter 1 7
P1.21 Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should
expect the area to be about A m m m2
≈ =30 50 1 500a fa f .
Categorize: We model the lot as a perfect rectangle to use Area = Length × Width. Use the
conversion: 1 m 3.281 ft= .
Analyze: A LW= =
F
HG I
KJ F
HG I
KJ = ×100
1
3 281
150
1
3 281
1 39 103
ft
m
ft
ft
m
ft
= 1 390 m m2 2
a f a f. .
. .
Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper
units of m2
. Unit conversion is a common technique that is applied to many problems.
P1.22 (a) V = = ×40.0 m 20.0 m 12.0 m . m3
a fa fa f 9 60 103
V = × = ×9 60 10 3 39 103 5 3
. m 3.28 ft 1 m ft3 3
b g .
(b) The mass of the air is
m V= = × = ×ρair
3 3
kg m 9.60 10 m . kg1 20 1 15 103 4
.e je j .
The student must look up weight in the index to find
F mgg = = × = ×1.15 10 kg 9.80 m s 1.13 10 N4 2 5
e je j .
Converting to pounds,
Fg = × = ×1 13 10 2 54 105 4
. N 1 lb 4.45 N lbe jb g . .
P1.23 (a) Seven minutes is 420 seconds, so the rate is
r = = × −30 0
420
7 14 10 2.
.
gal
s
gal s .
(b) Converting gallons first to liters, then to m3
,
r
r
= ×
F
HG
I
KJ
F
HG
I
KJ
= ×
−
−
−
7 14 10
3 786 10
2 70 10
2
3
4
.
.
. .
gal s
L
1 gal
m
1 L
m s
3
3
e j
(c) At that rate, to fill a 1-m3
tank would take
t =
×
F
HG
I
KJ
F
HG
I
KJ =−
1
2 70 10
1
1 034
m
m s
h
3 600
h
3
3
.
. .

*P1.24 (a) Length of Mammoth Cave =
F
HG I
KJ = = × = ×348
1 609
1
560 5 60 10 5 60 105 7
mi
km
mi
km m cm
.
. . .
(b) Height of Ribbon Falls =
F
HG I
KJ = = = ×1 612
0
1
491 m 0 491 4 91 104
ft
.304 8 m
ft
km cm. . .
(c) Height of Denali =
F
HG I
KJ = = × = ×20 320
0
1
6 6 19 10 6 19 103 5
ft
.304 8 m
ft
.19 km m cm. . .
(d) Depth of King’s Canyon =
F
HG I
KJ = = × = ×8 200
0
1
2 2 50 10 2 50 103 5
ft
.304 8 m
ft
.50 km m cm. . .
P1.25 From Table 1.5, the density of lead is 1 13 104
. kg m3
× , so we should expect our calculated value to
be close to this number. This density value tells us that lead is about 11 times denser than water,
which agrees with our experience that lead sinks.
Density is defined as mass per volume, in ρ =
m
V
. We must convert to SI units in the calculation.
ρ =
F
HG
I
KJF
HG I
KJ = ×
23 94
2 10
1
1 000
100
1
1 14 103
3
4.
.
g
cm
kg
g
cm
m
. kg m3
At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our
result is indeed close to the expected value. Since the last reported significant digit is not certain, the
difference in the two values is probably due to measurement uncertainty and should not be a
concern. One important common-sense check on density values is that objects which sink in water
must have a density greater than 1 g cm3
, and objects that float must be less dense than water.
P1.26 It is often useful to remember that the 1 600-m race at track and field events is approximately 1 mile
in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to
1
1
640
1 609
4 05 10
2
3
acre
mi
acres
m
mi
m
2
2
a fF
HG
I
KJF
HG I
KJ = ×. .
*P1.27 The weight flow rate is 1 200
2 000 1 1
667
ton
h
lb
ton
h
60 min
min
60 s
lb s
F
HG I
KJF
HG I
KJF
HG I
KJ = .
P1.28 1 1 609 1 609mi m km= = . ; thus, to go from mph to km h, multiply by 1.609.
(a) 1 1 609mi h km h= .
(b) 55 88 5mi h km h= .
(c) 65 104 6mi h km h= . . Thus, ∆v = 16 1. km h .

Chapter 1 9
P1.29 (a)
6 10 1 1 1
190
12
×F
HG
I
KJ
F
HG
I
KJF
HG I
KJF
HG
I
KJ =
$
1 000 $ s
h
3 600 s
day
24 h
yr
365 days
years
(b) The circumference of the Earth at the equator is 2 6 378 10 4 01 103 7
π . .× = ×m me j . The length
of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9 30 1011
. × m. Thus, the
6 trillion dollars would encircle the Earth
9 30 10
2 32 10
11
4.
.
×
×
= ×
m
4.01 0 m
times7
.
P1.30 N
m
m
atoms
Sun
atom
kg
1.67 kg
atoms= =
×
×
= ×−
1 99 10
10
1 19 10
30
27
57.
.
P1.31 V At= so t
V
A
= =
×
= ×
−
−3 78 10
25 0
1 51 10 151
3
4.
.
.
m
m
m or m
3
2
µb g
P1.32 V Bh= =
= ×
1
3
13 0 43 560
3
481
9 08 107
.
. ,
acres ft acre
ft
ft
2
3
a fe j a f
or
V = ×
×F
HG
I
KJ
= ×
−
9 08 10
2 83 10
1
2 57 10
7
2
6
.
.
.
ft
m
ft
m
3
3
3
3
e j B
h
B
h
FIG. P1.32
P1.33 Fg = × = ×2 50 2 00 10 2 000 1 00 106 10
. . .tons block blocks lb ton lbsb ge jb g
*P1.34 The area covered by water is
A A Rw = = = × = ×0 70 3 6 102 14
. 0.70 4 0.70 4 6.37 10 m . mEarth Earth
6 2 2
a fe j a fa fe jπ π .
The average depth of the water is
d = = ×2.3 miles 1 609 m l mile . ma fb g 3 7 103
.
The volume of the water is
V A dw= = × × = ×3 6 10 3 7 10 1 3 1014 2 3 18 3
. m . m . me je j
and the mass is
m V= = × = ×ρ 1 000 1 3 10 1 3 103 18 3 21
kg m . m kge je j . .

P1.35 (a) d d
d
d
nucleus, scale nucleus, real
atom, scale
atom, real
m
ft
1.06 10 m
ft=
F
HG
I
KJ = ×
×
F
HG I
KJ = ×−
−
−
2 40 10
300
6 79 1015
10
3
. .e j , or
dnucleus, scale ft mm 1 ft mm= × =−
6 79 10 304 8 2 073
. . .e jb g
(b)
V
V
r
r
d
d
r
r
atom
nucleus
atom
nucleus
atom
nucleus
atom
nucleus
m
m
times as large
= =
F
HG I
KJ =
F
HG I
KJ =
×
×
F
HG
I
KJ
= ×
−
−
4
3
4
3
3 3 10
15
3
13
3
3
1 06 10
2 40 10
8 62 10
π
π
.
.
.
*P1.36 scale distance
between
real
distance
scale
factor
km
m
m
km=
F
HG I
KJF
HG I
KJ = ×
×
×
F
HG
I
KJ =
−
4 0 10
7 0 10
1 4 10
20013
3
9
.
.
.
e j
P1.37 The scale factor used in the “dinner plate” model is
S =
×
= × −0 25
1 0 10
2 105
6.
.
m
lightyears
.5 m lightyears .
The distance to Andromeda in the scale model will be
D D Sscale actual
6 6
2.0 10 lightyears 2.5 10 m lightyears m= = × × =−
e je j 5 0. .
P1.38 (a)
A
A
r
r
r
r
Earth
Moon
Earth
Moon
2
Earth
Moon
m cm m
cm
= =
F
HG I
KJ =
×
×
F
H
GG
I
K
JJ =
4
4
6 37 10 100
1 74 10
13 4
2 2 6
8
2
π
π
.
.
.
e jb g
(b)
V
V
r
r
r
r
Earth
Moon
Earth
Moon
3
3
Earth
Moon
m cm m
cm
= =
F
HG I
KJ =
×
×
F
H
GG
I
K
JJ =
4
3
4
3
6
8
3
3
6 37 10 100
1 74 10
49 1
π
π
.
.
.
e jb g
P1.39 To balance, m mFe Al= or ρ ρFe Fe Al AlV V=
ρ π ρ π
ρ
ρ
Fe Fe Al Al
Al Fe
Fe
Al
cm cm
4
3
4
3
2 00
7 86
2 70
2 86
3 3
1 3 1 3
F
HG I
KJ =
F
HG I
KJ
=
F
HG I
KJ =
F
HG I
KJ =
r r
r r
/ /
.
.
.
. .a f

Chapter 1 11
P1.40 The mass of each sphere is
m V
r
Al Al Al
Al Al
= =ρ
π ρ4
3
3
and
m V
r
Fe Fe Fe
Fe Fe
= =ρ
π ρ4
3
3
.
Setting these masses equal,
4
3
4
3
3 3
π ρ π ρAl Al Fe Fer r
= and r rAl Fe
Fe
Al
=
ρ
ρ
3 .
Section 1.6 Estimates and Order-of-Magnitude Calculations
P1.41 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball
as a sphere of diameter 0.038 m. The volume of the room is 4 4 3 48× × = m3
, while the volume of
one ball is
4
3
0 038
2 87 10
3
5π .
.
m
2
m3F
HG I
KJ = × −
.
Therefore, one can fit about
48
2 87 10
105
6
.
~
× −
ping-pong balls in the room.
As an aside, the actual number is smaller than this because there will be a lot of space in the
room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best
packing fraction” is
1
6
2 0 74π = . so that at least 26% of the space will be empty. Therefore, the
above estimate reduces to 1 67 10 0 740 106 6
. . ~× × .
P1.42 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus,
the tire would make 50 000 5 280 1 3 107
mi ft mi rev 8 ft rev~ 10 rev7
b gb gb g= × .
P1.43 In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least
1
16
in 43 10 ft2 5 2
= × −
. Since 1 acre 43 560 ft2
= , the number of blades of grass to be expected on a
quarter-acre plot of land is about
n = =
×
= ×−
total area
area per blade
acre ft acre
ft blade
2.5 10 blades blades
2
2
7
0 25 43 560
43 10
105
7
.
~
a fe j .

P1.44 A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then
approximately 4 10 3
× −
in3
. Since 1 acre 43 560 ft2
= , the volume of water required to cover it to a
depth of 1 inch is
1 acre 1 inch 1 acre in
ft
1 acre
in
ft
6.3 10 in6 3
a fa f a f= ⋅
F
HG
I
KJ
F
HG
I
KJ ≈ ×
43 560 144
1
2 2
2
.
The number of raindrops required is
n = =
×
×
= ×−
volume of water required
volume of a single drop
in
in
.
6 3 10
4 10
1 6 10 10
6 3
3 3
9 9.
~ .
*P1.45 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then
V = =0 5 1 3 0 5 0 3 0 10. . . . .a fa fa fa fm m m m3
.
The mass of this volume of water is
m Vwater water
3 3
kg m m kg kg= = =ρ 1 000 0 10 100 102
e je j. ~ .
Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The
mass of copper required is
m Vcopper copper
3 3
kg m m kg kg= = =ρ 8 920 0 10 892 103
e je j. ~ .
P1.46 The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum
cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~250
million people, and 365 days in a year, so
250 10 365 106 11
× ≅cans day days year canse jb g
are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we
estimate this represents
10 0 1 1 1 3 1 1011 5
cans oz can lb 16 oz ton 2 000 lb tons yeare jb gb gb g. .≈ × . ~105
tons
P1.47 Assume: Total population = 107
; one out of every 100 people has a piano; one tuner can serve about
1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year).
Therefore,
# tuners ~
1 tuner
1 000 pianos
1 piano
100 people
people
F
HG
I
KJ
F
HG
I
KJ =( )10 1007
.

Chapter 1 13
Section 1.7 Significant Figures
*P1.48 METHOD ONE
We treat the best value with its uncertainty as a binomial 21 3 0 2 9 8 0 1. . . .± ±a f a fcm cm ,
A = ± ± ±21 3 9 8 21 3 0 1 0 2 9 8 0 2 0 1. . . . . . . .a f a f a f a fa f cm2
.
The first term gives the best value of the area. The cross terms add together to give the uncertainty
and the fourth term is negligible.
A = ±209 42 2
cm cm .
METHOD TWO
We add the fractional uncertainties in the data.
A = ± +
F
HG I
KJ = ± = ±21 3 9 8
0 2
21 3
0 1
9 8
209 2% 209 42 2 2
. .
.
.
.
.
cm cm cm cm cma fa f
P1.49 (a) π π
π
r2 2
2 2
2 2
10 5 0 2
10 5 2 10 5 0 2 0 2
346 13
= ±
= ± +
= ±
. m . m
m m m m
m m
a f
( . ) ( . )( . ) ( . )
(b) 2 2 10 5 0 2 66 0 1 3π πr = ± = ±. m . m m ma f . .
P1.50 (a) 3 (b) 4 (c) 3 (d) 2
P1.51 r
m
m
r
= ± = ± ×
= ±
=
−
6 50 0 20 6 50 0 20 10
1 85 0 02
2
4
3
3
. . cm . . m
. . kg
a f a f
a f
c hρ
π
also,
δ ρ
ρ
δ δ
= +
m
m
r
r
3
.
In other words, the percentages of uncertainty are cumulative. Therefore,
δ ρ
ρ
= + =
0 02
1 85
3 0 20
6 50
0 103
.
.
.
.
.
a f ,
ρ
π
=
×
= ×
−
1 85
6 5 10
1 61 10
4
3
2 3
3 3.
.
.
c h e jm
kg m
and
ρ δ ρ± = ± × = ± ×1 61 0 17 10 1 6 0 2 103 3
. . . .a f a fkg m kg m3 3
.

P1.52 (a) 756.??
37.2?
0.83
+ 2.5?
796./5/3 = 797
(b) 0 003 2 356 3 1 140 16 1 1. 2 s.f. . 4 s.f. . 2 s.f.a f a f a f× = = .
(c) 5.620 4 s.f. >4 s.f. 17.656= 4 s.f. 17.66a f a f a f× =π
*P1.53 We work to nine significant digits:
1 1
365 242 199 24 60 60
31 556 926 0yr yr
d
1 yr
h
1 d
min
1 h
s
1 min
s=
F
HG
I
KJF
HG I
KJF
HG I
KJF
HG I
KJ =
.
. .
P1.54 The distance around is 38.44 m 19.5 m 38.44 m 19.5 m 115.88 m+ + + = , but this answer must be
rounded to 115.9 m because the distance 19.5 m carries information to only one place past the
decimal. 115 9. m
P1.55 V V V V V
V
V
V
= + = +
= + + =
= =
= + =
2 2 2
17 0 1 0 1 0 1 0 0 09 1 70
10 0 1 0 0 090 0 900
2 1 70 0 900 5 2
1 2 1 2
1
2
3
b g
a fa fa f
a fa fa f
e j
. . . . . .
. . . .
. . .
m m m m m m
m m m m
m m m
3
3
3 3
δ
δ
δ
δ
1
1
1
1
1
1
0 12
0 0063
0 01
0 010
0 1
0 011
0 006 0 010 0 011 0 027 3%
= =
= =
= =
U
V
|||
W
|||
= + + = =
.
.
.
.
.
.
. . . .
m
19.0 m
m
1.0 m
cm
9.0 cm
w
w
t
t
V
V
FIG. P1.55
Additional Problems
P1.56 It is desired to find the distance x such that
x
x100
1 000
m
m
=
(i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that
x2 5
100 1 000 1 00 10= = ×m m m2
a fb g .
and therefore
x = × =1 00 10 3165
. m m2
.

