2000 by Harcourt College Publishers. All rights reserved.
Chapter 1 Even Answers 2. 623 kg/m3 4. 4 (r 3 2 r 3 1) 3 6. 7.69 cm
8. 8.72 1011 atoms/s 10. (a) 72.6 kg (b) 7.82 1026 atoms 12.
equation is dimensionally consistent 16. The units of G are: m3/kg
s2 18. 9.19 nm/s 20. (a) 3.39 105 ft3 (b) 2.54 104 lb 22. 8.32 104
m/s 24. 9.82 cm 26. (a) 6.31 104 AU (b) 1.33 1011 AU 28. (a) 1.609
km/h (b) 88.5 km/h (c) 16.1 km/h 30. (a) 3.16 107 s/yr (b) 6.05
1010 yr 32. 2.57 106 m3 34. 1.32 1021 kg 36. (a) 2.07 mm (b) 8.62
1013 times as large 38. (a) 13.4 (b) 49.1 40. rAl = rFe 3 (Fe/Al)
42. ~ 106 km 44. ~ 109 drops 46. time required 50 years or more;
advise against accepting the offer 48. ~ 105 tons 50. (a) 2 (b) 4
(c) 3 (d) 2 52. (a) 797 (b) 1.1 (c) 17.66 54. (a) 3 (b) 4 (c) 3 (d)
2 56. 5.2 m3, 2.7% 58. 1.79 109 m 60. 24.6 62. (b) Acylinder = R2,
Arectangular solid = lw 64. 0.141 nm 66. 289 m 68. (a) 1000 kg (b)
5.2 1016 kg 0.27 kg (d) 1.3 105 kg 70. Aluminum: 2.75 g cm3 (table
value is 2% smaller) Copper: 9.36 g cm3 (table value is 5% smaller)
Brass: 8.91 g cm3 Tin: 7.68 g cm3 Iron: 7.88 g cm3 (table value is
0.3% smaller)
2 Chapter 1 Even Answers 2000 by Harcourt College Publishers.
All rights reserved.
2000 by Harcourt College Publishers. All rights reserved.
Chapter 1 Solutions *1.1 With V = (base area) (height) V = r 2 h
and = m V , we have = m r 2 h = 1 kg (19.5 mm) 2 39.0 mm 10 9 mm 3
1 m 3 = 2.15 10 4 kg/m 3 1.2 = M V = M 4 3 R 3 = 3(5.64 10 26 kg) 4
(6.00 10 7 m) 3 = 623 kg/m 3 1.3 VCu = V0 Vi = 4 3 (r 3 o r 3 i )
VCu = 4 3 [ ](5.75 cm)3 (5.70 cm)3 = 20.6 cm3 = m V m = V = (8.92
g/cm3)(20.6 cm3) = 184 g 1.4 V = Vo Vi = 4 3 (r 3 2 r 3 1 ) = m V ,
so m = V = 4 3 (r 3 2 r 3 1) = 4 (r 3 2 r 3 1) 3 *1.5 (a) The
number of moles is n = m/M, and the density is = m/V. Noting that
we have 1 mole, V1 mol = mFe Fe = nFe MFe Fe = (1 mol)(55.8 g/mol)
7.86 g/cm3 = 7.10 cm3 5.7cm5.7cm 0.05 cm
2 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. (b) In 1 mole of iron are NA atoms: V1 atom = V1
mol NA = 7.10 cm3 6.02 1023 atoms/mol = 1.18 1023 cm3 = 1.18 10 -29
m 3 (c) datom = 3 1.18 1029 m3 = 2.28 1010 m = 0.228 nm (d) V1 mol
U = (1 mol)(238 g/mol) 18.7 g/cm3 = 12.7 cm3 V1 atom U = V1 mol U
NA = 12.7 cm3 6.02 1023 atoms/mol = 2.11 1023 cm3 = 2.11 10 -29 m 3
datom U = 3 V1 atom U = 3 2.11 1029 m3 = 2.77 1010 m = 0.277 nm
*1.6 r2 = r1 3 5 = (4.50 cm)(1.71) = 7.69 cm 1.7 Use m = molar
mass/NA and 1 u = 1.66 10 -24 g (a) For He, m = 4.00 g/mol 6.02 10
23 mol -1 = 6.64 10 -24 g = 4.00 u (b) For Fe, m = 55.9 g/mol 6.02
10 23 mol -1 = 9.29 10 -23 g = 55.9 u (c) For Pb, m = 207 g/mol
6.02 10 23 mol -1 = 3.44 10 -22 g = 207 u
Chapter 1 Solutions 3 2000 by Harcourt College Publishers. All
rights reserved. Goal Solution Calculate the mass of an atom of (a)
helium, (b) iron, and (c) lead. Give your answers in atomic mass
units and in grams. The molar masses are 4.00, 55.9, and 207 g/mol,
respectively, for the atoms given. Gather information: The mass of
an atom of any element is essentially the mass of the protons and
neutrons that make up its nucleus since the mass of the electrons
is negligible (less than a 0.05% contribution). Since most atoms
have about the same number of neutrons as protons, the atomic mass
is approximately double the atomic number (the number of protons).
We should also expect that the mass of a single atom is a very
small fraction of a gram (~1023 g) since one mole (6.02 1023) of
atoms has a mass on the order of several grams. Organize: An atomic
mass unit is defined as 1/12 of the mass of a carbon-12 atom (which
has a molar mass of 12.0 g/mol), so the mass of any atom in atomic
mass units is simply the numerical value of the molar mass. The
mass in grams can be found by multiplying the molar mass by the
mass of one atomic mass unit (u): 1 u = 1.66 1024 g. Analyze: For
He, m = 4.00 u = (4.00 u)(1.66 1024 g/u) = 6.64 1024 g For Fe, m =
55.9 u = (55.9 u)(1.66 1024g/u) = 9.28 1023 g For Pb, m = 207 u =
(207 u)(1.66 1024 g/u) = 3.44 1022 g Learn: As expected, the mass
of the atoms is larger for bigger atomic numbers. If we did not
know the conversion factor for atomic mass units, we could use the
mass of a proton as a close approximation: 1u mp = 1.67 1024 g.
*1.8 n = m M = 3.80 g 3.35 g 197 g/mol = 0.00228 mol N = (n)NA =
(0.00228 mol)(6.02 10 23 atoms/mol) = 1.38 10 21 atoms t = (50.0
yr)(365 d/yr)(24.0 hr/d)(3600 s/hr) = 1.58 10 9 s N t = 1.38 10 21
atoms 1.58 10 9 s = 8.72 10 11 atoms/s 1.9 (a) m = L 3 = (7.86 g/cm
3 )(5.00 10 -6 cm) 3 = 9.83 10 -16 g (b) N = m NA Molar mass =
(9.83 10 -16 g)(6.02 10 23 atoms/mol) 55.9 g/mol = 1.06 10 7
atoms
4 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 1.10 (a) The cross-sectional area is A = 2(0.150
m)(0.010 m) + (0.340 m)(0.010 m) = 6.40 10 -3 m 2 The volume of the
beam is V = AL = (6.40 10 -3 m 2 )(1.50 m) = 9.60 10 -3 m 3 Thus,
its mass is m = V = (7.56 10 3 kg/m 3 )(9.60 10 -3 m 3 ) = 72.6 kg
(b) Presuming that most of the atoms are of iron, we estimate the
molar mass as M = 55.9 g/mol = 55.9 10 -3 kg/mol. The number of
moles is then n = m M = 72.6 kg 55.9 10 -3 kg/mol = 1.30 10 3 mol
The number of atoms is N = nNA = (1.30 10 3 mol)(6.02 10 23
atoms/mol) = 7.82 10 26 atoms *1.11 (a) n = m M = 1.20 10 3 g 18.0
g/mol = 66.7 mol, and Npail = nNA = (66.7 mol)(6.02 10 23
molecules/mol) = 4.01 10 25 molecules (b) Suppose that enough time
has elapsed for thorough mixing of the hydrosphere. Nboth = Npail
mpail Mtotal = (4.01 10 25 molecules) 1.20 kg 1.32 10 21 kg , or
Nboth = 3.65 10 4 molecules 1.12 r, a, b, c and s all have units of
L. (s a)(s b)(s c) s = L L L L = L 2 = L Thus, the equation is
dimensionally consistent. 15.0 cm 36.0 cm36.0 cm 1.00 cm 1.00 cm
15.0 cm
Chapter 1 Solutions 5 2000 by Harcourt College Publishers. All
rights reserved. 1.13 The term s has dimensions of L, a has
dimensions of LT -2 , and t has dimensions of T. Therefore, the
equation, s = ka m t n has dimensions of L = (LT -2 ) m (T) n or L
1 T 0 = L m T n-2m The powers of L and T must be the same on each
side of the equation. Therefore, L 1 = L m and m = 1 Likewise,
equating terms in T, we see that n 2m must equal 0. Thus, n = 2m =
2 The value of k, a dimensionless constant, cannot be obtained by
dimensional analysis . 1.14 2 l g = L L/T 2 = T 2 = T 1.15 (a) This
is incorrect since the units of [ax] are m 2 /s 2 , while the units
of [v] are m/s. (b) This is correct since the units of [y] are m,
and cos(kx) is dimensionless if [k] is in m -1 . 1.16 Inserting the
proper units for everything except G, kg m s2 = G[kg]2 [m]2
Multiply both sides by [m] 2 and divide by [kg] 2 ; the units of G
are m 3 kg s 2 1.17 One month is 1 mo = (30 day)(24 hr/day)(3600
s/hr) = 2.592 10 6 s Applying units to the equation, V = (1.50 Mft
3 /mo)t + (0.00800 Mft 3 /mo 2 )t 2 Since 1 Mft 3 = 10 6 ft 3 , V =
(1.50 10 6 ft 3 /mo)t + (0.00800 10 6 ft 3 /mo 2 )t 2
6 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. Converting months to seconds, V = 1.50 10 6 ft 3
/mo 2.592 10 6 s/mo t + 0.00800 10 6 ft 3 /mo 2 (2.592 10 6 s/mo) 2
t 2 Thus, V[ft 3 ] = (0.579 ft3 /s)t + (1.19 10-9 ft3 /s2 )t2 *1.18
Apply the following conversion factors: 1 in = 2.54 cm, 1 d = 86400
s, 100 cm = 1 m, and 10 9 nm = 1 m 1 32 in/day (2.54 cm/in)(10 -2
m/cm)(10 9 nm/m) 86400 s/day = 9.19 nm/s This means the proteins
are assembled at a rate of many layers of atoms each second! 1.19
Area A = (100 ft)(150 ft) = 1.50 10 4 ft 2 , so A = (1.50 10 4 ft 2
)(9.29 10 -2 m 2 /ft 2 ) = 1.39 10 3 m 2 Goal Solution A
rectangular building lot is 100 ft by 150 ft. Determine the area of
this lot in m2. G: We must calculate the area and convert units.
