1. 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. 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 r =
= . .m me j . Its density is then = = = 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=
2 e j and = m V , we have = = 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 = m V for both. Then
iron kg= 9 35. V and gold gold = m V . Next, 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 e j = m V ,
so m V r r r r = = F HG I KJ = 4 3 4 3 2 3 1 3 2 3 1 3 e j e j
.
3. Chapter 1 3 P1.6 For either sphere the volume is V r= 4 3 3
and the mass is m V r= = 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 = = = 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.
4. 4 Physics and Measurement 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= = = = 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= = = 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 .
5. 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 LT2 , 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= = = .
6. 6 Physics and Measurement 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 m1 . *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 =
7. 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= = = air 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 . . .
8. 8 Physics and Measurement *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 Kings 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 = m V . We must convert to SI units
in the calculation. = 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
.
9. 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 . . = 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 b 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 j . 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= = = 1 000 1 3 10 1 3
103 18 3 21 kg m . m kge je j . .
10. 10 Physics and Measurement 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 . . . *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 . . . 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 . . . e jb g
P1.39 To balance, m mFe Al= or Fe Fe Al AlV V= 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
11. Chapter 1 11 P1.40 The mass of each sphere is m V r Al Al
Al Al Al = = 4 3 3 and m V r Fe Fe Fe Fe Fe = = 4 3 3 . Setting
these masses equal, 4 3 4 3 3 3 Al Al Fe Fer r = and r rAl Fe Fe Al
= 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 5 . . 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 74 = . 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 .
12. 12 Physics and Measurement 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= = = 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= = = 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 .
13. 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) 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 3 r = = . 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 h also, = + m m r r 3
. In other words, the percentages of uncertainty are cumulative.
Therefore, = + = 0 02 1 85 3 0 20 6 50 0 103 . . . . . a f , = = 1
85 6 5 10 1 61 10 4 3 2 3 3 3. . . c h e jm kg m and = = 1 61 0 17
10 1 6 0 2 103 3 . . . .a f a fkg m kg m3 3 .
14. 14 Physics and Measurement 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 = *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 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 .
15. 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 j 3 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 .
16. 16 Physics and Measurement P1.60 (deg) (rad) tan a f sin a
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 r r h r h = = = = = . m . m tan 55.0 . m (
. ) ma ftan . 5555 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 KJ 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. . 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 = 10 31 557 600 31 557 600 100% 0 449% 7 s yr s yr
s yr e j b g . .
17. 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 = = = 2 2 e j , where A R= 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 f
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 cma f . P1.66
(a) 1 cubic meter of water has a mass m V= = = 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 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 .
18. 18 Physics and Measurement 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 r 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 = = = 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 = = 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 e j . This would take 1 91 10 3 16
10 6 05 10 18 7 10. . . = micrometeorites micrometeorites yr yr
.
19. 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
KJ 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
20. 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 ANSWERS TO QUESTIONS 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
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.
22. 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. SOLUTIONS TO PROBLEMS 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
23. 24 Motion in One Dimension 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 .
24. 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 .
25. 26 Motion in One Dimension 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,
Secretariats 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
26. 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<
27. 28 Motion in One Dimension 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) continued
on next page
28. 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 . . .
29. 30 Motion in One Dimension *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<
30. 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 .
31. 32 Motion in One Dimension *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 .
32. 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
33. 34 Motion in One Dimension 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
34. 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.
35. 36 Motion in One Dimension *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
36. Chapter 2 37 P2.38 Take the original point to be when Sue
notices the van. Choose the origin of the x-axis at Sues 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
37. 38 Motion in One Dimension 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 workmans 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.
38. 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 . . . .
39. 40 Motion in One Dimension 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
40. 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+ =. .
.
41. 42 Motion in One Dimension 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 .
42. 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<
43. 44 Motion in One Dimension 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 .
44. 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 trains 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 . .
45. 46 Motion in One Dimension *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 rockets 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. continued on next page
46. 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
47. 48 Motion in One Dimension 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
48. 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
.