Chapter 1 15
*P1.57 Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass
m0
27
25
197 3 27 10=
×F
HG
I
KJ = ×
−
−
u
1.66 10 kg
1 u
kga f . .
So, the number of atoms in the cube is
N =
×
= ×−
19 300
5 90 1025
28kg
3.27 10 kg
. .
The imagined cubical volume of each atom is
d3
28
291
5 90 10
1 69 10=
×
= × −m
m
3
3
.
. .
So
d = × −
2 57 10 10
. m .
P1.58 A N A
V
V
A
V
r
rtotal drop
total
drop
drop
total
4
= =
F
HG
I
KJ =
F
H
GG
I
K
JJa fe j e j e jπ
π3
3
2
4
A
V
r
total
total
3
2m
m
m=
F
HG I
KJ =
×
×
F
HG
I
KJ =
−
−
3
3
30 0 10
2 00 10
4 50
6
5
.
.
.
P1.59 One month is
1 30 24 3 600 2 592 106
mo day h day s h s= = ×b gb gb g . .
Applying units to the equation,
V t t= +1 50 0 008 00 2
. .Mft mo Mft mo3 3 2
e j e j .
Since 1 106
Mft ft3 3
= ,
V t t= × + ×1 50 10 0 008 00 106 6 2
. .ft mo ft mo3 3 2
e j e j .
Converting months to seconds,
V t t=
×
×
+
×
×
1 50 10 0 008 00 106 6
2
2. .ft mo
2.592 10 s mo
ft mo
2.592 10 s mo
3
6
3 2
6
e j
.
Thus, V t tft ft s ft s3 3 3 2
[ ] . .= + × −
0 579 1 19 10 9 2
e j e j .

P1.60 ′α (deg) α(rad) tan αa f sin αa f difference
15.0 0.262 0.268 0.259 3.47%
20.0 0.349 0.364 0.342 6.43%
25.0 0.436 0.466 0.423 10.2%
24.0 0.419 0.445 0.407 9.34%
24.4 0.426 0.454 0.413 9.81%
24.5 0.428 0.456 0.415 9.87%
24.6 0.429 0.458 0.416 9.98% 24.6°
24.7 0.431 0.460 0.418 10.1%
P1.61 2 15 0
2 39
2 39 55 0 3 41
π r
r
h
r
h
=
=
= °
= ° =
. m
. m
tan 55.0
. m ( . ) ma ftan .
55°55°
h
rr
h
FIG. P1.61
*P1.62 Let d represent the diameter of the coin and h its thickness. The mass of the gold is
m V At
d
dh t= = = +
F
HG
I
KJρ ρ ρ
π
π
2
4
2
where t is the thickness of the plating.
m = +
L
N
MM
O
Q
PP ×
=
= × = =
−
19 3 2
2 41
4
2 41 0 178 0 18 10
0 003 64
0 003 64 036 4 3 64
2
4
.
.
. . .
.
. $10 $0. .
π π
a f a fa f e j
grams
cost grams gram cents
This is negligible compared to $4.98.
P1.63 The actual number of seconds in a year is
86 400 s day 365.25 day yr 31 557 600 s yrb gb g= .
The percent error in the approximation is
π × −
× =
10 31 557 600
31 557 600
100% 0 449%
7
s yr s yr
s yr
e j b g . .

Chapter 1 17
P1.64 (a) V = L3
, A = L2
, h = L
V A h=
L L L L3 2 3
= = . Thus, the equation is dimensionally correct.
(b) V R h R h Ahcylinder = = =π π2 2
e j , where A R= π 2
V wh w h Ahrectangular object = = =a f , where A w=
P1.65 (a) The speed of rise may be found from
v
D
= = =
Vol rate of flow
(Area:
cm s
cm s
3
cm
a f
a fπ π
2 2
4
6 30
4
16 5
0 529
)
.
.
.
.
(b) Likewise, at a 1.35 cm diameter,
v = =
16 5
11 5
1.35
4
2
.
.
cm s
cm s
3
cmπa f .
P1.66 (a) 1 cubic meter of water has a mass
m V= = × =−
ρ 1 00 10 1 00 10 1 0003 3 3 2 3
. kg cm . m cm m kge je je j
(b) As a rough calculation, we treat each item as if it were 100% water.
cell: kg m m
kg
kidney: . kg cm cm
kg
fly: kg cm mm mm cm mm
3
3
3 2
m V R D
m V R
m D h
= =
F
HG I
KJ =
F
HG I
KJ =
F
HG I
KJ ×
= ×
= =
F
HG I
KJ = ×
F
HG I
KJ
=
=
F
HG I
KJ = ×
F
HG I
KJ
−
−
−
− −
ρ ρ π ρ π π
ρ ρ π π
ρ
π π
4
3
1
6
1 000
1
6
1 0 10
5 2 10
4
3
1 00 10
4
3
4 0
0 27
4
1 10
4
2 0 4 0 10
3 3 6 3
16
3 3 3
2 3 1
e j e j
e j
e j a f a fe
.
.
( . )
.
. . j3
5
1 3 10= × −
. kg
P1.67 V20 mpg
10cars mi yr
mi gal
5.0 10 gal yr= = ×
( )( )10 10
20
8 4
V25 mpg
10cars mi yr
mi gal
4.0 10 gal yr= = ×
( )( )10 10
25
8 4
Fuel saved gal yr25 mpg 20 mpg= − = ×V V 1 0 1010
.

P1.68 v =
F
HG
I
KJ
F
HG
I
KJ
F
HG
I
KJ
F
HG
I
KJF
HG I
KJF
HG
I
KJ = × −
5 00
220
1
0 914 4
1
1
14
1
24
1
3 600
8 32 10 4
.
.
.
furlongs
fortnight
yd
furlong
m
yd
fortnight
days
day
hrs
hr
s
m s
This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth.
P1.69 The volume of the galaxy is
π πr t2 21 2 19 61
10 10 10= m m m3
e j e j~ .
If the distance between stars is 4 1016
× m, then there is one star in a volume on the order of
4 10 1016 3 50
× m m3
e j ~ .
The number of stars is about
10
10
10
61
50
11m
m star
stars
3
3
~ .
P1.70 The density of each material is ρ
π π
= = =
m
V
m
r h
m
D h2 2
4
.
Al:
g
cm cm
g
cm
The tabulated value
g
cm
is smaller.
Cu:
g
.23 cm .06 cm
g
cm
The tabulated value
g
cm
is smaller.
Brass:
.54 cm .69 cm
g
cm
Sn:
g
.75 cm .74 cm
g
cm
Fe:
.89 cm .77 cm
g
cm
3 3
3 3
3
3
3
ρ
π
ρ
π
ρ
π
ρ
π
ρ
π
= =
F
HG I
KJ
= =
F
HG I
KJ
= =
= =
= =
4 51 5
2 52 3 75
2 75 2 70 2%
4 56 3
1 5
9 36 8 92 5%
4 94.4 g
1 5
8 91
4 69 1
1 3
7 68
4 216.1 g
1 9
7 88
2
2
2
2
2
.
. .
. .
.
. .
.
.
.
.
b g
a f a f
b g
a f a f
b g
a f a f
b g
a f a f
b g
a f a f The tabulated value
g
cm
is smaller.3
7 86 0 3%. .
F
HG I
KJ
P1.71 (a) 3 600 s hr 24 hr day 365.25 days yr s yrb gb gb g= ×3 16 107
.
(b) V r
V
V
mm
cube
mm
18
. m . m
m
m
1.91 10 micrometeorites
= = × = ×
=
×
= ×
− −
−
4
3
4
3
5 00 10 5 24 10
1
5 24 10
3 7 3 19 3
3
19 3
π πe j
.
This would take
1 91 10
3 16 10
6 05 10
18
7
10.
.
.
×
×
= ×
micrometeorites
micrometeorites yr
yr .

Chapter 1 19
ANSWERS TO EVEN PROBLEMS
P1.2 5 52 103 3
. × kg m , between the densities
of aluminum and iron, and greater than
the densities of surface rocks.
P1.34 1 3 1021
. × kg
P1.36 200 km
P1.38 (a) 13.4; (b) 49.1P1.4 23.0 kg
P1.40 r rAl Fe
Fe
Al
=
F
HG I
KJρ
ρ
1 3
P1.6 7.69 cm
P1.8 (a) and (b) see the solution,
NA = ×6 022 137 1023
. ; (c) 18.0 g;
P1.42 ~10 rev7
(d) 44.0 g
P1.44 ~109
raindrops
P1.10 (a) 9 83 10 16
. × −
g ; (b) 1 06 107
. × atoms
P1.46 ~1011
cans; ~105
tons
P1.12 (a) 4 02 1025
. × molecules;
(b) 3 65 104
. × molecules P1.48 209 4 2
±a fcm
P1.14 (a) ii; (b) iii; (c) i P1.50 (a) 3; (b) 4; (c) 3; (d) 2
P1.16 (a)
M L
T2
⋅
; (b) 1 1newton kg m s2
= ⋅ P1.52 (a) 797; (b) 1.1; (c) 17.66
P1.54 115.9 m
P1.18 35 7. m2
P1.56 316 m
P1.20 1 39 10 4
. × −
m3
P1.58 4 50. m2
P1.22 (a) 3 39 105 3
. × ft ; (b) 2 54 104
. × lb
P1.60 see the solution; 24.6°
P1.24 (a) 560 5 60 10 5 60 105 7
km m cm= × = ×. . ;
P1.62 3 64. cents ; no(b) 491 m 0 491 4 91 104
= = ×. .km cm ;
(c) 6 6 19 10 6 19 103 5
.19 km m cm= × = ×. . ;
P1.64 see the solution
(d) 2 2 50 10 2 50 103 5
.50 km m cm= × = ×. .
P1.66 (a) 1 000 kg; (b) 5 2 10 16
. × −
kg ; 0 27. kg ;
1 3 10 5
. × −
kg
P1.26 4 05 103
. × m2
P1.28 (a) 1 1 609mi h km h= . ; (b) 88 5. km h ;
P1.68 8 32 10 4
. × −
m s ; a snail(c) 16 1. km h
P1.70 see the solution
P1.30 1 19 1057
. × atoms
P1.32 2 57 106 3
. × m

2
CHAPTER OUTLINE
2.1 Position, Velocity, and
Speed
2.2 Instantaneous Velocity and
Speed
2.3 Acceleration
2.4 Motion Diagrams
2.5 One-Dimensional Motion
with Constant Acceleration
2.6 Freely Falling Objects
2.7 Kinematic Equations
Derived from Calculus
Motion in One Dimension
Q2.1 If I count 5.0 s between lightning and thunder, the sound has
traveled 331 5 0 1 7m s s kmb ga f. .= . The transit time for the light
is smaller by
3 00 10
331
9 06 10
8
5.
.
×
= ×
m s
m s
times,
so it is negligible in comparison.
Q2.2 Yes. Yes, if the particle winds up in the +x region at the end.
Q2.3 Zero.
Q2.4 Yes. Yes.
Q2.5 No. Consider a sprinter running a straight-line race. His average velocity would simply be the
length of the race divided by the time it took for him to complete the race. If he stops along the way
to tie his shoe, then his instantaneous velocity at that point would be zero.
Q2.6 We assume the object moves along a straight line. If its average
velocity is zero, then the displacement must be zero over the time
interval, according to Equation 2.2. The object might be stationary
throughout the interval. If it is moving to the right at first, it must
later move to the left to return to its starting point. Its velocity must
be zero as it turns around. The graph of the motion shown to the
right represents such motion, as the initial and final positions are
the same. In an x vs. t graph, the instantaneous velocity at any time
t is the slope of the curve at that point. At t0 in the graph, the slope
of the curve is zero, and thus the instantaneous velocity at that time
is also zero.
x
tt0
FIG. Q2.6
Q2.7 Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the
velocity of the particle is unchanging, or is a constant.
21

22 Motion in One Dimension
Q2.8 Yes. If you drop a doughnut from rest v = 0a f, then its acceleration is not zero. A common
misconception is that immediately after the doughnut is released, both the velocity and acceleration
are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut
floating at rest in mid-air.
Q2.9 No: Car A might have greater acceleration than B, but they might both have zero acceleration, or
otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the
recent past.
Q2.10 Yes. Consider throwing a ball straight up. As the ball goes up, its
velocity is upward v > 0a f, and its acceleration is directed down
a < 0a f. A graph of v vs. t for this situation would look like the figure
to the right. The acceleration is the slope of a v vs. t graph, and is
always negative in this case, even when the velocity is positive.
v
t
v0
FIG. Q2.10
Q2.11 (a) Accelerating East (b) Braking East (c) Cruising East
(d) Braking West (e) Accelerating West (f) Cruising West
(g) Stopped but starting to move East
(h) Stopped but starting to move West
Q2.12 No. Constant acceleration only. Yes. Zero is a constant.
Q2.13 The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall,
and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken
as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is
taken as the bottom of the cliff, then the maximum height would be 30 m.
The velocity is independent of the origin. Since the change in position is used to calculate the
instantaneous velocity in Equation 2.5, the choice of origin is arbitrary.
Q2.14 Once the objects leave the hand, both are in free fall, and both experience the same downward
acceleration equal to the free-fall acceleration, –g.
Q2.15 They are the same. After the first ball reaches its apex and falls back downward past the student, it
will have a downward velocity equal to vi. This velocity is the same as the velocity of the second
ball, so after they fall through equal heights their impact speeds will also be the same.
Q2.16 With h gt=
1
2
2
,
(a) 0 5
1
2
0 707
2
. .h g t= a f . The time is later than 0.5t.
(b) The distance fallen is 0 25
1
2
0 5
2
. .h g t= a f . The elevation is 0.75h, greater than 0.5h.