Since a meter is about 3 feet, we should expect the area to be
about A (30 m)(50 m) = 1 500 m2. O: Area = Length Width. Use the
conversion: 1 m = 3.281 ft. A: A = L W = (100 ft) 1 m 3.281 ft (150
ft ) 1 m 3.281 ft = 1 390 m2 L: 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. 1.20 (a) V = (40.0 m)(20.0 m)(12.0 m) = 9.60 10 3 m
3 V = 9.60 10 3 m 3 (3.28 ft/1 m) 3 = 3.39 10 5 ft 3
Chapter 1 Solutions 7 2000 by Harcourt College Publishers. All
rights reserved. (b) The mass of the air is m = airV = (1.20 kg/m 3
)(9.60 10 3 m 3 ) = 1.15 10 4 kg The student must look up weight in
the index to find Fg = mg = (1.15 10 4 kg)(9.80 m/s 2 ) = 1.13 10 5
N Converting to pounds, Fg = (1.13 10 5 N)(1 lb/4.45 N) = 2.54 10 4
lb *1.21 (a) Seven minutes is 420 seconds, so the rate is r = 30.0
gal 420 s = 7.14 10 -2 gal/s (b) Converting gallons first to
liters, then to m 3 , r = 7.14 10 -2 gal s 3.786 L 1 gal 10 -3 m 3
1 L r = 2.70 10 -4 m 3 /s (c) At that rate, to fill a 1-m 3 tank
would take t = 1 m 3 2.70 10 -4 m 3 /s 1 hr 3600 s = 1.03 hr 1.22 v
= 5.00 furlongs fortnight 220 yd 1 furlong 0.9144 m 1 yd 1
fortnight 14 days 1 day 24 hrs 1 hr 3600 s = 8.32 10 -4 m/s This
speed is almost 1 mm/s; so we might guess the creature was a snail,
or perhaps a sloth. 1.23 It is often useful to remember that the
1600-m race at track and field events is approximately 1 mile in
length. To be precise, there are 1609 meters in a mile. Thus, 1
acre is equal in area to (1 acre) 1 mi 2 640 acres 1609 m mi 2 =
4.05 10 3 m 2
8 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 1.24 Volume of cube = L 3 = 1 quart (Where L =
length of one side of the cube.) Thus, L 3 = (1 quart) 1 gallon 4
quarts 3.786 liters 1 gallon 1000 cm3 1 liter = 946 cm 3 , and L =
9.82 cm 1.25 The mass and volume, in SI units, are m = (23.94 g) 1
kg 1000 g = 0.02394 kg V = (2.10 cm 3 )(10 -2 m/cm) 3 = 2.10 10 -6
m 3 Thus, the density is = m V = 0.02394 kg 2.10 10 -6 m 3 = 1.14
10 4 kg/m 3 Goal Solution A solid piece of lead has a mass of 23.94
g and a volume of 2.10 cm3. From these data, calculate the density
of lead in SI units (kg/m3). G: 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. O: Density is defined as mass per volume, in = m V
. We must convert to SI units in the calculation. A: = 23.94 g 2.10
cm3 1 kg 1000 g 100 cm 1 m 3 = 1.14 104 kg/m3 L: 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.
Chapter 1 Solutions 9 2000 by Harcourt College Publishers. All
rights reserved. 1.26 (a) We take information from Table 1.1: 1 LY
= (9.46 10 15 m) 1 AU 1.50 10 11 m = 6.31 10 4 AU (b) The distance
to the Andromeda galaxy is 2 10 22 m = (2 10 22 m) 1 AU 1.50 10 11
m = 1.33 10 11 AU 1.27 Natoms = mSun matom = 1.99 10 30 kg 1.67 10
-27 kg = 1.19 10 57 atoms 1.28 1 mi = 1609 m = 1.609 km; thus, to
go from mph to km/h, multiply by 1.609. (a) 1 mi/h = 1.609 km/h (b)
55 mi/h = 88.5 km/h (c) 65 mi/h = 104.6 km/h. Thus, v = 16.1 km/h
1.29 (a) 6 1012 $ 1000 $/s 1 hr 3600 s 1 day 24 hr 1 yr 365 days =
190 years (b) The circumference of the Earth at the equator is 2
(6378 10 3 m) = 4.01 10 7 m. The length of one dollar bill is 0.155
m so that the length of 6 trillion bills is 9.30 10 11 m. Thus, the
6 trillion dollars would encircle the Earth 9.30 10 11 m 4.01 10 7
m = 2.32 10 4 times Goal Solution At the time of this books
printing, the U.S. national debt is about $6 trillion. (a) If
payments were made at the rate of $1 000 per second, how many years
would it take to pay off a $6-trillion debt, assuming no interest
were charged? (b) A dollar bill is about 15.5 cm long. If six
trillion dollar bills were laid end to end around the Earths
equator, how many times would they encircle the Earth? Take the
radius of the Earth at the equator to be 6 378 km. (Note: Before
doing any of these calculations, try to guess at the answers. You
may be very surprised.) (a) G: $6 trillion is certainly a large
amount of money, so even at a rate of $1000/second, we might guess
that it will take a lifetime (~ 100 years) to pay off the debt. O:
Time to repay the debt will be calculated by dividing the total
debt by the rate at which it is repaid.
10 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. A: T = $6 trillion $1000/s = $6 1012
($1000/s)(3.16 107 s/yr) = 190 yr L: OK, so our estimate was a bit
low. $6 trillion really is a lot of money! (b) G: We might guess
that 6 trillion bills would encircle the Earth at least a few
hundred times, maybe more since our first estimate was low. O: The
number of bills can be found from the total length of the bills
placed end to end divided by the circumference of the Earth. A: N =
L C = (6 1012)(15.5 cm)(1 m/100 cm) 2 6.37 106 m = 2.32 104 times
L: OK, so again our estimate was low. Knowing that the bills could
encircle the earth more than 20 000 times, it might be reasonable
to think that 6 trillion bills could cover the entire surface of
the earth, but the calculated result is a surprisingly small
fraction of the earths surface area! 1.30 (a) (3600 s/hr)(24
hr/day)(365.25 days/yr) = 3.16 10 7 s/yr (b) Vmm = 4 3 r 3 = 4 3
(5.00 10 -7 m) 3 = 5.24 10 -19 m 3 Vcube Vmm = 1 m 3 5.24 10 -19 m
3 = 1.91 10 18 micrometeorites This would take 1.91 10 18
micrometeorites 3.16 10 7 micrometeorites/yr = 6.05 10 10 yr 1.31 V
= At, so t = V A = 3.78 10 -3 m 3 25.0 m 2 = 1.51 10 -4 m (or 151
m) 1.32 V = 1 3 Bh = [(13.0 acres)(43560 ft 2 /acre)] 3 (481 ft) =
9.08 10 7 ft 3 , or V = (9.08 10 7 ft 3 ) 2.83 10 -2 m 3 1 ft 3 =
2.57 10 6 m 3 h BBB hh
Chapter 1 Solutions 11 2000 by Harcourt College Publishers. All
rights reserved. 1.33 Fg = (2.50 tons/block)(2.00 10 6 blocks)(2000
lb/ton) = 1.00 10 10 lbs 1.34 The area covered by water is Aw =
0.700 AEarth = (0.700)(4 REarth 2 ) = (0.700)(4)(6.37 10 6 m) 2 =
3.57 10 14 m 2 The average depth of the water is d = (2.30
miles)(1609 m/l mile) = 3.70 10 3 m The volume of the water is V =
Awd = (3.57 10 14 m 2 )(3.70 10 3 m) = 1.32 10 18 m 3 and the mass
is m = V = (1000 kg/m 3 )(1.32 10 18 m 3 ) = 1.32 10 21 kg *1.35 SI
units of volume are in m 3 : V = (25.0 acre-ft) 43560 ft 2 1 acre
0.3048 m 1 ft 3 = 3.08 10 4 m 3 *1.36 (a) dnucleus, scale =
dnucleus, real datom, scale datom, real = (2.40 10 -15 m) 300 ft
1.06 10 -10 m = 6.79 10 -3 ft, or dnucleus, scale = (6.79 10 -3
ft)(304.8 mm/1 ft) = 2.07 mm (b) Vatom Vnucleus = 4 r 3 atom/3 4 r
3 nucleus/3 = ratom rnucleus 3 = datom dnucleus 3 = 1.06 10 -10 m
2.40 10 -15 m 3 = 8.62 10 13 times as large 1.37 The scale factor
used in the "dinner plate" model is S = 0.25 m 1.0 10 5 lightyears
= 2.5 10 -6 m/lightyears The distance to Andromeda in the scale
model will be Dscale = DactualS = (2.0 10 6 lightyears)(2.5 10 -6
m/lightyears) = 5.0 m
12 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 1.38 (a) AEarth AMoon = 4 rEarth 2 4rMoon 2 =
rEarth rMoon 2 = (6.37 10 6 m)(100 cm/m) 1.74 10 8 cm 2 = 13.4 (b)
VEarth VMoon = 4rEarth 3 /3 4rMoon 3 /3 = rEarth rMoon 3 = (6.37 10
6 m)(100 cm/m) 1.74 10 8 cm 3 = 49.1 1.39 To balance, mFe = mAl or
FeVFe = AlVAl Fe 4 3 r 3 Fe = Al 4 3 r 3 Al rAl = rFe Fe Al 1/3 rAl
= (2.00 cm) 7.86 2.70 1/3 = 2.86 cm 1.40 The mass of each sphere is
mA1 = AlVAl = 4AlrAl 3 3 and mFe = FeVFe = 4FerFe 3 3 Setting these
masses equal, 4Fer 3 Fe 3 = 4Fer 3 Fe 3 and rAl = rFe 3 Fe/Al 1.41
The volume of the room is 4 4 3 = 48 m 3 , while the volume of one
ball is 4 3 0.038 m 2 3 = 2.87 10-5 m 3 . Therefore, one can fit
about 48 2.87 10 -5 10 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 2
6 = 0.74 so that at least 26% of the space will be empty.