49. 50 Motion in One Dimension 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 xsin . The pack lifts at a rate x t sin . v x t v x v
x x h = = = + sin 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
50. Chapter 2 51 P2.72 h= 6 00. m, vboy m s= 2 00. v x t v x v
x x h = = = + sin 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
51. 52 Motion in One Dimension 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
52. 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 x L v x y FIG. P2.75 But y x = tan so v vB = F HG I KJ1
tan . When = 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
53. 54 Motion in One Dimension 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
54. 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 ANSWERS TO QUESTIONS 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 wont 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 books 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
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 = cos , 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. SOLUTIONS TO PROBLEMS
Section 3.1 Coordinate Systems P3.1 x r= = = = cos cos . 5 50 240 5
50 0 5 2 75. m . m . ma f a fa f y r= = = = sin sin . 5 50 240 5 50
0 866 4 76. m . m . ma f a fa f P3.2 (a) x r= cos and y r= sin ,
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) = F HG I KJ = tan .1 1 2 26 6 ; r
= 2 24 26 6. , .m
56. 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 1 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 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 = 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 . (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+ . (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 .
57. 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 E 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
58. 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 7 b 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 .
59. 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, , 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 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
60. 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 on the drawing and applying
the scale factor used in making the drawing. The results should be
d = = 420 3ft and (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 tan = A A y x . FIG. P3.19 So = 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: = = 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 .
61. 62 Vectors P3.21 x r= cos and y r= sin , 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 cos 50
0 120 25 0. m . ma f a f y d= = = = + sin sin . . . 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 = 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 unicyclists
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
crows 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
62. 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 h 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 tan = = d d DC
north DC east .1 11 and = 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 . . .
63. 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 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) A B+ = F HG I KJ= = tan .1 6 2 71 6 288 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 (b) E A B C i j= + = +6 6 E = + = = 6 6 8 49
1352 2 . m at
64. 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 = 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 (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
65. 66 Vectors P3.37 A = 3 00. m, A = 30 0. B= 3 00. m, B = 90
0. A Ax A= = =cos . cos . . 3 00 30 0 2 60 m A Ay A= = =sin . sin .
. 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 . . = 8 50 10 4 , giving = 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 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 = F HG I KJ= arctan . . . 7 60 10 1
21 10 32 2 3 4 above the horizontal .
66. 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 .
67. 68 Vectors P3.45 The hurricanes 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 . . . = =
( ) = A A y x 2 00 3 00 0 667, tan . . ( )= 1 0 667 33 7 is in the
2nd quadrant, so = + = 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. . .
68. Chapter 3 69 P3.48 Let + =x East, + =y North, x y 300 0 175
303 0 150 125 453 (a) = = 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 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+ + = .
69. 70 Vectors Additional Problems P3.51 Let 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 , 2 , and 2 . The magnitude of R is then R
A= F HG I KJ2 2 cos . [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 . Applying the law
of cosines and the identity 1 2 2 2 = F HG I KJcos sin a f gives
the magnitude of D as D A= F HG I KJ2 2 sin . The problem requires
that R D=100 . Thus, 2 2 200 2 A Acos sin F HG I KJ = F HG I KJ.
This gives tan . 2 0 010 F HG I KJ = and = 1 15. . A BR /2 A D B
FIG. P3.51 P3.52 Let 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 , 2 , and 2
. The magnitude of R is then R A= F HG I KJ2 2 cos . [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 . Applying the law of cosines and the identity 1 2 2 2 =
F HG I KJcos sin a f gives the magnitude of D as D A= F HG I KJ2 2
sin . The problem requires that R nD= or cos sin 2 2 F HG I KJ = F
HG I KJn giving = F HG I KJ 2 11 tan n . FIG. P3.52
70. 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 . x x x xR R x= = F HG I KJ= + R R 1 57 7 from cos
cos . y y y yR R y= = F HG I KJ= + R R 1 74 5 from cos cos . z 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 childs 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 .
71. 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