Chapter 2 23
Q2.17 Above. Your ball has zero initial speed and smaller average speed during the time of flight to the
passing point.
Section 2.1 Position, Velocity, and Speed
P2.1 (a) v = 2 30. m s
(b) v
x
t
= =
m m
s
= 16.1 m s
∆
∆
57 5 9 20
3 00
. .
.
−
(c) v
x
t
= =
−
=
∆
∆
57 5 0
11 5
.
.
m m
5.00 s
m s
*P2.2 (a) v
x
t
= =
F
HG I
KJ ×
F
HG I
KJ = × −∆
∆
20 1 1
3 156 10
2 107
7ft
1 yr
m
3.281 ft
yr
s
m s
.
or in particularly windy times
v
x
t
= =
F
HG I
KJ ×
F
HG I
KJ = × −∆
∆
100 1 1
3 156 10
1 107
6ft
1 yr
m
3.281 ft
yr
s
m s
.
.
(b) The time required must have been
∆
∆
t
x
v
= =
F
HG I
KJF
HG
I
KJ = ×
3 000 1 609 10
5 10
3
8mi
10 mm yr
m
1 mi
mm
1 m
yr .
P2.3 (a) v
x
t
= = =
∆
∆
10
5
m
2 s
m s
(b) v = =
5
1 2
m
4 s
m s.
(c) v
x x
t t
=
−
−
=
−
−
= −2 1
2 1
5 10
2
2 5
m m
4 s s
m s.
(d) v
x x
t t
=
−
−
=
− −
−
= −2 1
2 1
5 5
4
3 3
m m
7 s s
m s.
(e) v
x x
t t
=
−
−
=
−
−
=2 1
2 1
0 0
8 0
0 m s
P2.4 x t= 10 2
: For
t
x
s
m
a f
a f
=
=
2 0 2 1 3 0
40 44 1 90
. . .
.
(a) v
x
t
= = =
∆
∆
50
50 0
m
1.0 s
m s.
(b) v
x
t
= = =
∆
∆
4 1
41 0
.
.
m
0.1 s
m s

P2.5 (a) Let d represent the distance between A and B. Let t1 be the time for which the walker has
the higher speed in 5 00
1
. m s =
d
t
. Let t2 represent the longer time for the return trip in
− = −3 00
2
. m s
d
t
. Then the times are t
d
1
5 00
=
. m sb g and t
d
2
3 00
=
. m sb g. The average speed
is:
v
d d d
v
d d d
= =
+
+
=
= =
Total distance
Total time
m s
m s
m s
m s m s
m s
m s
2 2
2 25 00 3 00
8 00
15 0
2
2 15 0
8 00
3 75
. .
.
.
.
.
.
b g b g
b g
e j
e j
(b) She starts and finishes at the same point A. With total displacement = 0, average velocity
= 0 .
Section 2.2 Instantaneous Velocity and Speed
P2.6 (a) At any time, t, the position is given by x t= 3 00 2
. m s2
e j .
Thus, at ti = 3 00. s: xi = =3 00 3 00 27 0
2
. . .m s s m2
e ja f .
(b) At t tf = +3 00. s ∆ : x tf = +3 00 3 00
2
. .m s s2
e ja f∆ , or
x t tf = + +27 0 18 0 3 00
2
. . .m m s m s2
b g e ja f∆ ∆ .
(c) The instantaneous velocity at t = 3 00. s is:
v
x x
t
t
t
f i
t
=
−F
HG
I
KJ = + =
→ →
lim lim . . .
∆ ∆∆
∆
0 0
18 0 3 00 18 0m s m s m s2
e je j .
P2.7 (a) at ti = 1 5. s, xi = 8 0. m (Point A)
at tf = 4 0. s , x f = 2 0. m (Point B)
v
x x
t t
f i
f i
=
−
−
=
−
−
= − = −
2 0 8 0
4 1 5
6 0
2 4
. .
.
.
.
a f
a f
m
s
m
2.5 s
m s
(b) The slope of the tangent line is found from points C and
D. t xC C= =1 0 9 5. .s, mb g and t xD D= =3 5 0. s,b g,
v ≅ −3 8. m s .
FIG. P2.7
(c) The velocity is zero when x is a minimum. This is at t ≅ 4 s .

Chapter 2 25
P2.8 (a)
(b) At t = 5 0. s, the slope is v ≅ ≅
58
23
m
2.5 s
m s .
At t = 4 0. s, the slope is v ≅ ≅
54
18
m
3 s
m s .
At t = 3 0. s, the slope is v ≅ ≅
49 m
14
3.4 s
m s .
At t = 2 0. s , the slope is v ≅ ≅
36 m
9
4.0 s
.0 m s .
(c) a
v
t
= ≅ ≅
∆
∆
23
5 0
4 6
m s
s
m s2
.
.
(d) Initial velocity of the car was zero .
P2.9 (a) v =
−( )
−( )
=
5 0
1 0
5
m
s
m s
(b) v =
−( )
−( )
= −
5 10
4 2
2 5
m
s
m s.
(c) v =
−( )
−( )
=
5 5
5 4
0
m m
s s
(d) v =
− −( )
−( )
= +
0 5
8 7
5
m
s s
m s
FIG. P2.9
*P2.10 Once it resumes the race, the hare will run for a time of
t
x x
v
f i
x
=
−
=
−
=
1 000
25
m 800 m
8 m s
s .
In this time, the tortoise can crawl a distance
x xf i− = ( )=0 2 25 5 00. .m s s ma f .

Section 2.3 Acceleration
P2.11 Choose the positive direction to be the outward direction, perpendicular to the wall.
v v atf i= + : a
v
t
= =
− −
×
= ×−
∆
∆
22 0 25 0
3 50 10
1 34 103
4. .
.
.
m s m s
s
m s2a f .
P2.12 (a) Acceleration is constant over the first ten seconds, so at the end,
v v atf i= + = + ( )=0 2 00 10 0 20 0. . .m s s m s2
c h .
Then a = 0 so v is constant from t =10 0. s to t =15 0. s. And over the last five seconds the
velocity changes to
v v atf i= + = + ( )=20 0 3 00 5 00 5 00. . . .m s m s s m s2
c h .
(b) In the first ten seconds,
x x v t atf i i= + + = + + ( ) =
1
2
0 0
1
2
2 00 10 0 1002 2
. .m s s m2
c h .
Over the next five seconds the position changes to
x x v t atf i i= + + = + ( )+ =
1
2
100 20 0 5 00 0 2002
m m s s m. .a f .
And at t = 20 0. s,
x x v t atf i i= + + = + ( )+ − ( ) =
1
2
200 20 0 5 00
1
2
3 00 5 00 2622 2
m m s s m s s m2
. . . .a f c h .
*P2.13 (a) The average speed during a time interval ∆t is v
t
=
distance traveled
∆
. During the first
quarter mile segment, Secretariat’s average speed was
v1
0 250 1 320
52 4 35 6= = =
.
. .
mi
25.2 s
ft
25.2 s
ft s mi hb g.
During the second quarter mile segment,
v2
1 320
55 0 37 4= =
ft
24.0 s
ft s mi h. .b g.
For the third quarter mile of the race,
v3
1 320
55 5 37 7= =
ft
23.8 s
ft s mi h. .b g,
and during the final quarter mile,
v4
1 320
57 4 39 0= =
ft
23.0 s
ft s mi h. .b g.
continued on next page

Chapter 2 27
(b) Assuming that v vf = 4 and recognizing that vi = 0 , the average acceleration during the race
was
a
v vf i
=
−
=
−
+ + +( )
=
total elapsed time
ft s
s
ft s257 4 0
25 2 24 0 23 8 23 0
0 598
.
. . . .
. .
P2.14 (a) Acceleration is the slope of the graph of v vs t.
For 0 5 00< <t . s, a = 0 .
For 15 0 20 0. .s s< <t , a = 0 .
For 5 0 15 0. .s s< <t , a
v v
t t
f i
f i
=
−
−
.
a =
− −( )
−
=
8 00 8 00
15 0 5 00
1 60
. .
. .
. m s2
We can plot a t( ) as shown.
0.0
1.0
1050 15 20
t (s)
1.6
2.0
a (m/s2)
FIG. P2.14
(b) a
v v
t t
f i
f i
=
−
−
(i) For 5 00 15 0. .s s< <t , ti = 5 00. s , vi =−8 00. m s ,
t
v
a
v v
t t
f
f
f i
f i
=
=
=
−
−
=
− −
−
=
15 0
8 00
8 00 8 00
15 0 5 00
1 60
.
.
. .
. .
. .
s
m s
m s2a f
(ii) ti = 0 , vi =−8 00. m s , tf = 20 0. s, v f = 8 00. m s
a
v v
t t
f i
f i
=
−
−
=
− −( )
−
=
8 00 8 00
20 0 0
0 800
. .
.
. m s2
P2.15 x t t= + −2 00 3 00 2
. . , v
dx
dt
t= = −3 00 2 00. . , a
dv
dt
= =−2 00.
At t = 3 00. s :
(a) x = + −( ) =2 00 9 00 9 00 2 00. . . .m m
(b) v = −( ) = −3 00 6 00 3 00. . .m s m s
(c) a = −2 00. m s2

P2.16 (a) At t = 2 00. s, x = ( ) − ( )+ =3 00 2 00 2 00 2 00 3 00 11 0
2
. . . . . .m m.
At t = 3 00. s , x = − + =3 00 9 00 2 00 3 00 3 00 24 0
2
. . . . . .a f a f m m
so
v
x
t
= =
−
−
=
∆
∆
24 0 11 0
2 00
13 0
. .
.
.
m m
3.00 s s
m s .
(b) At all times the instantaneous velocity is
v
d
dt
t t t= − + = −( )3 00 2 00 3 00 6 00 2 002
. . . . .c h m s
At t = 2 00. s, v = ( )− =6 00 2 00 2 00 10 0. . . .m s m s .
At t = 3 00. s , v = ( )− =6 00 3 00 2 00 16 0. . . .m s m s .
(c) a
v
t
= =
−
−
=
∆
∆
16 0 10 0
3 00 2 00
6 00
. .
. .
.
m s m s
s s
m s2
(d) At all times a
d
dt
= −( )=6 00 2 00 6 00. . . m s2
. (This includes both t = 2 00. s and t = 3 00. s ).
P2.17 (a) a
v
t
= = =
∆
∆
8 00
6 00
1 3
.
.
.
m s
s
m s2
(b) Maximum positive acceleration is at t = 3 s, and is approximately 2 m s2
.
(c) a = 0 , at t = 6 s , and also for t >10 s .
(d) Maximum negative acceleration is at t = 8 s, and is approximately −1 5. m s2
.
Section 2.4 Motion Diagrams
P2.18 (a)
(b)
(c)
(d)
(e)

Chapter 2 29
(f) One way of phrasing the answer: The spacing of the successive positions would change
with less regularity.
Another way: The object would move with some combination of the kinds of motion shown
in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude
and direction.
Section 2.5 One-Dimensional Motion with Constant Acceleration
P2.19 From v v axf i
2 2
2= + , we have 10 97 10 0 2 2203 2
. × = + ( )m s mc h a , so that a = ×2 74 105
. m s2
which is a g= ×2 79 104
. times .
P2.20 (a) x x v v tf i i f− = +
1
2
c h becomes 40
1
2
2 80 8 50m m s s= + ( )vi . .a f which yields vi = 6 61. m s .
(b) a
v v
t
f i
=
−
=
−
= −
2 80 6 61
8 50
0 448
. .
.
.
m s m s
s
m s2
P2.21 Given vi =12 0. cm s when x ti = =( )3 00 0. cm , and at t = 2 00. s, x f =−5 00. cm,
x x v t atf i i− = +
1
2
2
: − − = ( )+ ( )5 00 3 00 12 0 2 00
1
2
2 00
2
. . . . .a
− = +8 00 24 0 2. . a a =− = −
32 0
2
16 0
.
. cm s2
.
*P2.22 (a) Let i be the state of moving at 60 mi h and f be at rest
v v a x x
a
a
xf xi x f i
x
x
2 2
2
2
0 60 2 121 0
1
3 600
242
5 280 1
21 8
21 8
1 609 1
9 75
= + −
= + −
F
HG
I
KJ
=
− F
HG I
KJF
HG
I
KJ = − ⋅
= − ⋅
F
HG I
KJF
HG
I
KJ = −
d i
b g a fmi h ft
mi
5 280 ft
mi
h
ft
1 mi
h
3 600 s
mi h s
mi h s
m
1 mi
h
3 600 s
m s
2
2
.
. . .
(b) Similarly,
0 80 2 211 0
6 400 5 280
422 3 600
22 2 9 94
2
= + −
= − ⋅ = − ⋅ = −
mi h ft
mi h s mi h s m s2
b g a f
b g
b g
a
a
x
x . . .
(c) Let i be moving at 80 mi h and f be moving at 60 mi h.
v v a x x
a
a
xf xi x f i
x
x
2 2
2 2
2
60 80 2 211 121
2 800 5 280
2 90 3 600
22 8 10 2
= + −
= + −
= − ⋅ = − ⋅ = −
d i
b g b g a f
b g
a fb g
mi h mi h ft ft
mi h s mi h s m s2
. . .

*P2.23 (a) Choose the initial point where the pilot reduces the throttle and the final point where the
boat passes the buoy:
xi = 0 , x f =100 m, vxi = 30 m s, vxf =?, ax =−3 5. m s2
, t =?
x x v t a tf i xi x= + +
1
2
2
:
100 0 30
1
2
3 5 2
m m s m s2
= + + −a f c ht t.
1 75 30 100 02
. m s m s m2
c h a ft t− + = .
We use the quadratic formula:
t
b b ac
a
=
− ± −2
4
2
t =
± − ( )
=
±
=
30 900 4 1 75 100
2 1 75
30 14 1
3 5
12 6
m s m s m s m
m s
m s m s
m s
s
2 2 2
2 2
.
.
.
.
.
c h
c h
or 4 53. s .
The smaller value is the physical answer. If the boat kept moving with the same acceleration,
it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s.
(b) v v a txf xi x= + = − =30 3 5 4 53 14 1m s m s s m s2
. . .e j
P2.24 (a) Total displacement = area under the v t,a f curve from t = 0
to 50 s.
∆
∆
x
x
= + −
+
=
1
2
50 15 50 40 15
1
2
50 10
1 875
m s s m s s
m s s
m
b ga f b ga f
b ga f
(b) From t =10 s to t = 40 s , displacement is
∆x = + + =
1
2
50 33 5 50 25 1 457m s m s s m s s mb ga f b ga f .
FIG. P2.24
(c) 0 15≤ ≤t s: a
v
t
1
50 0
15 0
3 3= =
−( )
−
=
∆
∆
m s
s
m s2
.
15 40s s< <t : a2 0=
40 50s s≤ ≤t : a
v
t
3
0 50
50 40
5 0= =
−( )
−
= −
∆
∆
m s
s s
m s2
.

Chapter 2 31
(d) (i) x a t t1 1
2 2
0
1
2
1
2
3 3= + = . m s2
c h or x t1
2
1 67= . m s2
c h
(ii) x t2
1
2
15 50 0 50 15= ( ) − + −( )s m s m s sa f or x t2 50 375= −m s ma f
(iii) For 40 50s s≤ ≤t ,
x
v t
t
a t t3 3
2
0
1
2
40 50 40=
=
F
HG I
KJ+ −( ) + −( )
area under vs
from to 40 s
s m s sa f
or
x t t3
2
375 1 250
1
2
5 0 40 50 40= + + − − + −m m m s s m s s2
.e ja f b ga f
which reduces to
x t t3
2
250 2 5 4 375= − −m s m s m2
b g e j. .
(e) v = = =
total displacement
total elapsed time
m
s
m s
1 875
50
37 5.
P2.25 (a) Compare the position equation x t t= + −2 00 3 00 4 00 2
. . . to the general form
x x v t atf i i= + +
1
2
2
to recognize that xi = 2 00. m, vi = 3 00. m s, and a =−8 00. m s2
. The velocity equation,
v v atf i= + , is then
v tf = −3 00 8 00. .m s m s2
c h .
The particle changes direction when v f = 0, which occurs at t =
3
8
s . The position at this
time is:
x = +
F
HG I
KJ−
F
HG I
KJ =2 00 3 00
3
8
4 00
3
8
2 56
2
. . . .m m s s m s s m2
a f c h .
(b) From x x v t atf i i= + +
1
2
2
, observe that when x xf i= , the time is given by t
v
a
i
=−
2
. Thus,
when the particle returns to its initial position, the time is
t =
−
−
=
2 3 00
8 00
3
4
.
.
m s
m s
s2
a f
and the velocity is v f = −
F
HG I
KJ= −3 00 8 00
3
4
3 00. . .m s m s s m s2
c h .