Therefore, the above estimate reduces to 1.67 10 6 0.740 10 6
.
Chapter 1 Solutions 13 2000 by Harcourt College Publishers. All
rights reserved. Goal Solution Estimate the number of Ping-Pong
balls that would fit into an average-size room (without being
crushed). In your solution state the quantities you measure or
estimate and the values you take for them. G: Since the volume of a
typical room is much larger than a Ping-Pong ball, we should expect
that a very large number of balls (maybe a million) could fit in a
room. O: Since we are only asked to find an estimate, we do not
need to be too concerned about how the balls are arranged.
Therefore, to find the number of balls we can simply divide the
volume of an average-size room by the volume of an individual
Ping-Pong ball. A: A typical room (like a living room) might have
dimensions 15 ft 20 ft 8 ft. Using the approximate conversion 1 ft
= 30 cm, we find Vroom 15 ft 20 ft 8 ft = 2400 ft3 30 cm 1 ft 3 = 7
107 cm3 A Ping-Pong ball has a diameter of about 3 cm, so we can
estimate its volume as a cube: Vball (3 3 3) cm3 = 30 cm3 The
number of Ping-Pong balls that can fill the room is N Vroom Vball =
7 107 cm3 30 cm3 = 2 106 balls ~ 106 balls L: So a typical room can
hold about a million Ping-Pong balls. This problem gives us a sense
of how big a million really is. *1.42 It might be reasonable to
guess that, on average, McDonalds sells a 3 cm 8 cm 10 cm = 240 cm
3 medium-sized box of fries, and that it is packed 3/4 full with
fries that have a cross section of 1/2 cm 1/2 cm. Thus, the typical
box of fries would contain fries that stretched a total of L = 3 4
V A = 3 4 240 cm 3 (0.5 cm) 2 = 720 cm = 7.2 m 250 million boxes
would stretch a total distance of (250 10 6 box)(7.2 m/box) = 1.8
10 9 m. But we require an order of magnitude, so our answer is 10 9
m = 1 million kilometers . *1.43 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 mi)(5280 ft/mi)(1 rev/8 ft) =
3 10 7 rev 10 7 rev
14 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 1.44 A typical raindrop is spherical and might
have a radius of about 0.1 inch. Its volume is then approximately 4
10 -3 in 3 . Since 1 acre = 43,560 ft 2 , the volume of water
required to cover it to a depth of 1 inch is (1 acre)(1 inch) = (1
acre in) 43,560 ft2 1 acre 144 in 2 1 ft 2 6.3 10 6 in 3 . The
number of raindrops required is n = volume of water required volume
of a single drop 6.3 10 6 in 3 4 10 -3 in 3 = 1.6 10 9 10 9 *1.45
In order to reasonably carry on photosynthesis, we might expect a
blade of grass to require at least 1/16 in 2 = 43 10 -5 ft 2 .
Since 1 acre = 43,560 ft 2 , the number of blades of grass to be
expected on a quarter-acre plot of land is about n= total area area
per blade = (0.25 acre)(43,560 ft 2 /acre) 43 10 -5 ft 2 /blade =
2.5 10 7 blades 10 7 blades 1.46 Since you have only 16 hours
(57,600 s) available per day, you can count only $57,600 per day.
Thus, the time required to count $1 billion dollars is t = 10 9
dollars 5.76 10 4 dollars/day 1 year 365 days = 47.6 years Since
you are at least 18 years old, you would be beyond age 65 before
you finished counting the money. It would provide a nice
retirement, but a very boring life until then. We would not advise
it. 1.47 Assume the tub measure 1.3 m by 0.5 m by 0.3 m. One-half
of its volume is then V = (0.5)(1.3 m)(0.5 m)(0.3 m) = 0.10 m 3 The
mass of this volume of water is mwater = waterV= (1000 kg/m 3
)(0.10 m 3 ) = 100 kg ~10 2 kg Pennies are now mostly zinc, but
consider copper pennies filling 50% of the volume of the tub. The
mass of copper required is mcopper = copperV = (8930 kg/m 3 )(0.10
m 3 ) = 893 kg ~10 3 kg
Chapter 1 Solutions 15 2000 by Harcourt College Publishers. All
rights reserved. *1.48 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 6 cans/day)(365 days/year) 10 10 cans are thrown away or
recycled each year. Guessing that each can weighs around 1/10 of an
ounce, we estimate this represents (10 10 cans)(0.1 oz/can)(1 lb/16
oz)(1 ton/2000 lb) 3.1 10 5 tons/year. 10 5 tons 1.49 Assume: Total
population = 10 7 ; 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 1000 pianos 1 piano 100 people (10 7 people) = 100
1.50 (a) 2 (b) 4 (c) 3 (d) 2 1.51 (a) r 2 = (10.5 m 0.2 m) 2 = [
](10.5 m) 2 2(10.5 m)(0.2 m) + (0.2 m) 2 = 346 m 2 13 m 2 (b) 2r =
2 (10.5 m 0.2 m) = 66.0 m 1.3 m 1.52 (a) 756.?? 37.2? 0.83 + 2.5?
796./5/3 = 797 (b) 0.0032 (2 s.f.) 356.3 (4 s.f.) = 1.14016 = (2
s.f.) 1.1 (c) 5.620 (4 s.f.) (> 4 s.f.) = 17.656 = (4 s.f.)
17.66
16 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 1.53 r = (6.50 0.20) cm = (6.50 0.20) 10 -2 m m =
(1.85 0.02) kg = m 4 3 r 3 also, = m m + 3r r In other words, the
percentages of uncertainty are cumulative. Therefore, = 0.02 1.85 +
3(0.20) 6.50 = 0.103 = 1.85 4 3 (6.5 10 -2 m) 3 = 1.61 10 3 kg/m 3
and = (1.61 0.17) 10 3 kg/m 3 1.54 (a) 3 (b) 4 (c) 3 (d) 2 1.55 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 1.56 V = 2V1 + 2V2 = 2(V1 + V2) V1 = (17.0 m + 1.0 m + 1.0
m)(1.0 m)(0.09 m) = 1.70 m 3 V2 = (10.0 m)(1.0 m)(0.090 m) = 0.900
m 3 V = 2(1.70 m 3 ) + 2(0.900 m 3 ) = 5.2 m 3 l1 l1 = 0.12 m 19.0
m = 0.0063 w1 w1 = 0.01 m 1.0 m = 0.010 t1 t1 = 0.1 cm 9.0 cm =
0.011 V V = 0.006 + 0.010 + 0.011 = 0.027 = 2.7% 19.0 m 36.0 cm10.0
m 19.0 m
Chapter 1 Solutions 17 2000 by Harcourt College Publishers. All
rights reserved. *1.57 It is desired to find the distance x such
that x 100 m = 1000 m x (i.e., such that x is the same multiple of
100 m as the multiple that 1000 m is of x) . Thus, it is seen that
x 2 = (100 m)(1000 m) = 1.00 10 5 m 2 , and therefore x = 1.00 10 5
m 2 = 316 m . 1.58 The volume of oil equals V = 9.00 10 -7 kg 918
kg/m 3 = 9.80 10 10 m 3 . If the diameter of a molecule is d, then
that same volume must equal d(r 2 ) = (thickness of slick)(area of
oil slick) where r = 0.418 m. Thus, d = 9.80 10 -10 m 3 (0.418 m) 2
= 1.79 10 -9 m 1.59 Atotal = (N)(Adrop) = Vtotal Vdrop (Adrop) =
Vtotal 4r 3 /3 (4r 2 ) = 3Vtotal r = 3 30.0 10 -6 m 3 2.00 10 -5 m
= 4.50 m 2 1.60 ' (deg) (rad) tan() sin() 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% 1.61 2r = 15.0 m r = 2.39 m h r =
tan55.0 h = (2.39 m)tan(55.0) = 3.41 m
18 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 5555 h rr h
Chapter 1 Solutions 19 2000 by Harcourt College Publishers. All
rights reserved. *1.62 (a) [V] = L 3 , [A] = L 2 , [h] = L [V] =
[A][h] L 3 = L 3 L = L 3 . Thus, the equation is dimensionally
correct. (b) Vcylinder = R 2 h = (R 2 )h = Ah, where A = R 2
Vrectangular object = lwh = ( lw)h = Ah, where A = lw 1.63 The
actual number of seconds in a year is (86,400 s/day)(365.25 day/yr)
= 31,557,600 s/yr The percentage error in the approximation is thus
( 107 s/yr) (31,557,600 s/yr) 31,557,600 s/yr 100% = 0.449% *1.64
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, Ldiag = L 2 + L 2 . Thus, since the atoms are
separated by a distance L = 0.200 nm, the diagonal planes are
separated 1 2 L 2 + L 2 = 0.141 nm *1.65 (a) The speed of flow may
be found from v = (Vol rate of flow) (Area: D 2 /4) = 16.5 cm 3 /s
(6.30 cm) 2 /4 = 0.529 cm/s (b) Likewise, at a 1.35 cm diameter, v
= 16.5 cm 3 /s (1.35 cm) 2 /4 = 11.5 cm/s *1.66 t = V A = V D 2 /4
= 4(12.0 cm 3 ) (23.0 cm) 2 = 0.0289 cm 1 m 100 cm 10 6 m 1 m = 289
m 1.67 V20 mpg = (10 8 cars)(10 4 mi/yr) 20 mi/gal = 5.0 10 10
gal/yr V25 mpg = (10 8 cars)(10 4 mi/yr) 25 mi/gal = 4.0 10 10
gal/yr Fuel saved = V25 mpg V20 mpg = 1.0 10 10 gal/yr
20 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved.
Chapter 1 Solutions 21 2000 by Harcourt College Publishers. All
rights reserved. 1.68 (a) 1 cubic meter of water has a mass m = V =
(1.00 10 -3 kg/cm 3 )(1.00 m 3 )(10 2 cm/m) 3 = 1000 kg (b) As a
rough calculation, we treat each item as if it were 100% water.
cell: m = V = Error! R 3 ) = Error! D 3 ) = ( )1000 kg/m 3 1 6 (1.0
10 -6 m) 3 = 5.2 10 -16 kg kidney: m = V = Error! R 3 ) = (1.00 10
-3 kg/cm 3 )Error! 3 = Error! fly: m = 4 D 2 h = (1 10 -3 kg/cm 3 )
4 (2.0 mm) 2 (4.0 mm)(10 -1 cm/mm) 3 = 1.3 10 -5 kg 1.69 The volume
of the galaxy is r 2 t = (10 21 m) 2 10 19 m ~ 10 61 m 3 If the
distance between stars is 4 10 16 m, then there is one star in a
volume on the order of (4 10 16 m) 3 ~ 10 50 m 3 . The number of
stars is about 10 61 m 3 10 50 m 3 /star ~ 10 11 stars
22 Chapter 1 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 1.70 The density of each material is = m V = m r 2
h = 4m D 2 h Al: = 4(51.5 g) (2.52 cm) 2 (3.75 cm) = 2.75 g cm 3
The tabulated value 2.70 g cm 3 is 2% smaller. Cu: = 4(56.3 g)
(1.23 cm) 2 (5.06 cm) = 9.36 g cm 3 The tabulated value 8.92 g cm 3
is 5% smaller. Brass: = 4(94.4 g) (1.54 cm) 2 (5.69 cm) = 8.91 g cm
3 Sn: = 4(69.1 g) (1.75 cm) 2 (3.74 cm) = 7.68 g cm 3 Fe: = 4(216.1
g) (1.89 cm) 2 (9.77 cm) = 7.88 g cm 3 The tabulated value 7.86 g
cm 3 is 0.3% smaller.
2000 by Harcourt College Publishers. All rights reserved.
Chapter 2 Even Answers 2. (a) 180 km (b) 63.4 km/h 4. (a) 50.0 m/s
(b) 41.0 m/s 6. (a) 2v1v2/(v1 + v2) (b) 0 8. (a) 27.0 m (b) xf =
27.0 m + (18.0 m/s)t + (3.00 m/s2)(t)2 (c) 18.0 m/s 10. (b) vt =
5.0 s = 23 m/s, vt = 4.0 s = 18 m/s, vt = 3.0 s = 14 m/s, vt = 2.0
s = 9.0 m/s (c) 4.6 m/s2 (d) 0 12. 4.00 m/s2, sign indicates that
acceleration is in negative x direction 14. (a) 20.0 m/s, 5.00 m/s
(b) 262 m 16. (c) 4 m/s2 (d) 34 m (e) 28 m 18. (a) 13.0 m/s (b)
10.0 m/s, 16.0 m/s (c) 6.00 m/s2 (d) 6.00 m/s2 20. (f) The spacing
of the successive positions would change with less regularity. 22.
(a) 5.25 m/s2 (b) 168 m (c) 52.5 m/s 24. 160 ft 26. (a) 1.87 km (b)
1.46 km (c) a1 = 3.3 m/s2 (0 t 15 s), a2 = 0 (15 s t 40 s), a3 =
5.0 m/s2 (40 s t 50 s) (d) (i) x1 = (1.67 m/s2)t2, (ii) x2 = (50
m/s)t 375 m, (iii) x3 = (250 m/s)t (2.5 m/s2)t2 4375 m (e) 37.5 m/s
28. (a) 12.7 m/s (b) -2.30 m/s 30. (a) x = (30.0t t2) m, v = (30.0
2.00t) m/s (b) 225 m 32. 3.10 m/s 34. (a) 4.90 105 m/s2 (b) 3.57
104 s (c) 18.0 cm 36. 200 m 38. (a) 4.98 109 s (b) 1.20 1015 m/s2
40. 11.4 s, 212 m 42. $99.4/h 44. 1.79 s 46. gh 48. (a) 96.0 ft/s
downward (b) 3.07 103 ft/s2 upward (c) 3.13 102 s 50. (a) 98.0 m/s
(b) 490 m 52. 7.96 s 54. (a) a = (10.0 107 m/s3)t + 3.00 105 m/s2;
x = (1.67 107 m/s3)t3 + (1.50 105 m/s2)t2 (b) 3.00 103 s (c) 450
m/s (d) 0.900 m 56. (a) 0.111 s (b) 5.53 m/s 58. 48.0 mm 60. (a)
15.0 s (b) 30.0 m/s (c) 225 m 62. 155 s, 129 s 64. ~ 103 m/s2 66.
(a) 26.4 m (b) 6.82% 68. 1.38 103 m 70. (c) v 2 boy h , 0 (d) vboy,
0 72. (b) a = 1.63 m/s2 downward
2 Chapter 2 Even Answers 2000 by Harcourt College Publishers.
All rights reserved.
2000 by Harcourt College Publishers. All rights reserved.
Chapter 2 Solutions *2.1 (a) v = 2.30 m/s (b) v = x t = 57.5 m 9.20
m 3.00 s = 16.1 m/s (c) v = x t = 57.5 m 0 m 5.00 s = 11.5 m/s 2.2
(a) Displacement = (8.50 104 m/h) 35.0 60.0 h + 130 103 m x = (49.6
+ 130) 103 m = 180 km (b) Average velocity = displacement time =
180 km (35.0 + 15.0) 60.0 + 2.00 h = 63.4 km/h 2.3 (a) vav = x t =
10 m 2 s = 5 m/s (b) vav = 5 m 4 s = 1.2 m/s (c) vav = x2 x1 t2 t1
= 5 m 10 m 4 s 2 s = 2.5 m/s (d) vav = x2 x1 t2 t1 = 5 m 5 m 7 s 4
s = 3.3 m/s (e) vav = x2 x1 t2 t1 = 0 0 8 0 = 0 m/s 2.4 x = 10t2
For t(s) = 2.0 2.1 3.0 x(m) = 40 44.1 90 (a) v = x t = 50m 1.0 s =
50.0 m/s (b) v = x t = 4.1 m 0.1 s = 41.0 m/s
2 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 2.5 (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 m/s = d t1 . Let t2 represent the longer time for the return
trip in 3.00 m/s = d t2 . Then the times are t1 = d (5.00 m/s) and
t2 = d (3.00 m/s) . The average speed is: v = Total distance Total
time = d + d d (5.00 m/s) + d (3.00 m/s) = 2d (8.00 m/s)d (15.0
m2/s2) v = 2(15.0 m2/s2) 8.00 m/s = 3.75 m/s (b) She starts and
finishes at the same point A. With total displacement = 0, average
velocity = 0 2.6 (a) v = Total distance Total time Let d be the
distance from A to B. Then the time required is d v1 + d v2 . And
the average speed is v = 2d d v1 + d v2 = 2v1v2 v1 + v2 (b) With
total displacement zero, her average velocity is 0 . 2.7 (a) 5 0 5
2 4 6 x (m) t (s) (b) v = slope = 5.00 m (3.00 m) (6.00 s 1.00 s) =
8.00 m 5.00 s = 1.60 m/s
Chapter 2 Solutions 3 2000 by Harcourt College Publishers. All
rights reserved. 2.8 (a) At any time, t, the displacement is given
by x = (3.00 m/s2)t2. Thus, at ti = 3.00 s: xi = (3.00 m/s2)(3.00
s)2 = 27.0 m (b) At tf = 3.00 s + t :xf = (3.00 m/s2)(3.00 s + t)2,
or xf = 27.0 m + (18.0 m/s)t + (3.00 m/s2)(t)2 (c) The
instantaneous velocity at t = 3.00 s is: v = lim t o xf xi t = lim
t 0 [(18.0 m/s) + (3.00 m/s2)t], or v = 18.0 m/s 2.9 (a) at ti =
1.5 s, xi = 8.0 m (Point A) at tf = 4.0 s, xf = 2.0 m (Point B) v =
xf xi tf ti = (2.0 8.0) m (4 1.5) s = 6.0 m 2.5 s = 2.4 m/s (b) The
slope of the tangent line is found from points C and D. (tC = 1.0
s, xC = 9.5 m) and (tD = 3.5 s, xD = 0), v 3.8 m/s (c) The velocity
is zero when x is a minimum. This is at t 4 s . 2.10 (b) At t = 5.0
s, the slope is v 58 m 2.5 s 23 m/s At t = 4.0 s, the slope is v 54
m 3 s 18 m/s At t = 3.0 s, the slope is v 49 m 3.4 s 14 m/s At t =
2.0 s, the slope is v 36 m 4.0 s 9.0 m/s (c) a = v t 23 m/s 5.0 s
4.6 m/s2 (d) Initial velocity of the car was zero . 4 2 0 8 6 3210
4 5 6 t (s) 10 12 x (m) A C B D A C B D 60 40 20 0 0 2 4 x (m) t
(s) 20 0 0 2 4 v (m/s) t (s)
4 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 2.11 (a) v = (5 0) m (1 0) s = 5 m/s (b) v = (5
10) m (4 2) s = 2.5 m/s (c) v = (5 m 5 m) (5 s 4 s) = 0 (d) v = 0
(5 m) (8 s 7 s) = +5 m/s 2.12 a = vf vi tf ti = 0 60.0 m/s 15.0 s 0
= 4.00 m/s2 The negative sign in the result shows that the
acceleration is in the negative x direction. *2.13 Choose the
positive direction to be the outward perpendicular to the wall. v =
vi + at a = v t = 22.0 m/s (25.0 m/s) 3.50 103 s = 1.34 104 m/s2
2.14 (a) Acceleration is constant over the first ten seconds, so at
the end v = vi + at = 0 + (2.00 m/s2)(10.0 s) = 20.0 m/s 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 = vi + at = 20.0 m/s (3.00
m/s2)(5.00 s) = 5.00 m/s (b) In the first ten seconds x = xi + vit