*P2.26 The time for the Ford to slow down we find from
x x v v t
t
x
v v
f i xi xf
xi xf
= + +
=
+
=
+
=
1
2
2 2 250
71 5 0
6 99
d i
a f∆ m
m s
s
.
. .
Its time to speed up is similarly
t =
( )
+
=
2 350
0 71 5
9 79
m
m s
s
.
. .
The whole time it is moving at less than maximum speed is 6 99 5 00 9 79 21 8. . . .s s s s+ + = . The
Mercedes travels
x x v v tf i xi xf= + + = + +
=
1
2
0
1
2
71 5 71 5 21 8
1 558
d i a fb ga f. . .m s s
m
while the Ford travels 250 350 600+ =m m, to fall behind by 1 558 600 958m m m− = .
P2.27 (a) vi =100 m s, a =−5 00. m s2
, v v atf i= + so 0 100 5= − t, v v a x xf i f i
2 2
2= + −c h so
0 100 2 5 00 0
2
=( ) − ( ) −. x fc h. Thus x f = 1 000 m and t = 20 0. s .
(b) At this acceleration the plane would overshoot the runway: No .
P2.28 (a) Take ti = 0 at the bottom of the hill where xi = 0 , vi = 30 0. m s, a =−2 00. m s2
. Use these
values in the general equation
x x v t atf i i= + +
1
2
2
to find
x t tf = + + −0 30 0
1
2
2 00 2
. .m s m s2
a f c h
when t is in seconds
x t tf = −30 0 2
.c hm .
To find an equation for the velocity, use v v at tf i= + = + −30 0 2 00. .m s m s2
e j ,
v tf = −( )30 0 2 00. . m s .
(b) The distance of travel x f becomes a maximum, xmax , when v f = 0 (turning point in the
motion). Use the expressions found in part (a) for v f to find the value of t when x f has its
maximum value:
From v tf = −( )3 00 2 00. . m s, v f = 0 when t =15 0. s. Then
x t tmax . . . .= − =( )( )−( ) =30 0 30 0 15 0 15 0 2252 2
c hm m .

Chapter 2 33
P2.29 In the simultaneous equations:
v v a t
x x v v t
xf xi x
f i xi xf
= +
− = +
R
S|
T|
U
V|
W|1
2
c h we have
v v
v v
xf xi
xi xf
= − ( )
= + ( )
R
S|
T|
U
V|
W|
5 60 4 20
62 4
1
2
4 20
. .
. .
m s s
m s
2
c h
c h
.
So substituting for vxi gives 62 4
1
2
56 0 4 20 4 20. . . .m m s s s2
= + ( )+ ( )v vxf xfc h
14 9
1
2
5 60 4 20. . .m s m s s2
= + ( )vxf c h .
Thus
vxf = 3 10. m s .
P2.30 Take any two of the standard four equations, such as
v v a t
x x v v t
xf xi x
f i xi xf
= +
− = +
R
S|
T|
U
V|
W|1
2
c h . Solve one for vxi , and
substitute into the other: v v a txi xf x= −
x x v a t v tf i xf x xf− = − +
1
2
c h .
Thus
x x v t a tf i xf x− = −
1
2
2
.
Back in problem 29, 62 4 4 20
1
2
5 60 4 20
2
. . . .m s m s s2
= ( )− − ( )vxf c h
vxf =
−
=
62 4 49 4
3 10
. .
.
m m
4.20 s
m s .
P2.31 (a) a
v v
t
f i
=
−
= = − = −
632
1 40
662 202
5 280
3 600e j
.
ft s m s2 2
(b) x v t atf i= + =
F
HG
I
KJ − = =
1
2
632
5 280
3 600
1 40
1
2
662 1 40 649 1982 2
a f a f a fa f. . ft m

P2.32 (a) The time it takes the truck to reach 20 0. m s is found from v v atf i= + . Solving for t yields
t
v v
a
f i
=
−
=
−
=
20 0 0
2 00
10 0
.
.
.
m s m s
m s
s2
.
The total time is thus
10 0 20 0 5 00 35 0. . . .s s s s+ + = .
(b) The average velocity is the total distance traveled divided by the total time taken. The
distance traveled during the first 10.0 s is
x vt1
0 20 0
2
10 0 100= =
+F
HG I
KJ( )=
.
. m.
With a being 0 for this interval, the distance traveled during the next 20.0 s is
x v t ati2
21
2
20 0 20 0 0 400= + =( )( )+ =. . m.
The distance traveled in the last 5.00 s is
x vt3
20 0 0
2
5 00 50 0= =
+F
HG I
KJ( )=
.
. . m.
The total distance x x x x= + + = + + =1 2 3 100 400 50 550 m, and the average velocity is
given by v
x
t
= = =
550
35 0
15 7
.
. m s .
P2.33 We have vi = ×2 00 104
. m s, v f = ×6 00 106
. m s, x xf i− = × −
1 50 10 2
. m.
(a) x x v v tf i i f− = +
1
2
c h : t
x x
v v
f i
i f
=
−
+
=
×
× + ×
= ×
−
−
2 2 1 50 10
2 00 10 6 00 10
4 98 10
2
4 6
9
c h c h.
. .
.
m
m s m s
s
(b) v v a x xf i x f i
2 2
2= + −d i:
a
v v
x x
x
f i
f i
=
−
−
=
× − ×
×
= ×−
2 2 6 2 4 2
2
15
2
6 00 10 2 00 10
2 1 50 10
1 20 10
( )
. .
( . )
.
m s m s
m
m s2e j e j

Chapter 2 35
*P2.34 (a) v v a x xxf xi x f i
2 2
2= + −c h: 0 01 3 10 0 2 408
2
. × = + ( )m s mc h ax
ax =
×
= ×
3 10
80
1 12 10
6 2
11
m s
m
m s2
c h .
(b) We must find separately the time t1 for speeding up and the time t2 for coasting:
x x v v t t
t
f i xf xi− = + = × +
= × −
1
2
40
1
2
3 10 0
2 67 10
1
6
1
1
5
d i e j: m m s
s.
x x v v t t
t
f i xf xi− = + = × + ×
= × −
1
2
60
1
2
3 10 3 10
2 00 10
2
6 6
2
2
5
d i e j:
.
m m s m s
s
total time = × −
4 67 10 5
. s .
*P2.35 (a) Along the time axis of the graph shown, let i = 0 and f tm= . Then v v a txf xi x= + gives
v a tc m m= +0
a
v
t
m
c
m
= .
(b) The displacement between 0 and tm is
x x v t a t
v
t
t v tf i xi x
c
m
m c m− = + = + =
1
2
0
1
2
1
2
2 2
.
The displacement between tm and t0 is
x x v t a t v t tf i xi x c m− = + = − +
1
2
02
0a f .
The total displacement is
∆x v t v t v t v t tc m c c m c m= + − = −
F
HG I
KJ1
2
1
2
0 0 .
(c) For constant vc and t0 , ∆x is minimized by maximizing tm to t tm = 0 . Then
∆x v t t
v t
c
c
min = −
F
HG I
KJ=0 0
01
2 2
.
(e) This is realized by having the servo motor on all the time.
(d) We maximize ∆x by letting tm approach zero. In the limit ∆x v t v tc c= − =0 00a f .
(e) This cannot be attained because the acceleration must be finite.

*P2.36 Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its
motion from entry to exit,
x x v t a t
v t a t v t
v v a t
f i xi x
i d d d d
d i d
= + +
= + + =
= +
1
2
0
1
2
1
2
2
2
∆ ∆ ∆
∆
(a) The speed halfway through the photogate in space is given by
v v a v av ths i i d d
2 2 2
2
2
= +
F
HG I
KJ= + ∆ .
v v av ths i d d= +2
∆ and this is not equal to vd unless a = 0 .
(b) The speed halfway through the photogate in time is given by v v a
t
ht i
d
= +
F
HG I
KJ∆
2
and this is
equal to vd as determined above.
P2.37 (a) Take initial and final points at top and bottom of the incline. If the ball starts from rest,
vi = 0 , a = 0 500. m s2
, x xf i− = 9 00. m.
Then
v v a x x
v
f i f i
f
2 2 2
2 0 2 0 500 9 00
3 00
= + − = +
=
d i e ja f. .
. .
m s m
m s
2
(b) x x v t atf i i− = +
1
2
2
9 00 0
1
2
0 500
6 00
2
. .
.
= +
=
m s
s
2
e jt
t
(c) Take initial and final points at the bottom of the planes and the top of the second plane,
respectively:
vi = 3 00. m s, v f = 0, x xf i− =15 00. m.
v v a x xf i f i
2 2
2= + −c h gives
a
v v
x x
f i
f i
=
−
−
=
−
( )
= −
2 2
2
2
0 3 00
2 15 0
0 300
c h
a f.
.
.
m s
m
m s2
.
(d) Take the initial point at the bottom of the planes and the final point 8.00 m along the second:
vi = 3 00. m s, x xf i− = 8 00. m, a =−0 300. m s2
v v a x x
v
f i f i
f
2 2 2
2 3 00 2 0 300 8 00 4 20
2 05
= + − = + − =
=
d i b g e ja f. . . .
. .
m s m s m m s
m s
2 2 2

Chapter 2 37
P2.38 Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sue’s car.
For her we have xis = 0, vis = 30 0. m s, as =−2 00. m s2
so her position is given by
x t x v t a t t ts is is s( )= + + = + −
1
2
30 0
1
2
2 002 2
. .m s m s2
a f c h .
For the van, xiv =155 m, viv = 5 00. m s , av = 0 and
x t x v t a t tv iv iv v( )= + + = + +
1
2
155 5 00 02
. m sa f .
To test for a collision, we look for an instant tc when both are at the same place:
30 0 155 5 00
0 25 0 155
2
2
. .
. .
t t t
t t
c c c
c c
− = +
= − +
From the quadratic formula
tc =
± ( ) − ( )
=
25 0 25 0 4 155
2
13 6
2
. .
. s or 11 4. s .
The smaller value is the collision time. (The larger value tells when the van would pull ahead again
if the vehicles could move through each other). The wreck happens at position
155 5 00 11 4 212m m s s m+ ( )=. .a f .
*P2.39 As in the algebraic solution to Example 2.8, we let t
represent the time the trooper has been moving. We graph
x tcar = +45 45
and
x ttrooper = 1 5 2
. .
They intersect at
t = 31 s .
x (km)
t (s)
10 20 30 40
0.5
1
1.5
car
police
officer
FIG. P2.39

Section 2.6 Freely Falling Objects
P2.40 Choose the origin y t= =0 0,a f at the starting point of the ball and take upward as positive. Then
yi = 0 , vi = 0 , and a g=− =−9 80. m s2
. The position and the velocity at time t become:
y y v t atf i i− = +
1
2
2
: y gt tf = − = −
1
2
1
2
9 802 2
. m s2
e j
and
v v atf i= + : v gt tf =− =− 9 80. m s2
c h .
(a) at t =1 00. s: y f =− ( ) = −
1
2
9 80 1 00 4 90
2
. . .m s s m2
c h
at t = 2 00. s: y f =− ( ) = −
1
2
9 80 2 00 19 6
2
. . .m s s m2
c h
at t = 3 00. s : y f =− ( ) = −
1
2
9 80 3 00 44 1
2
. . .m s s m2
c h
(b) at t =1 00. s: v f =− ( )= −9 80 1 00 9 80. . .m s s m s2
c h
at t = 2 00. s: v f =− ( )= −9 80 2 00 19 6. . .m s s m s2
c h
at t = 3 00. s : v f =− ( )= −9 80 3 00 29 4. . .m s s m s2
c h
P2.41 Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is
that due to gravity, a g=− =−9 80. m s2
. During the flight, Goff went 1 mile (1 609 m) up and then
1 mile back down. Determine his speed just after launch by considering his upward flight:
v v a y y v
v
f i f i i
i
2 2 2
2 0 2 9 80 1 609
178
= + − = −
=
d i e jb g: .
.
m s m
m s
2
His time in the air may be found by considering his motion from just after launch to just before
impact:
y y v t atf i i− = +
1
2
2
: 0 178
1
2
9 80 2
= − −m s m s2
a f c ht t. .
The root t = 0 describes launch; the other root, t = 36 2. s, describes his flight time. His rate of pay
may then be found from
pay rate = = =
$1.
.
. $99.
00
36 2
0 027 6 3 600 3
s
$ s s h hb gb g .
We have assumed that the workman’s flight time, “a mile”, and “a dollar”, were measured to three-
digit precision. We have interpreted “up in the sky” as referring to the free fall time, not to the
launch and landing times. Both the takeoff and landing times must be several seconds away from
the job, in order for Goff to survive to resume work.

Chapter 2 39
P2.42 We have y gt v t yf i i=− + +
1
2
2
0 4 90 8 00 30 02
=− − +. . .m s m s m2
c h a ft t .
Solving for t,
t =
± +
−
8 00 64 0 588
9 80
. .
.
.
Using only the positive value for t, we find that t = 1 79. s .
P2.43 (a) y y v t atf i i− = +
1
2
2
: 4 00 1 50 4 90 1 50
2
. . . .=( ) −( )( )vi and vi = 10 0. m s upward .
(b) v v atf i= + = −( )( )=−10 0 9 80 1 50 4 68. . . . m s
v f = 4 68. m s downward
P2.44 The bill starts from rest vi = 0 and falls with a downward acceleration of 9 80. m s2
(due to gravity).
Thus, in 0.20 s it will fall a distance of
∆y v t gti= − = − ( ) =−
1
2
0 4 90 0 20 0 202 2
. . .m s s m2
c h .
This distance is about twice the distance between the center of the bill and its top edge ≅ 8 cma f.
Thus, David will be unsuccessful .
*P2.45 (a) From ∆y v t ati= +
1
2
2
with vi = 0 , we have
t
y
a
= =
−( )
−
=
2 2 23
9 80
2 17
∆a f m
m s
s2
.
. .
(b) The final velocity is v f = + − ( )= −0 9 80 2 17 21 2. . .m s s m s2
c h .
(c) The time take for the sound of the impact to reach the spectator is
t
y
v
sound
sound
m
340 m s
s= = = × −∆ 23
6 76 10 2
. ,
so the total elapsed time is ttotal s s s= + × ≈−
2 17 6 76 10 2 232
. . . .