+ 1 2 at2 = 0 + 0 + 1 2 (2.00 m/s2)(10.0 s) 2 = 100 m Over the next
five seconds the position changes to x = xi + vit + 1 2 at2 = 100 m
+ 20.0 m/s (5.00 s) + 0 = 200 m And at t = 20.0 s x = xi + vit + 1
2 at2 = 200 m + 20.0 m/s (5.00 s) + 1 2 (3.00 m/s2)(5.00 s) 2 = 262
m 4 2 0 2 4 8 6 321 4 5 6321 4 5 6 t (s) 7 87 8 10 x (m) 6
Chapter 2 Solutions 5 2000 by Harcourt College Publishers. All
rights reserved. *2.15 (a) Acceleration is the slope of the graph
of v vs t. For 0 < t < 5.00 s, a = 0 For 15.0 s < t <
20.0 s, a = 0 For 5.0 s < t < 15.0 s, a = vf vi tf ti a =
8.00 (8.00) 15.0 5.00 = 1.60 m/s2 We can plot a(t) as shown. (b) a
= vf vi tf ti (i) For 5.00 s < t < 15.0 s, ti = 5.00 s, vi =
8.00 m/s tf = 15.0 s, vf = 8.00 m/s; a = vf vi tf ti = 8.00 (8.00)
15.0 5.00 = 1.60 m/s2 (ii) ti = 0, vi = 8.00 m/s, tf = 20.0 s, vf =
8.00 m/s a = vf vi tf ti = 8.00 (8.00) 20.0 0 = 0.800 m/s2 0.0 1.0
1050 15 20 t (s) 1.6 2.0 a (m/s2 )
6 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 2.16 (a) See the Graphs at the right. Choose x = 0
at t = 0 At t = 3 s, x = 1 2 (8 m/s)(3 s) = 12 m At t = 5 s, x = 12
m + (8 m/s)(2 s) = 28 m At t = 7 s, x = 28 m + 1 2 (8 m/s)(2 s) =
36 m (b) For 0 < t < 3 s, a = (8 m/s)/3 s = 2.67 m/s2 For 3
< t < 5 s, a = 0 (c) For 5 s < t < 9 s, a = (16 m/s)/4
s = 4 m/s2 (d) At t = 6 s, x = 28 m + (6 m/s)(1 s) = 34 m (e) At t
= 9 s, x = 36 m + 1 2 ( 8 m/s) 2 s = 28 m 2.17 x = 2.00 + 3.00t t2,
v = dx d t = 3.00 2.00t, a = dv d t = 2.00 At t = 3.00 s: (a) x =
(2.00 + 9.00 9.00) m = 2.00 m (b) v = (3.00 6.00) m/s = 3.00 m/s
(c) a = 2.00 m/s2 2.18 (a) At t = 2.00 s, x = [3.00(2.00)2
2.00(2.00) + 3.00] m = 11.0 m At t = 3.00 s, x = [3.00(9.00)2
2.00(3.00) + 3.00] m = 24.0 m so v = x t = 24.0 m 11.0 m 3.00 s
2.00 s = 13.0 m/s 20 0 40 50 5 t (s) 1010 x (m) 0 10 10 55 t (s)
1010 v (m/s) 0 5 5 t (s) 1010 a (m/s2 ) 5
Chapter 2 Solutions 7 2000 by Harcourt College Publishers. All
rights reserved. (b) At all times the instantaneous velocity is v =
d d t (3.00t2 2.00t + 3.00) = (6.00t 2.00) m/s At t = 2.00 s, v =
[6.00(2.00) 2.00] m/s = 10.0 m/s At t = 3.00 s, v = [6.00(3.00)
2.00] m/s = 16.0 m/s (c) a = v t = 16.0 m/s 10.0 m/s 3.00 s 2.00 s
= 6.00 m/s2 (d) At all times a = d d t (6.00 2.00) = 6.00 m/s2
(This includes both t= 2.00 s and t= 3.00 s). 2.19 (a) a = v t =
8.00 m/s 6.00 s = 4 3 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
8 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. *2.20 a = reading order = velocity = acceleration
b c d e 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
acceleration vectors would vary in magnitude and direction. *2.21
From v 2 f = vi 2 + 2ax, we have (10.97 103 m/s)2 = 0 + 2a(220 m),
so that a = 2.74 105 m/s2 which is 2.79 104 times g 2.22 (a)
Assuming a constant acceleration: a = vf vi t = 42.0 m/s 8.00 s =
5.25 m/s2 (b) Taking the origin at the original position of the
car, x = 1 2 (vi + vf) t = 1 2 (42.0 m/s)(8.00 s) = 168 m (c) From
vf = vi + at, the velocity 10.0 s after the car starts from rest
is: vf = 0 + (5.25 m/s2)(10.0 s) = 52.5 m/s
Chapter 2 Solutions 9 2000 by Harcourt College Publishers. All
rights reserved. *2.23 (a) x xi = 1 2 (vi + v) t becomes 40 m = 1 2
(vi + 2.80 m/s)(8.50 s) which yields vi = 6.61 m/s (b) a = v vi t =
2.80 m/s 6.61 m/s 8.50 s = 0.448 m/s2 2.24 Suppose the unknown
acceleration is constant as a car moving at vi = 35.0 mi/h comes to
a v = 0 stop in x xi = 40.0 ft. We find its acceleration from v2 =
v 2 i + 2a(x xi ) a = (v2 v 2 i ) 2(x xi ) = 0 (35.0 mi/h)2 2(40.0
ft) 5280 ft 1 mi 2 1 h 3600 s 2 = 32.9 ft/s2 Now consider a car
moving at vi = 70.0 mi/h and stopping to v = 0 with a = 32.9 ft/s2.
From the same equation its stopping distance is x xi = v2 vi 2 2a =
0 (70.0 mi/h)2 2(32.9 ft/s2) 5280 ft 1 mi 2 1 h 3600 s 2 = 160 ft
2.25 Given vi = 12.0 cm/s when xi = 3.00 cm (t = 0), and at t =
2.00 s, x = 5.00 cm x = vit + 1 2 at2; x xi = vit + 1 2 at2; 5.00
3.00 = 12.0(2.00) + 1 2 a (2.00)2 ; 8.00 = 24.0 + 2a a = 32.0 2 =
16.0 cm/s2
10 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. Goal Solution A body moving with uniform
acceleration has a velocity of 12.0 cm/s in the positive x
direction when its x coordinate is 3.00 cm. If its x coordinate
2.00s later is 5.00 cm, what is the magnitude of its acceleration?
G: Since the object must slow down as it moves to the right and
then speeds up to the left, the acceleration must be negative and
should have units of cm/s2. O: First we should sketch the problem
to see what is happening: x (cm) 5 0 5 5 0 5 initial final Here we
can see that the object travels along the x-axis, first to the
right, slowing down, and then speeding up as it travels to the left
in the negative x direction. We can show the position as a function
of time with the notation: x(t) x(0) = 3.00 cm, x(2.00) = 5.00 cm,
and v(0) = 12.0 cm/s A: Use the kinematic equation x xi = vit + 1 2
at2, and solve for a. a = 2(x xi vit) t2 a = 2[5.00 cm 3.00 cm
(12.0 cm/s)(2.00 s)] (2.00 s)2 a = 16.0 cm/s2 L: The acceleration
is negative as expected and it has the correct units of cm/s2. It
also makes sense that the magnitude of the acceleration must be
greater than 12 cm/s2 since this is the acceleration that would
cause the object to stop after 1 second and just return the object
to its starting point after 2 seconds.
Chapter 2 Solutions 11 2000 by Harcourt College Publishers. All
rights reserved. 2.26 (a) Total displacement = area under the (v,
t) curve from t = 0 to 50 s. x = 1 2 (50 m/s)(15 s) + (50 m/s)(40
15)s + 1 2 (50 m/s)(10 s) = 1875 m (b) From t = 10 s to t = 40 s,
displacement (area under the curve) is x = 1 2 (50 m/s + 33 m/s)(5
s) + (50 m/s)(25 s) = 1457 m (c) 0 t 15 s: a1 = v t = (50 0) m/s 15
s 0 = 3.3 m/s2 15 s < t < 40 s: a2 = 0 40 s t 50 s: a3 = v t
= (0 50) m/s 50 s 40 s = 5.0 m/s2 (d) (i) x1 = 0 + 1 2 a1t2 = 1 2
(3.3 m/s2) t2, or x1 = (1.67 m/s2)t2 (ii) x2 = 1 2 (15 s) [50 m/s
0] + (50 m/s)(t 15 s), or x2 = (50 m/s)t 375 m (iii) For 40 s t 50
s, x3 = area under v vs t from t = 0 to 40 s + 1 2 a3(t 40 s)2 +
(50 m/s)(t 40 s) or x3 = 375 m + 1250 m + 1 2 (5.0 m/s2)(t 40 s) 2
+ (50 m/s)(t 40 s) which reduces to x3 = (250 m/s)t (2.5 m/s2)t2
4375 m (e) v = total displacement total elapsed time = 1875 m 50 s
= 37.5 m/s *2.27 (a) Compare the position equation x = 2.00 + 3.00t
4.00t2 to the general form x = xi + vit + 1 2 at2 to recognize
that: xi = 2.00 m, vi = 3.00 m/s, and a = 8.00 m/s2 The velocity
equation, v = vi + at, is then v = 3.00 m/s (8.00 m/s2)t. The
particle changes direction when v = 0, which occurs at t = 3 8 s.