P2.46 At any time t, the position of the ball released from rest is given by y h gt1
21
2
= − . At time t, the
position of the ball thrown vertically upward is described by y v t gti2
21
2
= − . The time at which the
first ball has a position of y
h
1
2
= is found from the first equation as
h
h gt
2
1
2
2
= − , which yields
t
h
g
= . To require that the second ball have a position of y
h
2
2
= at this time, use the second
equation to obtain
h
v
h
g
g
h
g
i
2
1
2
= −
F
HG
I
KJ. This gives the required initial upward velocity of the second
ball as v ghi = .
P2.47 (a) v v gtf i= − : v f = 0 when t = 3 00. s , g = 9 80. m s2
. Therefore,
v gti = = ( )=9 80 3 00 29 4. . .m s s m s2
c h .
(b) y y v v tf i f i− = +
1
2
c h
y yf i− = =
1
2
29 4 3 00 44 1. . .m s s mb ga f
*P2.48 (a) Consider the upward flight of the arrow.
v v a y y
y
y
yf yi y f i
2 2
2
2
0 100 2 9 8
10 000
19 6
510
= + −
= + −
= =
d i
b g e jm s m s
m s
m s
m
2
2 2
2
.
.
∆
∆
(b) Consider the whole flight of the arrow.
y y v t a t
t t
f i yi y= + +
= + + −
1
2
0 0 100
1
2
9 8
2
2
m s m s2
b g e j.
The root t = 0 refers to the starting point. The time of flight is given by
t = =
100
4 9
20 4
m s
m s
s2
.
. .
P2.49 Time to fall 3.00 m is found from Eq. 2.12 with vi = 0 , 3 00
1
2
9 80 2
. .m m s2
= c ht , t = 0 782. s.
(a) With the horse galloping at 10 0. m s, the horizontal distance is vt = 7 82. m .
(b) t = 0 782. s

Chapter 2 41
P2.50 Take downward as the positive y direction.
(a) While the woman was in free fall,
∆y =144 ft , vi = 0 , and a g= = 32 0. ft s2
.
Thus, ∆y v t at ti= + → = +
1
2
144 0 16 02 2
ft ft s2
.c h giving tfall s= 3 00. . Her velocity just
before impact is:
v v gtf i= + = + ( )=0 32 0 3 00 96 0. . .ft s s ft s2
c h .
(b) While crushing the box, vi = 96 0. ft s, v f = 0, and ∆y = =18 0 1 50. .in. ft. Therefore,
a
v v
y
f i
=
−
=
−
( )
=− ×
2 2 2
3
2
0 96 0
2 1 50
3 07 10
∆a f
a f.
.
.
ft s
ft
ft s2
, or a = ×3 07 103
. ft s upward2
.
(c) Time to crush box: ∆
∆ ∆
t
y
v
y
v vf i
= = =
( )
++
2
2 1 50
0 96 0
.
.
ft
ft s
or ∆t = × −
3 13 10 2
. s .
P2.51 y t= 3 00 3
. : At t = 2 00. s, y = =3 00 2 00 24 0
3
. . .a f m and
v
dy
dt
ty = = = A9 00 36 02
. . m s .
If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is
y y v t gt t tb bi i= + − = + − ( )
1
2
24 0 36 0
1
2
9 802 2
. . . .
Setting yb = 0,
0 24 0 36 0 4 90 2
= + −. . .t t .
Solving for t, (only positive values of t count), t = 7 96. s .
*P2.52 Consider the last 30 m of fall. We find its speed 30 m above the ground:
y y v t a t
v
v
f i yi y
yi
yi
= + +
= + + −
=
− +
= −
1
2
0 30 1 5
1
2
9 8 1 5
30 11 0
12 6
2
2
m s m s s
m m
1.5 s
m s
2
. . .
.
. .
a f e ja f
Now consider the portion of its fall above the 30 m point. We assume it starts from rest
v v a y y
y
y
yf yi y f i
2 2
2
2
12 6 0 2 9 8
160
19 6
8 16
= + −
− = + −
=
−
= −
d i
b g e j. .
.
. .
m s m s
m s
m s
m
2
2 2
2
∆
∆
Its original height was then 30 8 16 38 2m m m+ − =. . .

Section 2.7 Kinematic Equations Derived from Calculus
P2.53 (a) J
da
dt
= = constant
da Jdt=
a J dt Jt c= = +z 1
but a ai= when t = 0 so c ai1 = . Therefore, a Jt ai= +
a
dv
dt
dv adt
v adt Jt a dt Jt a t ci i
=
=
= = + = + +z zb g 1
2
2
2
but v vi= when t = 0, so c vi2 = and v Jt a t vi i= + +
1
2
2
v
dx
dt
dx vdt
x vdt Jt a t v dt
x Jt a t v t c
x x
i i
i i
i
=
=
= = + +
F
HG I
KJ
= + + +
=
z z 1
2
1
6
1
2
2
3 2
3
when t = 0, so c xi3 = . Therefore, x Jt a t v t xi i i= + + +
1
6
1
2
3 2
.
(b) a Jt a J t a Ja ti i i
2 2 2 2 2
2= + = + +a f
a a J t Ja ti i
2 2 2 2
2= + +c h
a a J Jt a ti i
2 2 2
2
1
2
= + +
F
HG I
KJ
Recall the expression for v: v Jt a t vi i= + +
1
2
2
. So v v Jt a ti i− = +a f 1
2
2
. Therefore,
a a J v vi i
2 2
2= + −a f .

Chapter 2 43
P2.54 (a) See the graphs at the right.
Choose x = 0 at t = 0.
At t = 3 s, x = ( )=
1
2
8 3 12m s s ma f .
At t = 5 s, x = + ( )=12 8 2 28m m s s ma f .
At t = 7 s, x = + ( )=28
1
2
8 2 36m m s s ma f .
(b) For 0 3< <t s, a = =
8
3
2 67
m s
s
m s2
. .
For 3 5< <t s, a = 0 .
(c) For 5 9s s< <t , a =− = −
16
4
4
m s
s
m s2
.
(d) At t = 6 s, x = + ( )=28 6 1 34m m s s ma f .
(e) At t = 9 s, x = + − ( )=36
1
2
8 2 28m m s s ma f .
FIG. P2.54
P2.55 (a) a
dv
dt
d
dt
t t= = − × + ×5 00 10 3 00 107 2 5
. .
a t=− × + ×10 0 10 3 00 107 5
. .m s m s3 2
c h
Take xi = 0 at t = 0. Then v
dx
dt
=
x vdt t t dt
x
t t
x t t
t t
− = = − × + ×
= − × + ×
= − × + ×
z z0 5 00 10 3 00 10
5 00 10
3
3 00 10
2
1 67 10 1 50 10
0
7 2 5
0
7
3
5
2
7 3 5 2
. .
. .
. . .
e j
e j e jm s m s3 2
(b) The bullet escapes when a = 0 , at − × + × =10 0 10 3 00 10 07 5
. .m s m s3 2
c ht
t =
×
×
= × −3 00 10
3 00 10
5
3.
.
s
10.0 10
s7
.
(c) New v = − × × + × ×− −
5 00 10 3 00 10 3 00 10 3 00 107 3 2 5 3
. . . .c hc h c hc h
v =− + =450 900 450m s m s m s .
(d) x =− × × + × ×− −
1 67 10 3 00 10 1 50 10 3 00 107 3 3 5 3 2
. . . .c hc h c hc h
x =− + =0 450 1 35 0 900. . .m m m

P2.56 a
dv
dt
v= =−3 00 2
. , vi =1 50. m s
Solving for v,
dv
dt
v=−3 00 2
.
v dv dt
v v
t t
v v
v v
v
t
t
i i
i
−
= =
z z= −
− + = − = −
2
0
3 00
1 1
3 00 3 00
1 1
.
. . .or
When v
vi
=
2
, t
vi
= =
1
3 00
0 222
.
. s .
Additional Problems
*P2.57 The distance the car travels at constant velocity, v0 , during the reaction time is ∆ ∆x v tra f1 0= . The
time for the car to come to rest, from initial velocity v0 , after the brakes are applied is
t
v v
a
v
a
v
a
f i
2
0 00
=
−
=
−
=−
and the distance traveled during this braking period is
∆x vt
v v
t
v v
a
v
a
f i
a f2 2 2
0 0 0
2
2
0
2 2
= =
+F
HG
I
KJ =
+F
HG I
KJ −
F
HG I
KJ = − .
Thus, the total distance traveled before coming to a stop is
s x x v t
v
a
rstop = + = −∆ ∆ ∆a f a f1 2 0
0
2
2
.
*P2.58 (a) If a car is a distance s v t
v
a
rstop = −0
0
2
2
∆ (See the solution to Problem 2.57) from the
intersection of length si when the light turns yellow, the distance the car must travel before
the light turns red is
∆ ∆x s s v t
v
a
si r i= + = − +stop 0
0
2
2
.
Assume the driver does not accelerate in an attempt to “beat the light” (an extremely
dangerous practice!). The time the light should remain yellow is then the time required for
the car to travel distance ∆x at constant velocity v0 . This is
∆
∆ ∆
∆t
x
v
v t s
v
t
v
a
s
v
r
v
a i
r
i
light = =
− +
= − +
0
0 2
0
0
0
0
2
2
.
(b) With si =16 m, v = 60 km h, a = −2 0. m s2
, and ∆tr = 1 1. s,
∆tlight 2
s
km h
m s
m s
km h
m
60 km h
km h
m s
s= −
−
F
HG
I
KJ+
F
HG
I
KJ =1 1
60
2 2 0
0 278
1
16 1
0 278
6 23.
.
.
.
.
e j
.

Chapter 2 45
*P2.59 (a) As we see from the graph, from about −50 s to 50 s
Acela is cruising at a constant positive velocity in
the +x direction. From 50 s to 200 s, Acela
accelerates in the +x direction reaching a top speed
of about 170 mi/h. Around 200 s, the engineer
applies the brakes, and the train, still traveling in
the +x direction, slows down and then stops at
350 s. Just after 350 s, Acela reverses direction (v
becomes negative) and steadily gains speed in the
−x direction.
t(s)
100 200 300
–100
100
200
400
∆v
∆t
–50
0
0
FIG. P2.59(a)
(b) The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangent
to the v versus t curve in this interval. From the tangent line shown, we find
a
v
t
= = =
−( )
−( )
= =slope
mi h
s
mi h s m s2∆
∆
155 45
100 50
2 2 0 98. .a f .
(c) Let us use the fact that the area under the v versus
t curve equals the displacement. The train’s
displacement between 0 and 200 s is equal to the
area of the gray shaded region, which we have
approximated with a series of triangles and
rectangles.
∆x0 200
50 50 50 50
160 100
1
2
100
1
2
100 170 160
24 000
→ = + + + +
≈ +
+
+
+ −
=
s 1 2 3 4 5area area area area area
mi h s mi h s
mi h s
50 s mi h
s mi h mi h
mi h s
b ga f b ga f
b ga f
a fb g
a fb g
b ga f
t(s)
100 200 300
100
200
400
1 2
4 3
5
0
0
FIG. P2.59(c)
Now, at the end of our calculation, we can find the displacement in miles by converting
hours to seconds. As 1 3 600h s= ,
∆x0 200
24 000
6 7→ ≈
F
HG
I
KJ =s
mi
3 600 s
s mia f . .

*P2.60 Average speed of every point on the train as the first car passes Liz:
∆
∆
x
t
= =
8 60
5 73
.
.
m
1.50 s
m s.
The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway
through the next 1.10 s, the speed of the train is
8 60
7 82
.
.
m
1.10 s
m s= . The time required for the speed
to change from 5.73 m/s to 7.82 m/s is
1
2
1 50
1
2
1 10 1 30. . .s s s( )+ ( )=
so the acceleration is: a
v
t
x
x
= =
−
=
∆
∆
7 82 5 73
1 30
1 60
. .
.
.
m s m s
s
m s2
.
P2.61 The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then
vxi =1 04. mm d and ax =
F
HG I
KJ0 132.
mm d
w
. The increase in the length of the hair (i.e., displacement)
during a time of t = =5 00 35 0. .w d is
∆
∆
x v t a t
x
xi x= +
= + ⋅
1
2
1 04 35 0
1
2
0 132 35 0 5 00
2
. . . . .mm d d mm d w d wb ga f b ga fa f
or ∆x = 48 0. mm .
P2.62 Let point 0 be at ground level and point 1 be at the end of the engine burn. Let
point 2 be the highest point the rocket reaches and point 3 be just before
impact. The data in the table are found for each phase of the rocket’s motion.
(0 to 1) v f
2 2
80 0 2 4 00 1 000− =. .a f a fb g so v f =120 m s
120 80 0 4 00= +( ). . t giving t =10 0. s
(1 to 2) 0 120 2 9 80
2
−( ) = −( ) −. x xf ic h giving x xf i− = 735 m
0 120 9 80− =− . t giving t =12 2. s
This is the time of maximum height of the rocket.
(2 to 3) v f
2
0 2 9 80 1 735− = − −.a fb g
v tf =− = −( )184 9 80. giving t =18 8. s
FIG. P2.62
(a) ttotal s= + + =10 12 2 18 8 41 0. . .
(b) x xf i− =c htotal
km1 73.

Chapter 2 47
(c) vfinal m s= −184
t x v a
0 Launch 0.0 0 80 +4.00
#1 End Thrust 10.0 1 000 120 +4.00
#2 Rise Upwards 22.2 1 735 0 –9.80
#3 Fall to Earth 41.0 0 –184 –9.80
P2.63 Distance traveled by motorist = 15 0. m sa ft
Distance traveled by policeman =
1
2
2 00 2
. m s2
c ht
(a) intercept occurs when 15 0 2
. t t= , or t = 15 0. s
(b) v tofficer m s m s2
( )= =2 00 30 0. .c h
(c) x tofficer m s m2
( )= =
1
2
2 00 2252
.c h
P2.64 Area A1 is a rectangle. Thus, A hw v txi1 = = .
Area A2 is triangular. Therefore A bh t v vx xi2
1
2
1
2
= = −b g.
The total area under the curve is
A A A v t
v v t
xi
x xi
= + = +
−
1 2
2
b g
and since v v a tx xi x− =
A v t a txi x= +
1
2
2
.
The displacement given by the equation is: x v t a txi x= +
1
2
2
, the
same result as above for the total area.
vx
vx
vxi
0 t
t
A2
A1
FIG. P2.64

P2.65 (a) Let x be the distance traveled at acceleration a until maximum speed v is reached. If this is
achieved in time t1 we can use the following three equations:
x v v ti= +
1
2
1a f , 100 10 2 1− = −x v t.a fand v v ati= + 1 .
The first two give
100 10 2
1
2
10 2
1
2
200
20 4
1 1 1
1 1
= −
F
HG I
KJ = −
F
HG I
KJ
=
−
. .
.
.
t v t at
a
t tb g
For Maggie: m s
For Judy: m s
2
2
a
a
= =
= =
200
18 4 2 00
5 43
200
17 4 3 00
3 83
. .
.
. .
.
a fa f
a fa f
(b) v a t= 1
Maggie: m s
Judy: m s
v
v
= =
= =
5 43 2 00 10 9
3 83 3 00 11 5
. . .
. . .
a fa f
a fa f
(c) At the six-second mark
x at v t= + −
1
2
6 001
2
1.a f
Maggie: m
Judy: m
x
x
= + =
= + =
1
2
5 43 2 00 10 9 4 00 54 3
1
2
3 83 3 00 11 5 3 00 51 7
2
2
. . . . .
. . . . .
a fa f a fa f
a fa f a fa f
Maggie is ahead by 2 62. m .
P2.66 a1 0 100= . m s2
a2 0 500=− . m s2
x a t v t a t= = + +1 000
1
2
1
2
1 1
2
1 2 2 2
2
m t t t= +1 2 and v a t a t1 1 1 2 2= =−
1 000
1
2
1
2
1 1
2
1 1
1 1
2
2
1 1
2
2
= + −
F
HG I
KJ+
F
HG I
KJa t a t
a t
a
a
a t
a
1 000
1
2
11
1
2
1
2
= −
F
HG I
KJa
a
a
t
t1
20 000
1 20
129= =
.
s
t
a t
a
2
1 1
2
12 9
0 500
26=
−
= ≈
.
.
s Total time = =t 155 s

Chapter 2 49
P2.67 Let the ball fall 1.50 m. It strikes at speed given by
v v a x xxf xi f i
2 2
2= + −c h:
vxf
2
0 2 9 80 1 50= + − −( ). .m s m2
c h
vxf =−5 42. m s
and its stopping is described by
v v a x x
a
a
xf xi x f i
x
x
2 2
2 2
2
3
2
0 5 42 2 10
29 4
2 00 10
1 47 10
= + −
= − + −
=
−
− ×
= + ×
−
−
d i
b g e j.
.
.
. .
m s m
m s
m
m s
2 2
2
Its maximum acceleration will be larger than the average acceleration we estimate by imagining
constant acceleration, but will still be of order of magnitude ~103
m s2
.
*P2.68 (a) x x v t a tf i xi x= + +
1
2
2
. We assume the package starts from rest.
− = + + −145 0 0
1
2
9 80 2
m m s2
.c ht
t =
−( )
−
=
2 145
9 80
5 44
m
m s
s2
.
.
(b) x x v t a tf i xi x= + + = + + − ( ) =−
1
2
0 0
1
2
9 80 5 18 1312 2
. .m s s m2
c h
distance fallen = =x f 131 m
(c) speed = = + = + − =v v a txf xi x 0 9 8 5 18 50 8. . .m s s m s2
e j
(d) The remaining distance is
145 131 5 13 5m m m− =. . .
During deceleration,
vxi =−50 8. m s, vxf = 0, x xf i− =−13 5. m
v v a x xxf xi x f i
2 2
2= + −c h:
0 50 8 2 13 5
2
= − + −( ). .m s ma f ax
ax =
−
−
= + =
2 580
2 13 5
95 3 95 3
m s
m
m s m s upward
2 2
2 2
.
. .
a f .