The position at this time is: x = 2.00 m + (3.00 m/s)Error! s)
(4.00 m/s2)Error! s) 2 = Error! 5 0 5 10 20 30 40 50 a (m/s2) t
(s)
12 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. (b) From x = xi + vit + 1 2 at2, observe that when
x = xi, the time is given by t = 2vi a . Thus, when the particle
returns to its initial position, the time is t = 2(3.00 m/s) 8.00
m/s2 = 3 4 s and the velocity is v = 3.00 m/s (8.00 m/s2) 3 4 s =
3.00 m/s 2.28 vi = 5.20 m/s (a) v(t = 2.50 s) = vi + at = 5.20 m/s
+ (3.00 m/s2)(2.50 s) = 12.7 m/s (b) v(t = 2.50 s) = vi + at = 5.20
m/s + (3.00 m/s2)(2.50 s) = 2.30 m/s 2.29 (a) x = 1 2 at2 (Eq 2.11)
400 m = 1 2 (10.0 m/s2) t2 t = 8.94 s (b) v = at (Eq 2.8) v = (10.0
m/s2)(8.94 s) = 89.4 m/s 2.30 (a) Take ti = 0 at the bottom of the
hill where xi = 0, vi = 30.0 m/s, and a = 2.00 m/s2. Use these
values in the general equation x = xi + vi t + 1 2 at2 to find x =
0 + 30.0t m/s + 1 2 (2.00 m/s2) t2 when t is in seconds x = (30.0t
t2)m To find an equation for the velocity, use v = vi + at = 30.0
m/s + (2.00 m/s2)t v = (30.0 2.00t) m/s
Chapter 2 Solutions 13 2000 by Harcourt College Publishers. All
rights reserved. (b) The distance of travel x becomes a maximum,
xmax, when v = 0 (turning point in the motion). Use the expressions
found in part (a) for v to find the value of t when x has its
maximum value: From v = (30.0 2.00t) m/s , v = 0 when t = 15.0 s
Then xmax = (30.0t t2) m = (30.0)(15.0) (15.0)2 = 225 m 2.31 (a) vi
= 100 m/s, a = 5.00 m/s2 v2 = v 2 i + 2ax 0 = (100)2 2(5.00)x x =
1000 m and t = 20.0 s (b) No, at this acceleration the plane would
overshoot the runway. *2.32 In the simultaneous equations vx = vxi
+ axt x xi = 1 2 (vxi + vx)t we have vx = vxi (5.60 m/s2)(4.20 s)
62.4 m = 1 2 (vxi+ vx)4.20 s So substituting for vxi gives 62.4 m =
1 2 [vx + (5.60 m/s2)(4.20 s) + vx]4.20 s 14.9 m/s = vx + 1 2 (5.60
m/s2)(4.20 s) vx = 3.10 m/s
14 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. *2.33 Take any two of the standard four equations,
such as vx = vxi + axt x xi = 1 2 (vxi + vx)t solve one for vxi,
and substitute into the other: vxi = vx axt x xi = 1 2 (vx axt +
vx) t Thus x xi = vxt 1 2 axt2 Back in problem 32, 62.4 m = vx(4.20
s) 1 2 (5.60 m/s2)(4.20 s) 2 vx = 62.4 m 49.4 m 4.20 s = 3.10 m/s
2.34 We assume the bullet is a cylinder. It slows down just as its
front end pushes apart wood fibers. (a) a = v2 v 2 i 2x = (280
m/s)2 (420 m/s)2 2(0.100 m) = 4.90 105 m/s2 (b) t = 0.100 350 +
0.020 280 = 3.57 10 -4 s (c) vi = 420 m/s, v = 0; a = 4.90 105
m/s2; v2 = vi 2 + 2ax x = v2 vi 2 2a = vi 2 2a = (420 m/s)2 (2 4.90
105 m/s2) x = 0.180 m *2.35 (a) The time it takes the truck to
reach 20.0 m/s is found from v = vi + at, solving for t yields t =
v vi a = 20.0 m/s 0 m/s 2.00 m/s2 = 10.0 s The total time is thus
10.0 s + 20.0 + 5.00 s = 35.0 s
Chapter 2 Solutions 15 2000 by Harcourt College Publishers. All
rights reserved. (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 x1 = v t = 0 + 20.0 2 (10.0) = 100 m The
distance traveled during the next 20.0 s is x2 = vit + 1 2 at2 =
(20.0)(20.0) + 0 = 400 m, a being 0 for this interval. The distance
traveled in the last 5.00 s is x3 = v t = 20.0 + 0 2 (5.00) = 50.0
m The total distance x = x1 + x2 + x3 = 100 + 400 + 50.0 = 550 m,
and the average velocity is given by v = x t = 550 35.0 = 15.7 m/s
*2.36 Using the equation x = vit + 1 2 at2 yields x = 20.0(40.0)
1.00(40.0)2/2 = 0, which is obviously wrong. The error occurs
because the equation used is for uniformly accelerated motion,
which this is not. The acceleration is 1.00 m/s2 for the first 20.0
s and 0 for the last 20.0 s. The distance traveled in the first
20.0 s is: x = vit + 1 2 at2 = (20.0)(20.0) 1.00(20.02)/2 = 200 m
During the last 20.0 s, the train is at rest. Thus, the total
distance traveled in the 40.0 s interval is 200 m . 2.37 (a) a = v
vi t = 632(5280/3600) 1.40 = 662 ft/s2 = 202 m/s2 (b) x = vit + 1 2
at2 = (632)(5280/3600)(1.40) 1 2 662(1.40)2 = 649 ft = 198 m 2.38
We have vi = 2.00 104 m/s, v = 6.00 106 m/s, x xi = 1.50 102 m (a)
x xi = 1 2 (vi + v) t t = 2(x xi) vi + v = 2(1.50 102 m) 2.00 104
m/s + 6.00 106 m/s = 4.98 109 s
16 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. (b) v2 = v 2 i + 2a(x xi ) a = v2 v 2 i 2(x xi) =
(6.00 106 m/s)2 (2.00 104 m/s)2 2(1.50 102 m) = 1.20 1015 m/s2 2.39
(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 xi = 9.00 m
Then v2 = v 2 i + 2a(x xi) = 02 + 2(0.5 00 m/s2) 9.00 m v = 3.00
m/s (b) x xi = vit + 1 2 at2 9.00 m = 0 + 1 2 (0.500 m/s2) t2 t =
6.00 s (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 = 0 x xi = 15.00 m v2 = v 2 i + 2a(x xi) gives a = (v2 v 2 i )
2(x xi ) = [0 (3.00 m/s)2] 2(15.0 m) = 0.300 m/s2 (d) Take initial
point at the bottom of the planes and final point 8.00 m along the
second: vi = 3.00 m/s x xi = 8.00 m a = 0.300 m/s2 v2 = v 2 i +
2a(x xi) = (3.00 m/s)2 + 2( 0.300 m/s2)(8.00 m) = 4.20 m2/s2 v =
2.05 m/s
Chapter 2 Solutions 17 2000 by Harcourt College Publishers. All
rights reserved. 2.40 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 xs(t) = xis + vis t + 1 2 ast2 = (30.0 m/s)t + 1 2
(2.00 m/s2) t2 For the van, xiv = 155 m viv = 5.00 m/s av = 0 and
xv(t) = xiv + vivt + 1 2 avt2 = 155 m + (5.00 m/s)t + 0 To test for
a collision, we look for an instant tc when both are at the same
place: 30.0tc t 2 c = 155 + 5.00tc 0 = t 2 c 25.0tc + 155 From the
quadratic formula tc = 25.0 (25.0)2 4(155) 2 = 13.6 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 m + (5.00 m/s)(11.4
s) = 212 m . 2.41 Choose the origin (y = 0, t = 0) 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 yi = vit + 1 2 at2 y = 1 2 gt2 = 1 2 (9.80 m/s2) t2 and v
= vi + at v = gt = (9.80 m/s2)t (a) at t = 1.00 s: y = 1 2 (9.80
m/s2)(1.00 s) 2 = 4.90 m at t = 2.00 s: y = 1 2 (9.80 m/s2)(2.00 s)
2 = 19.6 m at t = 3.00 s: y = 1 2 (9.80 m/s2)(3.00 s) 2 = 44.1
m
18 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. (b) at t = 1.00 s: v = (9.80 m/s2)(1.00 s) = 9.80
m/s at t = 2.00 s: v = (9.80 m/s2)(2.00 s) = 19.6 m/s at t = 3.00
s: v = (9.80 m/s2)(3.00 s) = 29.4 m/s *2.42 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 (1609 m) up and then 1 mile back down.
Determine his speed just after launch by considering his upward
flight: v2 = v 2 i + 2a(y yi) 0 = v 2 i 2(9.80 m/s2)(1609 m) vi =
178 m/s His time in the air may be found by considering his motion
from just after launch to just before impact: y yi = vit + 1 2 at2
0 = (178 m/s)t 1 2 (9.80 m/s2) t2 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 = $1.00 36.2 s = 0.0276 $ s
3600 s 1 h = $99.4/h 2.43 (a) y = vit + 1 2 at2 4.00 = (1.50)vi
(4.90)(1.50)2 and vi = 10.0 m/s upward (b) v = vi + at = 10.0
(9.80)(1.50) = 4.68 m/s v = 4.68 m/s downward 2.44 We have y = 1 2
gt2 + vit + yi 0 = (4.90 m/s2)t2 (8.00 m/s)t + 30.0 m Solving for
t, t = 8.00 64.0 + 588 9.80 Using only the positive value for t, we
find t = 1.79 s
Chapter 2 Solutions 19 2000 by Harcourt College Publishers. All
rights reserved. *2.45 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 = vit 1 2 gt2 = 0 (4.90
m/s2)(0.20 s)2 = 0.20 m This distance is about twice the distance
between the center of the bill and its top edge ( 8 cm). Thus,
David will be unsuccessful . Goal Solution Emily challenges her
friend David to catch a dollar bill as follows. She holds the bill
vertically, as in Figure P2.45, with the center of the bill between
Davids index finger and thumb. David must catch the bill after
Emily releases it without moving his hand downward. If his reaction
time is 0.2 s, will he succeed? Explain your reasoning. G: David
will be successful if his reaction time is short enough that he can
catch the bill before it falls half of its length (about 8 cm).
Anyone who has tried this challenge knows that this is a difficult
task unless the catcher cheats by anticipating the moment the bill
is released. Since Davids reaction time of 0.2 s is typical of most
people, we should suspect that he will not succeed in meeting
Emilys challenge. O: Since the bill is released from rest and
experiences free fall, we can use the equation y = 1 2 gt2 to find
the distance y the bill falls in t = 0.2 s A: y = 1 2 (9.80
m/s2)(0.2 s) 2 = 0.196 m > 0.08 m Since the bill falls below
Davids fingers before he reacts, he will not catch it. L: It
appears that even if David held his fingers at the bottom of the
bill (about 16 cm below the top edge), he still would not catch the
bill unless he reduced his reaction time by tensing his arm muscles
or anticipating the drop. *2.46 At any time t, the position of the
ball released from rest is given by y1 = h 1 2 gt2. At time t, the
position of the ball thrown vertically upward is described by y2 =
vit 1 2 gt2. The time at which the first ball has a position of y1
= h/2 is found from the first equation as h/2 = h 1 2 gt2, which
yields t = h/g . To require that the second ball have a position of
y2 = h/2 at this time, use the second equation to obtain h/2 = vi
h/g 1 2 g(h/g). This gives the required initial upward velocity of
the second ball as vi = g h .
20 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 2.47 (a) v = vi gt (Eq. 2.8) v = 0 when t = 3.00
s, g = 9.80 m/s2, vi = gt = (9.80 m/s2)(3.00 s) = 29.4 m/s (b) y =
1 2 (v + vi) t = 1 2 (29.4 m/s)(3.00 s) = 44.1 m Goal Solution A
baseball is hit such that it travels straight upward after being
struck by the bat. A fan observes that it requires 3.00 s for the
ball to reach its maximum height. Find (a) its initial velocity and
(b) the maximum height it reaches. G: We can expect the initial
speed of the ball to be somewhat greater than the speed of the
pitch, which might be about 60 mph (~30 m/s), so an initial upward
velocity off the bat of somewhere between 20 and 100 m/s would be
reasonable. We also know that the length of a ball field is about
300 ft. (~100m), and a pop-fly usually does not go higher than this
distance, so a maximum height of 10 to 100 m would be reasonable
for the situation described in this problem. O: Since the balls
motion is entirely vertical, we can use the equation for free fall
to find the initial velocity and maximum height from the elapsed
time. A: Choose the initial point when the ball has just left
contact with the bat. Choose the final point at the top of its
flight. In between, the ball is in free fall for t = 3.00 s and has
constant acceleration, a = -g = -9.80 m/s2. Solve the equation vyf
= vyi gt for vyi when vyf = 0 (when the ball reaches its maximum
height). (a) vyi = vyf + gt = 0 + (9.80 m/s2)(3.00 s) = 29.4 m/s
(upward) (b) The maximum height in the vertical direction is yf =
vyi t + 1 2 at2 = (29.4 m/s)(3.00 s) + 1 2 (9.80 m/s2)(3.00 s) 2 =
44.1 m L: The calculated answers seem reasonable since they lie
within our expected ranges, and they have the correct units and
direction. We must remember that it is possible to solve a problem
like this correctly, yet the answers may not seem reasonable simply
because the conditions stated in the problem may not be physically
possible (e.g. a time of 10 seconds for a pop fly would not be
realistic).