P2.69 (a) y v t at t tf i= + = = + ( )1
2 21
2
50 0 2 00
1
2
9 80. . . ,
4 90 2 00 50 0 02
. . .t t+ − =
t =
− + − ( ) −( )
( )
2 00 2 00 4 4 90 50 0
2 4 90
2
. . . .
.
Only the positive root is physically meaningful:
t = 3 00. s after the first stone is thrown.
(b) y v t atf i= +2
21
2
and t = − =3 00 1 00 2 00. . . s
substitute 50 0 2 00
1
2
9 80 2 002
2
. . . .= ( )+ ( )( )vi :
vi2 15 3= . m s downward
(c) v v atf i1 1 2 00 9 80 3 00 31 4= + = +( )( )=. . . . m s downward
v v atf i2 2 15 3 9 80 2 00 34 8= + = +( )( )=. . . . m s downward
P2.70 (a) d t= ( )
1
2
9 80 1
2
. d t= 336 2
t t1 2 2 40+ = . 336 4 90 2 402 2
2
t t= −. .a f
4 90 359 5 28 22 02
2
2. . .t t− + = t2
2
359 5 359 5 4 4 90 28 22
9 80
=
± − ( )( ). . . .
.
t2
359 5 358 75
9 80
0 076 5=
±
=
. .
.
. s so d t= =336 26 42 . m
(b) Ignoring the sound travel time, d = ( )( ) =
1
2
9 80 2 40 28 2
2
. . . m, an error of 6 82%. .
P2.71 (a) In walking a distance ∆x, in a time ∆t, the length
of rope is only increased by ∆xsinθ .
∴ The pack lifts at a rate
∆
∆
x
t
sinθ .
v
x
t
v
x
v
x
x h
= = =
+
∆
∆
sinθ boy boy
2 2
(b) a
dv
dt
v dx
dt
v x
d
dt
= = +
F
HG I
KJboy
boy
1
a v
v v x d
dt
= −boy
boy boy
2
, but
d
dt
v v
x
= = boy
∴ = −
F
HG
I
KJ= =
+
a
v x v h h v
x h
boy
2
boy
2
boy
2
1
2
2
2
2
2
2 2 3 2
c h
(c)
v
h
boy
2
, 0
(d) vboy , 0
FIG. P2.71

Chapter 2 51
P2.72 h= 6 00. m, vboy m s= 2 00. v
x
t
v
x v x
x h
= = =
+
∆
∆
sinθ boy
boy
2 2 1 2
c h
.
However, x v t= boy : ∴ =
+
=
+
v
v t
v t h
t
t
boy
2
boy
2 2 2 1 2 2 1 2
4
4 36c h c h
.
(a) t vs m s
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
0.32
0.63
0.89
1.11
1.28
1.41
1.52
1.60
1.66
1.71
a f b g
FIG. P2.72(a)
(b) From problem 2.71 above, a
h v
x h
h v
v t h t
=
+
=
+
=
+
2
2 2 3 2
2
2 2 3 2 2 3 2
144
4 36
boy
2
boy
2
boy
2
c h c h c h
.
t as m s
0
0.5
1
1.5
2
2.5
3.
3.5
4.
4.5
5
0.67
0.64
0.57
0.48
0.38
0.30
0.24
0.18
0.14
0.11
0.09
2
a f e j
FIG. P2.72(b)
P2.73 (a) We require x xs k= when t ts k= + 1 00.
x t t x
t t
t
s k k k
k k
k
= + = =
+ =
=
1
2
3 50 1 00
1
2
4 90
1 00 1 183
5 46
2 2
. . .
. .
. .
m s m s
s
2 2
e jb g e jb g
(b) xk = =
1
2
4 90 5 46 73 0
2
. . .m s s m2
e ja f
(c) vk = =4 90 5 46 26 7. . .m s s m s2
e ja f
vs = =3 50 6 46 22 6. . .m s s m s2
e ja f

P2.74 Time
t (s)
Height
h (m)
∆h
(m)
∆t
(s)
v
(m/s)
midpt time
t (s)
0.00 5.00
0.75 0.25 3.00 0.13
FIG. P2.74
0.25 5.75
0.65 0.25 2.60 0.38
0.50 6.40
0.54 0.25 2.16 0.63
0.75 6.94
0.44 0.25 1.76 0.88
1.00 7.38
0.34 0.25 1.36 1.13
1.25 7.72
0.24 0.25 0.96 1.38
1.50 7.96
0.14 0.25 0.56 1.63
1.75 8.10
0.03 0.25 0.12 1.88
2.00 8.13
–0.06 0.25 –0.24 2.13
2.25 8.07
–0.17 0.25 –0.68 2.38
2.50 7.90
–0.28 0.25 –1.12 2.63
2.75 7.62
–0.37 0.25 –1.48 2.88
3.00 7.25
–0.48 0.25 –1.92 3.13
3.25 6.77
–0.57 0.25 –2.28 3.38
3.50 6.20
–0.68 0.25 –2.72 3.63
3.75 5.52
–0.79 0.25 –3.16 3.88
4.00 4.73
–0.88 0.25 –3.52 4.13
4.25 3.85
–0.99 0.25 –3.96 4.38
4.50 2.86
–1.09 0.25 –4.36 4.63
4.75 1.77
–1.19 0.25 –4.76 4.88
5.00 0.58
TABLE P2.74
acceleration = slope of line is constant.
a =− =1 63 1 63. .m s m s downward2 2

Chapter 2 53
P2.75 The distance x and y are always related by x y L2 2 2
+ = .
Differentiating this equation with respect to time, we have
2 2 0x
dx
dt
y
dy
dt
+ =
Now
dy
dt
is vB , the unknown velocity of B; and
dx
dt
v=− .
From the equation resulting from differentiation, we have
dy
dt
x
y
dx
dt
x
y
v=−
F
HG I
KJ=− −( ).
B
O
y
A
α
x
L
v
x
y
FIG. P2.75
But
y
x
= tanα so v vB =
F
HG I
KJ1
tanα
. When α = °60 0. , v
v v
vB =
°
= =
tan .
.
60 0
3
3
0 577 .
ANSWERS TO EVEN PROBLEMS
P2.2 (a) 2 10 7
× −
m s; 1 10 6
× −
m s; P2.24 (a) 1.88 km; (b) 1.46 km;
(c) see the solution;(b) 5 108
× yr
(d) (i) x t1
2
1 67= . m s2
e j ;
P2.4 (a) 50 0. m s ; (b) 41 0. m s (ii) x t2 50 375= −m s mb g ;
(iii) x t t3
2
250 2 5 4 375= − −m s m s m2
b g e j. ;P2.6 (a) 27 0. m;
(e) 37 5. m s(b) 27 0 18 0 3 00
2
. . .m m s m s2
+ +b g e ja f∆ ∆t t ;
(c) 18 0. m s
P2.26 958 m
P2.8 (a), (b), (c) see the solution; 4 6. m s2
; (d) 0
P2.28 (a) x t tf = −30 0 2
.e jm; v tf = −30 0 2.a f m s ;
(b) 225 mP2.10 5.00 m
P2.12 (a) 20 0. m s; 5 00. m s ; (b) 262 m
P2.30 x x v t a tf i xf x− = −
1
2
2
; 3 10. m s
P2.14 (a) see the solution;
P2.32 (a) 35.0 s; (b) 15 7. m s(b) 1 60. m s2
; 0 800. m s2
P2.34 (a) 1 12 1011
. × m s2
; (b) 4 67 10 5
. × −
sP2.16 (a) 13 0. m s; (b) 10 0. m s; 16 0. m s;
(c) 6 00. m s2
; (d) 6 00. m s2
P2.36 (a) False unless the acceleration is zero;
see the solution; (b) TrueP2.18 see the solution
P2.38 Yes; 212 m; 11.4 sP2.20 (a) 6 61. m s; (b) −0 448. m s2
P2.40 (a) −4 90. m; −19 6. m; −44 1. m;
P2.22 (a) − ⋅ = −21 8 9 75. .mi h s m s2
;
(b) −9 80. m s; −19 6. m s; −29 4. m s
(b) − ⋅ = −22 2 9 94. .mi h s m s2
;
(c) − ⋅ = −22 8 10 2. .mi h s m s2
P2.42 1.79 s

P2.44 No; see the solution P2.60 1 60. m s2
P2.46 The second ball is thrown at speed
v ghi =
P2.62 (a) 41.0 s; (b) 1.73 km; (c) −184 m s
P2.64 v t a txi x+
1
2
2
; displacements agree
P2.48 (a) 510 m; (b) 20.4 s
P2.66 155 s; 129 sP2.50 (a) 96 0. ft s;
(b) a = ×3 07 103
. ft s upward2
;
P2.68 (a) 5.44 s; (b) 131 m; (c) 50 8. m s ;
(c) ∆t = × −
3 13 10 2
. s
(d) 95 3. m s upward2
P2.52 38.2 m
P2.70 (a) 26.4 m; (b) 6.82%
P2.54 (a) and (b) see the solution; (c) −4 m s2
;
(d) 34 m; (e) 28 m P2.72 see the solution
P2.74 see the solution; ax = −1 63. m s2
P2.56 0.222 s
P2.58 (a) see the solution; (b) 6.23 s

3
CHAPTER OUTLINE
3.1 Coordinate Systems
3.2 Vector and Scalar
Quantities
3.3 Some Properties of Vectors
3.4 Components of a Vector
and Unit Vectors
Vectors
Q3.1 No. The sum of two vectors can only be zero if they are in
opposite directions and have the same magnitude. If you walk
10 meters north and then 6 meters south, you won’t end up
where you started.
Q3.2 No, the magnitude of the displacement is always less than or
equal to the distance traveled. If two displacements in the same
direction are added, then the magnitude of their sum will be
equal to the distance traveled. Two vectors in any other
orientation will give a displacement less than the distance
traveled. If you first walk 3 meters east, and then 4 meters
south, you will have walked a total distance of 7 meters, but
you will only be 5 meters from your starting point.
Q3.3 The largest possible magnitude of R A B= + is 7 units, found when A and B point in the same
direction. The smallest magnitude of R A B= + is 3 units, found when A and B have opposite
directions.
Q3.4 Only force and velocity are vectors. None of the other quantities requires a direction to be described.
Q3.5 If the direction-angle of A is between 180 degrees and 270 degrees, its components are both
negative. If a vector is in the second quadrant or the fourth quadrant, its components have opposite
signs.
Q3.6 The book’s displacement is zero, as it ends up at the point from which it started. The distance
traveled is 6.0 meters.
Q3.7 85 miles. The magnitude of the displacement is the distance from the starting point, the 260-mile
mark, to the ending point, the 175-mile mark.
Q3.8 Vectors A and B are perpendicular to each other.
Q3.9 No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction,
not magnitude.
55

56 Vectors
Q3.10 Any vector that points along a line at 45° to the x and y axes has components equal in magnitude.
Q3.11 A Bx x= and A By y= .
Q3.12 Addition of a vector to a scalar is not defined. Think of apples and oranges.
Q3.13 One difficulty arises in determining the individual components. The relationships between a vector
and its components such as A Ax = cosθ , are based on right-triangle trigonometry. Another problem
would be in determining the magnitude or the direction of a vector from its components. Again,
A A Ax y= +2 2
only holds true if the two component vectors, Ax and Ay , are perpendicular.
Q3.14 If the direction of a vector is specified by giving the angle of the vector measured clockwise from the
positive y-axis, then the x-component of the vector is equal to the sine of the angle multiplied by the
magnitude of the vector.
Section 3.1 Coordinate Systems
P3.1 x r= = °= − = −cos cos .θ 5 50 240 5 50 0 5 2 75. m . m . ma f a fa f
y r= = °= − = −sin sin .θ 5 50 240 5 50 0 866 4 76. m . m . ma f a fa f
P3.2 (a) x r= cosθ and y r= sinθ , therefore
x1 2 50 30 0= °. m .a fcos , y1 2 50 30 0= °. m .a fsin , and
x y1 1 2 17 1 25, . , . mb g a f=
x2 3 80 120= °. cosma f , y2 3 80 120= °. sinma f , and
x y2 2 1 90 3 29, . , . mb g a f= − .
(b) d x y= + = + =( ) ( ) . . .∆ ∆2 2
16 6 4 16 4 55 m
P3.3 The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m.
(a) We can use the Pythagorean theorem to find the distance from the origin to the fly.
distance m m m m2
= + = + = =x y2 2 2 2
2 00 1 00 5 00 2 24. . . .a f a f
(b) θ =
F
HG I
KJ = °−
tan .1 1
2
26 6 ; r = °2 24 26 6. , .m

Chapter 3 57
P3.4 (a) d x x y y= − + − = − − + − −2 1
2
2 1
2 2 2
2 00 3 00 4 00 3 00b g b g c h a f. . . .
d = + =25 0 49 0 8 60. . . m
(b) r1
2 2
2 00 4 00 20 0 4 47= + − = =. . . .a f a f m
θ1
1 4 00
2 00
63 4= −
F
HG I
KJ = − °−
tan
.
.
.
r2
2 2
3 00 3 00 18 0 4 24= − + = =. . . .a f a f m
θ 2 135= ° measured from the +x axis.
P3.5 We have 2 00 30 0. .= °r cos
r =
°
=
2 00
30 0
2 31
.
cos .
.
and y r= °= °=sin sin .30 0 2 31 30 0 1 15. . . .
P3.6 We have r x y= +2 2
and θ =
F
HG I
KJ−
tan 1 y
x
.
(a) The radius for this new point is
− + = + =x y x y ra f2 2 2 2
and its angle is
tan−
−
F
HG I
KJ = °−1
180
y
x
θ .
(b) ( ) ( )− + − =2 2 22 2
x y r . This point is in the third quadrant if x y,b gis in the first quadrant
or in the fourth quadrant if x y,b gis in the second quadrant. It is at an angle of 180°+θ .
(c) ( ) ( )3 3 32 2
x y r+ − = . This point is in the fourth quadrant if x y,b gis in the first quadrant
or in the third quadrant if x y,b gis in the second quadrant. It is at an angle of −θ .

58 Vectors
Section 3.2 Vector and Scalar Quantities
Section 3.3 Some Properties of Vectors
P3.7 tan .
tan . .
35 0
100
100 35 0 70 0
°=
= °=
x
x
m
m ma f
FIG. P3.7
P3.8 R =
= °
14
65
km
N of Eθ
θ
R 13 km
6 km1 km
FIG. P3.8
P3.9 − = °R 310 km at 57 S of W
(Scale: 1 20unit km= )
FIG. P3.9
P3.10 (a) Using graphical methods, place the tail of
vector B at the head of vector A. The new
vector A B+ has a magnitude of
6.1 at 112° from the x-axis.
(b) The vector difference A B− is found by
placing the negative of vector B at the
head of vector A. The resultant vector
A B− has magnitude 14 8. units at an
angle of 22° from the + x-axis.
y
x
A + B
A
A — B
B —B
O
FIG. P3.10

Chapter 3 59
P3.11 (a) d i= − =10 0 10 0. . m since the displacement is in a
straight line from point A to point B.
(b) The actual distance skated is not equal to the straight-line
displacement. The distance follows the curved path of the
semi-circle (ACB).
s r= = =
1
2
2 5 15 7π πb g . m
C
B A
5.00 m
d
FIG. P3.11
(c) If the circle is complete, d begins and ends at point A. Hence, d = 0 .
P3.12 Find the resultant F F1 2+ graphically by placing the tail of F2 at the head of F1 . The resultant force
vector F F1 2+ is of magnitude 9 5. N and at an angle of 57° above the -axisx .
0 1 2 3 N
x
y
F2
F1
F1
F2+
FIG. P3.12
P3.13 (a) The large majority of people are standing or sitting at this hour. Their instantaneous foot-to-
head vectors have upward vertical components on the order of 1 m and randomly oriented
horizontal components. The citywide sum will be ~105
m upward .
(b) Most people are lying in bed early Saturday morning. We suppose their beds are oriented
north, south, east, west quite at random. Then the horizontal component of their total vector
height is very nearly zero. If their compressed pillows give their height vectors vertical
components averaging 3 cm, and if one-tenth of one percent of the population are on-duty
nurses or police officers, we estimate the total vector height as ~ . m m10 0 03 10 15 2
a f a f+
~103
m upward .

60 Vectors
P3.14 Your sketch should be drawn to scale, and
should look somewhat like that pictured to
the right. The angle from the westward
direction, θ, can be measured to be
4° N of W , and the distance R from the
sketch can be converted according to the
scale to be 7 9. m .
15.0 meters
N
EW
S
8.20
meters
3.50
meters
1 m
30.0°
R
θ
FIG. P3.14
P3.15 To find these vector expressions graphically, we
draw each set of vectors. Measurements of the
results are taken using a ruler and protractor.
(Scale: 1 0 5unit m= . )
(a) A + B = 5.2 m at 60°
(b) A – B = 3.0 m at 330°
(c) B – A = 3.0 m at 150°
(d) A – 2B = 5.2 m at 300°.
FIG. P3.15
*P3.16 The three diagrams shown below represent the graphical solutions for the three vector sums:
R A B C1 = + + , R B C A2 = + + , and R C B A3 = + + . You should observe that R R R1 2 3= = ,
illustrating that the sum of a set of vectors is not affected by the order in which the vectors are
added.
100 m
C A
B
R2
B
A
R1
C
B
A
R3
C
FIG. P3.16

Chapter 3 61
P3.17 The scale drawing for the graphical solution
should be similar to the figure to the right. The
magnitude and direction of the final displacement
from the starting point are obtained by measuring
d and θ on the drawing and applying the scale
factor used in making the drawing. The results
should be
d = = − °420 3ft and θ
(Scale: 1 20unit ft= )
FIG. P3.17
Section 3.4 Components of a Vector and Unit Vectors
P3.18 Coordinates of the super-hero are:
x
y
= − ° =
= − ° = −
100 30 0 86 6
100 30 0 50 0
m m
m m
a f a f
a f a f
cos . .
sin . .
FIG. P3.18
P3.19 A
A
A A A
x
y
x y
= −
=
= + = − + =
25 0
40 0
25 0 40 0 47 22 2 2 2
.
.
. . .a f a f units
We observe that
tanφ =
A
A
y
x
.
FIG. P3.19
So
φ =
F
HG
I
KJ = = = °− −
tan tan
.
.
tan . .1 140 0
25 0
1 60 58 0
A
A
y
x
a f .
The diagram shows that the angle from the +x axis can be found by subtracting from 180°:
θ = °− °= °180 58 122 .
P3.20 The person would have to walk 3 10 1 31. 25.0 km northsin .° =a f , and
3 10 25 0 2 81. . km eastcos .° =a f .

62 Vectors
P3.21 x r= cosθ and y r= sinθ , therefore:
(a) x = °12 8 150. cos , y = °12 8 150. sin , and x y, . . mb g e j= − +11 1 6 40i j
(b) x = °3 30 60 0. cos . , y = °3 30 60 0. sin . , and x y, cmb g e j= +1 65 2 86. .i j
(c) x = °22 0 215. cos , y = °22 0 215. sin , and x y, inb g e j= − −18 0 12 6. .i j
P3.22 x d= = = −cos cosθ 50 0 120 25 0. m . ma f a f
y d= = =
= − +
sin sin .
. .
θ 50 0 120 43 3
25 0 43 3
. m m
m m
a f a f
a f a fd i j
*P3.23 (a) Her net x (east-west) displacement is − + + = +3 00 0 6 00 3 00. . . blocks, while her net y (north-
south) displacement is 0 4 00 0 4 00+ + = +. . blocks. The magnitude of the resultant
displacement is
R x y= + = + =net netb g b g a f a f2 2 2 2
3 00 4 00 5 00. . . blocks
and the angle the resultant makes with the x-axis (eastward direction) is
θ =
F
HG I
KJ = = °− −
tan
.
.
tan . .1 14 00
3 00
1 33 53 1a f .
The resultant displacement is then 5 00 53 1. .blocks at N of E° .
(b) The total distance traveled is 3 00 4 00 6 00 13 0. . . .+ + = blocks .
*P3.24 Let i = east and j = north. The unicyclist’s displacement is, in meters
280 220 360 300 120 60 40 90 70j i j i j i j i j+ + − − + − − + .
R i j= − +
= +
= °
−
110 550
110 550
110
550
561 11 3
2 2 1
tan
. .
m m at
m
m
west of north
m at west of north
a f a f
The crow’s velocity is
v
x
= =
°
= °
∆
∆t
561 11 3
40
14 0 11 3
m at W of N
s
m s at west of north
.
. . .
R
N
E
FIG. P3.24

Chapter 3 63
P3.25 +x East, +y North
x
y
d x y
y
x
∑
∑
∑ ∑
∑
∑
= °=
= °− = −
= + = + − =
= = − = −
= − °
= °
250 125 30 358
75 125 30 150 12 5
358 12 5 358
12 5
358
0 0349
2 00
358 2 00
2 2 2 2
+ m
+ . m
m
.
.
m at S of E
cos
sin
.
tan
.
.
c h c h a f a f
c h
c hθ
θ
d
P3.26 The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of
the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to
Chicago. In equation form:
d d d
d d d
DC east DA east AC east
DC north DA north AC north
+ . . miles.
+ . +560 . miles.
= = °− °=
= = ° °=
730 5 00 560 21 0 527
730 5 00 21 0 586
cos sin
sin cos
By the Pythagorean theorem, d d d= + =( ) ( )DC east DC north mi2 2
788 .
Then tanθ = =
d
d
DC north
DC east
.1 11 and θ = °48 0. .
Thus, Chicago is 788 48 0miles at northeast of Dallas. ° .
P3.27 (a) See figure to the right.
(b) C A B i j i j i j= + = + + − = +2 00 6 00 3 00 2 00 5 00 4 00. . . . . .
C = +
F
HG I
KJ = °−
25 0 16 0 6 401
. . tan .at
4
5
at 38.7
D A B i j i j i j= − = + − + = − +2 00 6 00 3 00 2 00 1 00 8 00. . . . . .
D = − +
−
F
HG I
KJ−
1 00 8 00
8 00
1 00
2 2 1
. . tan
.
.
a f a f at
D = °− ° = °8 06 180 82 9 8 06 97 2. . . .at atb g
FIG. P3.27
P3.28 d x x x y y y= + + + + +
= − + + + + = =
=
F
HG I
KJ = °−
1 2 3
2
1 2 3
2
2 2
1
3 00 5 00 6 00 2 00 3 00 1 00 52 0 7 21
6 00
4 00
56 3
b g b g
a f a f m. . . . . . . .
tan
.
.
.θ

64 Vectors
P3.29 We have B R A= − :
A
A
R
R
x
y
x
y
= °= −
= °=
= °=
= °=
150 75 0
150 120 130
140 35 0 115
140 35 0 80 3
cos120 cm
cm
cm
cm
.
sin
cos .
sin . .
Therefore,
FIG. P3.29
B i j i j
B
= − − + − = −
= + =
= −
F
HG I
KJ = − °−
115 75 80 3 130 190 49 7
190 49 7 196
49 7
190
14 7
2 2
1
a f e j. .
.
tan
.
. .
cm
cm
θ
P3.30 A i j= − +8 70 15 0. . and B i j= −13 2 6 60. .
A B C− + =3 0 :
3 21 9 21 6
7 30 7 20
C B A i j
C i j
= − = −
= −
. .
. .
or
Cx = 7 30. cm ; Cy = −7 20. cm
P3.31 (a) A B i j i j i j+ = − + − − = −a f e j e j3 2 4 2 6
(b) A B i j i j i j− = − − − − = +a f e j e j3 2 4 4 2
(c) A B+ = + =2 6 6 322 2
.
(d) A B− = + =4 2 4 472 2
.
(e) θ A B+
−
= −
F
HG I
KJ=− °= °tan .1 6
2
71 6 288
θ A B−
−
=
F
HG I
KJ= °tan .1 2
4
26 6
P3.32 (a) D A B C i j= + + = +2 4
D = + = = °2 4 4 47 63 42 2
. .m at θ
(b) E A B C i j= − − + = − +6 6
E = + = = °6 6 8 49 1352 2
. m at θ

Chapter 3 65
P3.33 d1 3 50= − . je jm
d2 8 20 45 0 8 20 45 0 5 80 5 80= ° + ° = +. cos . . sin . . .i j i je jm
d3 15 0= − . ie jm
R i j i j= + + = − + + − = − +d d d1 2 3 15 0 5 80 5 80 3 50 9 20 2 30. . . . . .a f a f e jm
(or 9.20 m west and 2.30 m north)
The magnitude of the resultant displacement is
R = + = − + =R Rx y
2 2 2 2
9 20 2 30 9 48. . .a f a f m .
The direction is θ =
−
F
HG I
KJ = °arctan
.
.
2 30
9 20
166 .
P3.34 Refer to the sketch
R A B C i j i
i j
R
= + + = − − +
= −
= + − =
10 0 15 0 50 0
40 0 15 0
40 0 15 0 42 7
2 2 1 2
. . .
. .
. . .a f a f yards
A = 10 0.
B = 15 0.
C = 50 0.
R
FIG. P3.34
P3.35 (a) F F F
F i j i j
F i j i j i j
F
= +
= ° + ° − ° + °
= + − + = +
= + =
=
F
HG I
KJ = °−
1 2
2 2
1
120 60 0 120 60 0 80 0 75 0 80 0 75 0
60 0 104 20 7 77 3 39 3 181
39 3 181 185
181
39 3
77 8
cos . sin . . cos . . sin .
. . . .
.
tan
.
.
a f a f a f a f
e jN
N
θ
(b) F F i j3 39 3 181= − = − −.e jN
P3.36 East West
x y
0 m 4.00 m
1.41 1.41
–0.500 –0.866
+0.914 4.55
R = + = °x y
2 2
4 64. m at 78.6 N of E

66 Vectors
P3.37 A = 3 00. m, θA = °30 0. B= 3 00. m, θB = °90 0.
A Ax A= = °=cos . cos . .θ 3 00 30 0 2 60 m A Ay A= = °=sin . sin . .θ 3 00 30 0 1 50 m
A i j i j= + = +A Ax y . .2 60 1 50e jm
Bx = 0, By = 3 00. m so B j= 3 00. m
A B i j j i j+ = + + = +2 60 1 50 3 00 2 60 4 50. . . . .e j e jm
P3.38 Let the positive x-direction be eastward, the positive y-direction be vertically upward, and the
positive z-direction be southward. The total displacement is then
d i j j k i j k= + + − = + −4 80 4 80 3 70 3 70 4 80 8 50 3 70. . . . . . .e j e j e jcm cm cm.
(a) The magnitude is d = ( ) +( ) + −( ) =4 80 8 50 3 70 10 4
2 2 2
. . . .cm cm .
(b) Its angle with the y-axis follows from cos
.
.
θ =
8 50
10 4
, giving θ = °35 5. .
P3.39 B i j k i j k
B
= + + = + +
= + + =
=
F
HG I
KJ = °
=
F
HG I
KJ = °
=
F
HG I
KJ = °
−
−
−
B B Bx y z . . .
. . . .
cos
.
.
.
cos
.
.
.
cos
.
.
.
4 00 6 00 3 00
4 00 6 00 3 00 7 81
4 00
7 81
59 2
6 00
7 81
39 8
3 00
7 81
67 4
2 2 2
1
1
1
α
β
γ
P3.40 The y coordinate of the airplane is constant and equal to 7 60 103
. × m whereas the x coordinate is
given by x v ti= where vi is the constant speed in the horizontal direction.
At t = 30 0. s we have x = ×8 04 103
. , so vi = 268 m s. The position vector as a function of time
is
P i j= + ×268 7 60 103
m s mb g e jt . .
At t = 45 0. s, P i j= × + ×1 21 10 7 60 104 3
. . m. The magnitude is
P = × + × = ×1 21 10 7 60 10 1 43 104 2 3 2 4
. . .c h c h m m
and the direction is
θ =
×
×
F
HG
I
KJ= °arctan
.
.
.
7 60 10
1 21 10
32 2
3
4
above the horizontal .