Chapter 2 Solutions 21 2000 by Harcourt College Publishers. All
rights reserved. 2.48 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 = vit + 1 2 at2 144 ft = 0 + (16.0 ft/s2)t2
giving tfall = 3.00 s. Her velocity just before impact is: v = vi +
gt = 0 + (32.0 ft/s2)(3.00 s) = 96.0 ft/s . (b) While crushing the
box, vi = 96.0 ft/s, v = 0, and y = 18.0 in = 1.50 ft. Therefore, a
= v2 v 2 i 2(y) = 0 (96.0 ft/s)2 2(1.50 ft) = 3.07 103 ft/s2, or a
= 3.07 103 ft/s2 upward (c) Time to crush box: t = y v = y (v +
vi)/2 = 2(1.50 ft) 0 + 96.0 ft/s or t = 3.13 102 s 2.49 Time to
fall 3.00 m is found from Eq. 2.11 with vi = 0, 3.00 m = 1 2 (9.80
m/s2) t2; 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
22 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 2.50 Time to top = 10.0 s. v = vi gt (a) At the
top, v = 0. Then, t = vi g = 10.0 s vi = 98.0 m/s (b) h = vit 1 2
gt2 At t = 10.0 s, h = (98.0)(10.0) 1 2 (9.80)(10.0) 2 = 490 m 2.51
vi = 15.0 m/s (a) v = vi gt = 0 t = vi g = 15.0 m/s 9.80 m/s2 =
1.53 s (b) h = vit 1 2 gt2 = v 2 i 2g = 225 19.6 m= 11.5 m (c) At t
= 2.00 s v = vi gt = 15.0 19.6 = 4.60 m/s a = g = 9.80 m/s2 2.52 y
= 3.00t3 At t = 2.00 s, y = 3.00(2.00)3 = 24.0 m, and vy = dy d t =
9.00t2 = 36.0 m/s If the helicopter releases a small mailbag at
this time, the equation of motion of the mailbag is yb = ybi + vit
1 2 gt2 = 24.0 + 36.0t 1 2 (9.80) t2 Setting yb = 0, 0 = 24.0 +
36.0t 4.90t2 Solving for t, (only positive values of t count), t =
7.96 s
Chapter 2 Solutions 23 2000 by Harcourt College Publishers. All
rights reserved. 2.53 (a) J = da d t = constant da = Jdt a = J dt =
Jt + c2 but a = ai when t = 0 so c1 = ai, Therefore, a = Jt + ai a
= dv d t ; dv = adt v = adt = (Jt + ai)dt = 1 2 Jt2 + ait + c2 but
v = vi when t = 0, so c2 = vi and v = 1 2 Jt2 + ait + vi v = dx d t
; dx = vdt x = vdt = 1 2 Jt2 + ait + vi dt x = 1 6 Jt3 + 1 2 ait2 +
vit + c3 x = xi when t = 0, so c3 = xi Therefore, x = 1 6 Jt3 + 1 2
ait2 + vit + xi (b) a2 = (Jt + ai)2 = J2t2 + a 2 i + 2Jait a2 = a 2
i + (J2t2 + 2Jait) a2 = a 2 i + 2J 1 2 Jt2 + ait Recall the
expression for v: v = 1 2 Jt2 + ait + vi So (v vi) = 1 2 Jt2 + ait
Therefore, a2 = a 2 i + 2J(v vi)
24 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 2.54 (a) a = dv d t = d d t [ 5.00 107 t2 + 3.00
105 t] Error! t + 3.00 105 m/s2 ) Take xi = 0 at t = 0. Then v = dx
d t x 0 = 0 t vdt = 0 t (5.00 107 t2 + 3.00 105t) dt x = 5.00 107
t3 3 + 3.00 105 t2 2 x = (1.67 107 m/s3)t3 + (1.50 105 m/s2)t2 (b)
The bullet escapes when a = 0, at (10.0 107 m/s3)t + 3.00 105 m/s2
= 0 t = 3.00 105 s 10.0 107 = 3.00 10-3 s (c) New v = ( 5.00
107)(3.00 10 -3)2 + (3.00 105)(3.00 10 -3) v = 450 m/s + 900 m/s =
450 m/s (d) x = (1.67 107)(3.00 10 -3)3 + (1.50 105)(3.00 10 -3)2 x
= 0.450 m + 1.35 m = 0.900 m 2.55 a = dv d t = 3.00v2, vi = 1.50
m/s Solving for v, dv d t = 3.00v2 v = vi v v2dv = 3.00 t = 0 0 dt
1 v + 1 vi = 3.00t or 3.00t = 1 v 1 vi When v = vi 2 , t = 1 3.00
vi = 0.222 s
Chapter 2 Solutions 25 2000 by Harcourt College Publishers. All
rights reserved. 2.56 (a) The minimum distance required for the
motorist to stop, from an initial speed of 18.0 m/s, is x = v2 v 2
i 2a = 0 (18.0 m/s)2 2(4.50 m/s2) = 36.0 m Thus, the motorist can
travel at most (38.0 m 36.0 m) = 2.00 m before putting on the
brakes if he is to avoid hitting the deer. The maximum acceptable
reaction time is then tmax = 2.00 m vi = 2.00 m 18.0 m/s = 0.111 s
(b) In 0.300 s, the distance traveled at 18.0 m/s is x = vit1 =
(18.0 m/s)(0.300) = 5.40 m The displacement for an acceleration
4.50 m/s2 is 38.0 5.40 = 32.6 m. v2 = v 2 i + 2ax = (18.0 m/s)2
2(4.50 m/s2)(32.6 m) = 30.6 m2/s2 v = 30.6 = 5.53 m/s 2.57 The
total time to reach the ground is given by y yi = vit + 1 2 at2 0
25.0 m = 0 + 1 2 (9.80 m/s2) t2 t = 2(25.0 m) 9.80 m/s2 = 2.26 s
The time to fall the first fifteen meters is found similarly: 15.0
m = 0 1 2 (9.80 m/s2) t 2 1 t1 = 1.75 s So t t1 = 2.26 s 1.75 s =
0.509 s suffices for the last ten meters.
26 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. *2.58 The rate of hair growth is a velocity and
the rate of its increase is an acceleration. Then vi = 1.04 mm/d
and a = 0.132 mm/d w . The increase in the length of the hair
(i.e., displacement) during a time of t = 5.00 w = 35.0 d is x =
vit + 1 2 at2 x = (1.04 mm/d)(35.0 d) + 1 2 (0.132 mm/d w)(35.0
d)(5.00 w) or x = 48.0 mm 2.59 Let path (#1) correspond to the
motion of the rocket accelerating under its own power. Path (#2) is
the motion of the rocket under the influence of gravity with the
rocket still rising. Path (#3) is the motion of the rocket under
the influence of gravity, but with the rocket falling. The data in
the table is found for each phase of the rocket's motion. (#1): v2
(80.0)2 = 2(4.00)(1000); therefore v = 120 m/s 120 = 80.0 + (4.00)t
giving t = 10.0 s (#2): 0 (120)2 = 2(9.80)x giving x = 735 m 0 120
= 9.80t giving t = 12.2 s This is the time of maximum height of the
rocket. (#3): v2 0 = 2(9.80)(1735) v = 184 = (9.80)t giving t =
18.8 s (a) ttotal = 10 + 12.2 + 18.8 = 41.0 s (b) xtotal = 1.73 km
(c) vfinal = 184 m/s t x v a 0 Launch 0 0 80 +4.00 #1 End Thrust
10.0 1000 120 +4.00 #2 Rise Upwards 22.2 1735 0 9.80 #3 Fall to
Earth 41.0 0 184 9.80 0 3 1 2
Chapter 2 Solutions 27 2000 by Harcourt College Publishers. All
rights reserved. 2.60 Distance traveled by motorist = (15.0 m/s)t
Distance traveled by policeman = 1 2 (2.00 m/s2) t2 (a) intercept
occurs when 15.0t =t2 t = 15.0 s (b) v (officer) = (2.00 m/s2)t =
30.0 m/s (c) x (officer) = 1 2 (2.00 m/s2) t2 = 225 m *2.61 Area A1
is a rectangle. Thus, A1 = hw = vit. Area A2 is triangular.
Therefore A2 = 1 2 bh = 1 2 t(v vi). The total area under the curve
is A = A1 + A2 = vit + (v vi)t/2 and since v vi = at A = vit + 1 2
at2 The displacement given by the equation is: x = vit + 1 2 at2,
the same result as above for the total area. 2.62 a1 = 0.100 m/s2,
a2 = 0.500 m/s2 x = 1000 m = 1 2 a1t 2 1 + v1t2 + 1 2 a2t 2 2 t =
t1 + t2 and v1 = a1t1 = a2t2 1000 = 1 2 a1t 2 1 + a1t1 a1t1 a2 + 1
2 a2 a1t1 a2 2 0 vi t t v v A2 A1
28 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 1000 = 1 2 a1 1 a1 a2 t 2 1 t1 = 20,000 1.20 = 129
s t2 = a1t1 a2 = 12.9 0.500 26 s Total time = t = 155 s 2.63 (a)
Let x be the distance traveled at acceleration a until maximum
speed v is reached. If this achieved in time t1 we can use the
following three equations: x = (v + vi) 2 t1, 100 x = v(10.2 t1)
and v = vi + at1 The first two give 100 = 10.2 1 2 t1 v = 10.2 1 2
t1 at1 a = 200 (20.4 t1)t1 . For Maggie a = 200 (18.4)(2.00) = 5.43
m/s2 For Judy a = 200 (17.4)(3.00) = 3.83 m/s2 (b) v = at1 Maggie:
v = (5.43)(2.00) = 10.9 m/s Judy: v = (3.83)(3.00) = 11.5 m/s (c)
At the six-second mark x = 1 2 at 2 1 + v(6.00 t1) Maggie: x = 1 2
(5.43)(2.00) 2 + (10.9)(4.00) = 54.3 m Judy: x = 1 2 (3.83)(3.00) 2
+ (11.5)(3.00) = 51.7 m Maggie is ahead by 2.62 m .