Chapter 3 67
P3.41 (a) A i j k= + −8 00 12 0 4 00. . .
(b) B
A
i j k= = + −
4
2 00 3 00 1 00. . .
(c) C A i j k= − = − − +3 24 0 36 0 12 0. . .
P3.42 R i j i j i j= ° + ° + ° + ° + ° + °75 0 240 75 0 240 125 135 125 135 100 160 100 160. cos . sin cos sin cos sin
R i j i j i j= − − − + − +37 5 65 0 88 4 88 4 94 0 34 2. . . . . .
R i j= − +220 57 6.
R = −( ) +
F
HG I
KJ220 57 6
57 6
220
2 2
. arctan
.
at above the –x-axis
R = °227 paces at 165
P3.43 (a) C A B i j k= + = − −5 00 1 00 3 00. . .e jm
C = ( ) +( ) +( ) =5 00 1 00 3 00 5 92
2 2 2
. . . .m m
(b) D A B i j k= − = − +2 4 00 11 0 15 0. . .e jm
D = ( ) +( ) +( ) =4 00 11 0 15 0 19 0
2 2 2
. . . .m m
P3.44 The position vector from radar station to ship is
S i j i j= ° + ° = −17 3 136 17 3 136 12 0 12 4. sin . cos . .e j e jkm km.
From station to plane, the position vector is
P i j k= ° + ° +19 6 153 19 6 153 2 20. sin . cos .e jkm,
or
P i j k= − +8 90 17 5 2 20. . .e jkm.
(a) To fly to the ship, the plane must undergo displacement
D S P i j k= − = + −3 12 5 02 2 20. . .e jkm .
(b) The distance the plane must travel is
D = = ( ) +( ) +( ) =D 3 12 5 02 2 20 6 31
2 2 2
. . . .km km .

68 Vectors
P3.45 The hurricane’s first displacement is
41 0
3 00
.
.
km
h
h
F
HG I
KJ( ) at 60 0. ° N of W, and its second displacement
is
25 0
1 50
.
.
km
h
h
F
HG I
KJ( ) due North. With i representing east and j representing north, its total
displacement is:
41 0 60 0 3 00 41 0 60 0 3 00 25 0 1 50 61 5
144
. cos . . . sin . . . . .
km
h
h
km
h
h
km
h
h km
km
°
F
HG I
KJ − + °
F
HG I
KJ +
F
HG I
KJ = −
+
a fe j a f a f e ji j j i
j
with magnitude 61 5 144 157
2 2
. km km km( ) +( ) = .
P3.46 (a) E i j= ° + °17 0 27 0 17 0 27 0. cos . . sin .cm cma f a f
E i j= +15 1 7 72. .e jcm
(b) F i j= − ° + °17 0 27 0 17 0 27 0. sin . . cos .cm cma f a f
F i j= − +7 72 15 1. .e jcm
(c) G i j= + ° + °17 0 27 0 17 0 27 0. sin . . cos .cm cma f a f
G i j= + +7 72 15 1. .e jcm
F
y
x
27.0°
G
27.0°
E
27.0°
FIG. P3.46
P3.47 Ax =−3 00. , Ay = 2 00.
(a) A i j i j= + = − +A Ax y . .3 00 2 00
(b) A = + = −( ) +( ) =A Ax y
2 2 2 2
3 00 2 00 3 61. . .
tan
.
.
.θ = =
−( )
=−
A
A
y
x
2 00
3 00
0 667, tan . .−
−( )=− °1
0 667 33 7
θ is in the 2nd
quadrant, so θ = °+ − ° = °180 33 7 146.a f .
(c) Rx = 0, Ry =−4 00. , R A B= + thus B R A= − and
B R Ax x x= − = − −( )=0 3 00 3 00. . , B R Ay y y= − =− − =−4 00 2 00 6 00. . . .
Therefore, B i j= −3 00 6 00. . .

Chapter 3 69
P3.48 Let + =x East, + =y North,
x y
300 0
–175 303
0 150
125 453
(a) θ = = °−
tan .1
74 6
y
x
N of E
(b) R = + =x y2 2
470 km
P3.49 (a)
(b)
R
R
x
y
= °+ °=
= °− °+ =
= +
40 0 45 0 30 0 45 0 49 5
40 0 45 0 30 0 45 0 20 0 27 1
49 5 27 1
. cos . . cos . .
. sin . . sin . . .
. .R i j
R = + =
=
F
HG I
KJ = °−
49 5 27 1 56 4
27 1
49 5
28 7
2 2
1
. . .
tan
.
.
.
a f a f
θ
A
y
x
B
45°
C
45°
O
FIG. P3.49
P3.50 Taking components along i and j , we get two equations:
6 00 8 00 26 0 0. . .a b− + =
and
− + + =8 00 3 00 19 0 0. . .a b .
Solving simultaneously,
a b= =5 00 7 00. , . .
Therefore,
5 00 7 00 0. .A B C+ + = .

70 Vectors
Additional Problems
P3.51 Let θ represent the angle between the directions of A and B. Since
A and B have the same magnitudes, A, B, and R A B= + form an
isosceles triangle in which the angles are 180°−θ ,
θ
2
, and
θ
2
. The
magnitude of R is then R A=
F
HG I
KJ2
2
cos
θ
. [Hint: apply the law of
cosines to the isosceles triangle and use the fact that B A= .]
Again, A, –B, and D A B= − form an isosceles triangle with apex
angle θ. Applying the law of cosines and the identity
1 2
2
2
− =
F
HG I
KJcos sinθ
θ
a f
gives the magnitude of D as D A=
F
HG I
KJ2
2
sin
θ
.
The problem requires that R D=100 .
Thus, 2
2
200
2
A Acos sin
θ θF
HG I
KJ =
F
HG I
KJ. This gives tan .
θ
2
0 010
F
HG I
KJ = and
θ = °1 15. .
A
BR
θ/2
θ
A
D –B
θ
FIG. P3.51
P3.52 Let θ represent the angle between the directions of A and B. Since
A and B have the same magnitudes, A, B, and R A B= + form an
isosceles triangle in which the angles are 180°−θ ,
θ
2
, and
θ
2
. The
magnitude of R is then R A=
F
HG I
KJ2
2
cos
θ
. [Hint: apply the law of
cosines to the isosceles triangle and use the fact that B A= . ]
Again, A, –B, and D A B= − form an isosceles triangle with apex
angle θ. Applying the law of cosines and the identity
1 2
2
2
− =
F
HG I
KJcos sinθ
θ
a f
gives the magnitude of D as D A=
F
HG I
KJ2
2
sin
θ
.
The problem requires that R nD= or cos sin
θ θ
2 2
F
HG I
KJ =
F
HG I
KJn giving
θ =
F
HG I
KJ−
2
11
tan
n
.
FIG. P3.52

Chapter 3 71
P3.53 (a) Rx = 2 00. , Ry = 1 00. , Rz = 3 00.
(b) R = + + = + + = =R R Rx y z
2 2 2
4 00 1 00 9 00 14 0 3 74. . . . .
(c) cos cos .θ θx
x
x
xR R
x= ⇒ =
F
HG
I
KJ= ° +−
R R
1
57 7 from
cos cos .θ θy
y
y
yR R
y= ⇒ =
F
HG
I
KJ= ° +−
R R
1
74 5 from
cos cos .θ θz
z
z
zR R
z= ⇒ =
F
HG
I
KJ= ° +−
R R
1
36 7 from
*P3.54 Take the x-axis along the tail section of the snake. The displacement from tail to head is
240 420 240 180 105 180 75 287m + m m m 174 mcos sini i j i j− °− ° − ° = −a f a f .
Its magnitude is 287 174 335
2 2
( ) +( ) =m m. From v
t
=
distance
∆
, the time for each child’s run is
Inge:
distance m h km s
km m h
s
Olaf:
m s
3.33 m
s
∆
∆
t
v
t
= = =
=
⋅
=
335 1 3 600
12 1 000 1
101
420
126
a fa fb g
a fb ga f
.
Inge wins by 126 101 25 4− = . s .
*P3.55 The position vector from the ground under the controller of the first airplane is
r i j k
i j k
1 19 2 25 19 2 25 0 8
17 4 8 11 0 8
= ° + ° +
= + +
. cos . sin .
. . . .
km km km
km
a fa f a fa f a f
e j
The second is at
r i j k
i j k
2 17 6 20 17 6 20 1
16 5 6 02 1 1
= ° + ° +
= + +
. cos . sin
. . . .
km km .1 km
km
a fa f a fa f a f
e j
Now the displacement from the first plane to the second is
r r i j k2 1 0 863 2 09 0 3− = − − +. . .e jkm
with magnitude
0 863 2 09 0 3 2 29
2 2 2
. . . .( ) +( ) +( ) = km .

72 Vectors
*P3.56 Let A represent the distance from island 2 to island 3. The
displacement is A = A at 159°. Represent the displacement from 3
to 1 as B= B at 298°. We have 4.76 km at 37° + + =A B 0.
For x-components
4 76 37 159 298 0
3 80 0 934 0 469 0
8 10 1 99
. cos cos cos
. . .
. .
km
km
km
a f °+ °+ °=
− + =
= − +
A B
A B
B A
For y-components,
4 76 37 159 298 0
2 86 0 358 0 883 0
. sin sin sin
. . .
km
km
a f °+ °+ °=
+ − =
A B
A B
N
B28°
A
C
69°
37°
1
2
3
E
FIG. P3.56
(a) We solve by eliminating B by substitution:
2 86 0 358 0 883 8 10 1 99 0
2 86 0 358 7 15 1 76 0
10 0 1 40
7 17
. . . . .
. . . .
. .
.
km km
km km
km
km
+ − − + =
+ + − =
=
=
A A
A A
A
A
a f
(b) B =− + ( )=8 10 1 99 7 17 6 15. . . .km km km
*P3.57 (a) We first express the corner’s position vectors as sets of components
A i j i j
B i j i j
= ° + ° =
= ° + ° =
10 50 10 50 6 43
12 30 12 30 10 4
m m m +7.66 m
m m m +6.00 m
a f a f
a f a f
cos sin .
cos sin . .
The horizontal width of the rectangle is
10 4 6 43 3 96. . .m m m− = .
Its vertical height is
7 66 6 00 1 66. . .m m m− = .
Its perimeter is
2 3 96 1 66 11 2. . .+( ) =m m .
(b) The position vector of the distant corner is B Ax y . .i j i j+ = +10 4 7 662
m +7.66 m = 10.4 m2
at
tan
.
.−
= °1 7 66
12 9
m
10.4 m
m at 36.4 .

Chapter 3 73
P3.58 Choose the +x-axis in the direction of the first force. The total force,
in newtons, is then
12 0 31 0 8 40 24 0 3 60 7 00. . . . . .i j i j i j+ − − = +e j e jN .
The magnitude of the total force is
3 60 7 00 7 87
2 2
. . .( ) +( ) =N N
and the angle it makes with our +x-axis is given by tan
.
.
θ =
( )
( )
7 00
3 60
,
θ = °62 8. . Thus, its angle counterclockwise from the horizontal is
35 0 62 8 97 8. . .°+ °= ° .
R
35.0°
y
24 N
horizontal
31 N
8.4 N
12 N
x
FIG. P3.58
P3.59 d i
d j
d i j i j
d i j i j
R d d d d i j
R
1
2
3
4
1 2 3 4
2 2
1
100
300
150 30 0 150 30 0 130 75 0
200 60 0 200 60 0 100 173
130 202
130 202 240
202
130
57 2
180 237
=
= −
= − ° − ° = − −
= − ° + ° = − +
= + + + = − −
= − + − =
=
F
HG I
KJ = °
= + = °
−
cos . sin . .
cos . sin .
tan .
a f a f
a f a f
e j
a f a f
m
m
φ
θ φ
FIG. P3.59
P3.60
d
dt
d t
dt
r i j j
j j=
+ −
= + − = −
4 3 2
0 0 2 2 00.
e j b gm s
The position vector at t = 0 is 4 3i j+ . At t =1 s, the position is 4 1i j+ , and so on. The object is
moving straight downward at 2 m/s, so
d
dt
r
represents its velocity vector .
P3.61 v i j i j
v i j
v
= + = + ° + °
= +
= °
v vx y cos . sin .
.
300 100 30 0 100 30 0
387 50 0
390
a f a f
e j mi h
mi h at 7.37 N of E

74 Vectors
P3.62 (a) You start at point A: r r i j1 30 0 20 0= = −A . .e jm.
The displacement to B is
r r i j i j i jB A− = + − + = +60 0 80 0 30 0 20 0 30 0 100. . . . . .
You cover half of this, 15 0 50 0. .i j+e j to move to r i j i j i j2 30 0 20 0 15 0 50 0 45 0 30 0= − + + = +. . . . . . .
Now the displacement from your current position to C is
r r i j i j i jC − = − − − − = − −2 10 0 10 0 45 0 30 0 55 0 40 0. . . . . . .
You cover one-third, moving to
r r r i j i j i j3 2 23 45 0 30 0
1
3
55 0 40 0 26 7 16 7= + = + + − − = +∆ . . . . . .e j .
The displacement from where you are to D is
r r i j i j i jD − = − − − = −3 40 0 30 0 26 7 16 7 13 3 46 7. . . . . . .
You traverse one-quarter of it, moving to
r r r r i j i j i j4 3 3
1
4
26 7 16 7
1
4
13 3 46 7 30 0 5 00= + − = + + − = +Db g e j. . . . . . .
The displacement from your new location to E is
r r i j i j i jE − = − + − − = − +4 70 0 60 0 30 0 5 00 100 55 0. . . . .
of which you cover one-fifth the distance, − +20 0 11 0. .i j, moving to
r r i j i j i j4 45 30 0 5 00 20 0 11 0 10 0 16 0+ = + − + = +∆ . . . . . . .
The treasure is at 10 0. m, 16.0 m( ) .
(b) Following the directions brings you to the average position of the trees. The steps we took
numerically in part (a) bring you to
r r r
r r
A B A
A B
+ − =
+F
HG I
KJ1
2 2
a f
then to
r r r r r r
r r
A B C A B C
A B
+
+
−
=
+ +
+
a f a f
2 3 3
2
then to
r r r r r r r r
r r r
A B C D A B C D
A B C
+ +
+
−
=
+ + +
+ +
a f a f
3 4 4
3
and last to
r r r r r r r r r r
r r r r
A B C D E A B C D E
A B C D
+ + +
+
−
=
+ + + +
+ + +
a f a f
4 5 5
4
.
This center of mass of the tree distribution is the same location whatever order we take the
trees in.

Chapter 3 75
*P3.63 (a) Let T represent the force exerted by each child. The x-
component of the resultant force is
T T T T T Tcos cos cos . .0 120 240 1 0 5 0 5 0+ °+ °= + − + − =a f a f a f .
The y-component is
T T T T Tsin sin sin . .0 120 240 0 0 866 0 866 0+ + = + − = .
Thus,
F∑ = 0.
FIG. P3.63
(b) If the total force is not zero, it must point in some direction. When each child moves one
space clockwise, the total must turn clockwise by that angle,
360°
N
. Since each child exerts
the same force, the new situation is identical to the old and the net force on the tire must still
point in the original direction. The contradiction indicates that we were wrong in supposing
that the total force is not zero. The total force must be zero.
P3.64 (a) From the picture, R i j1 = +a b and R1
2 2
= +a b .
(b) R i j k2 = + +a b c ; its magnitude is
R1
2 2 2 2 2
+ = + +c a b c .
FIG. P3.64

Serway, raymond a physics for scientists and engineers (6e) solutions

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