Chapter 2 Solutions 29 2000 by Harcourt College Publishers. All
rights reserved. *2.64 Let the ball fall 1.50 m. It strikes at
speed given by: v 2 x = v 2 xi + 2a(x xi ) v 2 x = 0 + 2(9.80
m/s2)(1.50 m) vx = 5.42 m/s and its stopping is described by v 2 x
= v 2 xi + 2ax(x xi) 0 = (5.42 m/s)2 + 2ax(102 m) ax = 29.4 m2/s2
2.00 102 m = +1.47 103 m/s2 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 .
2.65 Acceleration a = 3.00 m/s2 Deceleration a' = 4.50 m/s2 (a)
Keeping track of speed and time for each phase of motion, v0 = 0,
v1 = 12.0 m/s t01 = 4.00 s v1 = 12.0 m/s t1 = 5.00 s v1 = 12.0 m/s,
v2 = 0 t12 = 2.67 s v2 = 0 m/s, v3 = 18.0 m/s t23 = 6.00 s v3 =
18.0 m/s t3 = 20 .0 s v3 = 18.0 m/s, v4 = 6.00 m/s t34 = 2.67 s v4
= 6.00 m/s t4 = 4 .00 s v4 = 6.00 m/s, v5 = 0 t45 = 1.33 s t = 45.7
s (b) x = vi ti = 6.00(4.00) + 12.0(5.00) + 6.00(2.67) + 9.00(6.00)
+ 18.0(20.0) + 12.0(2.67) + 6.00(4.00) + 3.00(1.33) = 574 m (c) v =
x t = 574 m 45.7 s = 12.6 m/s
30 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. (d) tWALK = 2x vWALK = 2(574 m) (1.50 m/s) = 765 s
(about 13 minutes, and better exercise!) 2.66 (a) d = 1 2 (9.80) t
2 1 d = 336t2 t1 + t2 = 2.40 336t2 = 4.90(2.40 t2)2 4.90t 2 2
359.5t2 + 28.22 = 0 t2 = 359.5 (359.5)2 4(4.90)(28.22) 9.80 t2 =
359.5 358.75 9.80 = 0.0765 s d = 336t2 = 26.4 m (b) Ignoring the
sound travel time, d = 1 2 (9.80)(2.40) 2 = 28.2 m, an error of
6.82% . 2.67 (a) y = vi1t + 1 2 at2 = 50.0 = 2.00t + 1 2 (9.80) t2
t = 2.99 s after the first stone is thrown. (b) y = vi2t + 1 2 at2
and t = 2.99 1.00 = 1.99 s substitute 50.0 = vi2(1.99) + 1 2
(9.80)(1.99) 2 vi2 = 15.4 m/s downward (c) v1 = vi1 + at = 2.00 +
(9.80)(2.99) = 31.3 m/s v2 = vi2 + at = 15.3 + (9.80)(1.99) = 34.9
m/s 2.68 The time required for the car to come to rest and the time
required to regain its original speed of 25.0 m/s are both given by
t = |v| |a| = 25.0 m/s 2.50 m/s2 . The total distance the car
travels in these two intervals is xcar = x1 + x2 = (25.0 m/s + 0) 2
(10.0 s) + (0 + 25.0 m/s) 2 (10.0 s) = 250 m
Chapter 2 Solutions 31 2000 by Harcourt College Publishers. All
rights reserved. The total elapsed time when the car regains its
original speed is ttotal = 10.0 s + 45.0 s + 10.0 s = 65.0 s The
distance the train has traveled in this time is xtrain = (25.0
m/s)(65.0 s) = 1.63 103 m Thus, the train is 1.63 103 m 250 m =
1.38 103 m ahead of the car. 2.69 (a) We require xs = xk when ts =
tk + 1.00 xs = 1 2 (3.50 m/s2)(tk + 1.00) 2 = 1 2 (4.90 m/s2)(tk) 2
= xk tk + 1.00 = 1.183tk tk = 5.46 s (b) xk = 1 2 (4.90 m/s2)(5.46
s) 2 = 73.0 m (c) vk = (4.90 m/s2)(5.46 s) = 26.7 m/s vs = (3.50
m/s2)(6.46 s) = 22.6 m/s 2.70 (a) In walking a distance x, in a
time t, the length of rope l is only increased by x sin . The pack
lifts at a rate x t sin . v = x t sin = vboy x l = vboy x x2 + h2
(b) a = dv d t = vboy l dx d t + vboyx d d t 1 l a = vboy vboy l
vboyx l2 dl d t , but dl d t = v a = v 2 boy l 1 x2 l2 = v 2 boy l
h2 l2 = h2v 2 boy (x2 + h2)3/2 (c) v 2 boy h , 0 (d) vboy, 0 x h m
l vboy av
32 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 2.71 h = 6.00 m, vboy = 2.00 m/s v = x t sin =
vboy x l = vboyx (x2 + h2)1/2 However x = vboyt v = v 2 boyt (v 2
boyt2 + h2)1/2 = 4t (4t2 + 36)1/2 (a) t(s) v(m/s) 0 0 0.5 0.32 1
0.63 1.5 0.89 2 1.11 2.5 1.28 3 1.41 3.5 1.52 4 1.60 4.5 1.66 5
1.71 (b) From problem 2.70 above, a = h2v 2 boy (x2 + h2)3/2 = h2v
2 boy (v 2 boyt2 + h2)3/2 = 144 (4t2 + 36)3/2 t(s) a(m/s2) 0 0.67
0.5 0.64 1 0.57 1.5 0.48 2 0.38 2.5 0.30 3 0.24 3.5 0.18 4 0.14 4.5
0.11 5 0.09 0.6 0.3 0.0 1.2 0.9 3210 4 5 t (s) 1.5 1.8 v (m/s) 0.2
0.1 0.0 0.4 0.3 3210 4 5 t (s) 0.5 0.6 0.7 a (m/s2 )
Chapter 2 Solutions 33 2000 by Harcourt College Publishers. All
rights reserved. 2.72 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 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 acceleration = slope of
line is constant. a = 1.63 m/s2 = 1.63 m/s2 downward 4.00 2.00 0.00
2.00 4.00 321 4 5321 4 5 t (s) 6.00 v (m/s)
34 Chapter 2 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 2.73 The distance x and y are always related by x2
+ y2 = L2. Differentiating this equation with respect to time, we
have 2x dx d t + 2y dy d t = 0 Now dy d t is vB, the unknown
velocity of B; and dx d t = v. From the equation resulting from
differentiation, we have d y d t = x y dx d t = x y (v) But y x =
tan so vB = 1 tan v When = 60.0, vB = v tan 60.0 = v 3 3 = 0.577v
Goal Solution Two objects, A and B, are connected by a rigid rod
that has a length L. The objects slide along perpendicular guide
rails, as shown in Figure P2.73. If A slides to the left with a
constant speed v, find the velocity of B when = 60.0. G: The
solution to this problem may not seem obvious, but if we consider
the range of motion of the two objects, we realize that B will have
the same speed as A when = 45, and when = 90, then vB = 0.
Therefore when = 60, we should expect vB to be between 0 and v. O:
Since we know a distance relationship and we are looking for a
velocity, we might try differentiating with respect to time to go
from what we know to what we want. We can express the fact that the
distance between A and B is always L, with the relation: x2+ y2 =
L2. By differentiating this equation with respect to time, we can
find vB = dy/dt in terms of dx/dt = vA = v. A: Differentiating x2 +
y2 = L2 gives us 2x dx d t + 2y dy d t = 0 Substituting and solving
for the speed of B: vB = dy d t = x y dx d t = x y (v) Now from the
geometry of the figure, we notice that y x = tan , so vB = v tan
When = 60.0, vB = v tan 60 = v 3 = 0.577v (B is moving up) L: Our
answer seems reasonable since we have specified both a magnitude
and direction for the velocity of B, and the speed is between 0 and
v in agreement with our earlier prediction. In this and many other
physics problems, we can find it helpful to examine the limiting
cases that define boundaries for the answer. L y x v A B x O y
Chapter 2 Solutions 35 2000 by Harcourt College Publishers. All
rights reserved.
2000 by Harcourt College Publishers. All rights reserved.
Chapter 3 Even Answers 2. (a) 8.60 m (b) 4.47 m, -63.4, 4.24 m, 135
4. (a) (2.17, 1.25) m and (1.90, 3.29) m (b) 4.55 m 6. (a) r, 180
(b) 2r, 180 + (c) 3r, 8. 14 km, 65 N of E 10. 310 km at 57 S of W
12. 9.54 N, 57.0 above the x-axis 14. 7.92 m at 4.34 N of W 16. (a)
~105 m upward (b) ~103 m upward 18. 5.24 km at 25.9 N of W 20. 86.6
m, - 50.0 m 22. 358 m at 2.00 S of E 24. |B| = 7.81, = 59.2, =
39.8, = 67.4 26. 788 miles at 48.0 NE of Dallas 28. (b) 5.00i +
4.00j, 6.40 at 38.7, 1.00i + 8.00j, 8.06 at 97.2 30. Cx = 7.30 cm,
Cy = 7.20 cm 32. 6.22 blocks at 110 counterclockwise from east 34.
(a) 4.47 m at = 63.4 (b) 8.49 m at = 135 36. 42.7 yards 38. 4.64 m
at 78.6 N of E 40. 1.43 104 m at 32.2 above the horizontal 42. 106
44. - 220i + 57.6j, 227 paces at 165 46. (a) (3.12i + 5.02j 2.20k)
km (b) 6.31 km 48. (a) (15.1i + 7.72j) cm (b) (7.72i + 15.1j) cm
(c) (+7.72i + 15.1j) cm 50. (a) 74.6 N of E (b) 470 km 52. a =
5.00, b = 7.00 54. 2 tan1(1/n) 56. (3.60i + 7.00j) N, 7.87 N at
97.8counterclockwise from horizontal 58. 2.00 m/s j, it is the
velocity vector 60. (a) (10.0 m, 16.0 m)
2 Chapter 3 Even Answers 2000 by Harcourt College Publishers.
All rights reserved.
2000 by Harcourt College Publishers. All rights reserved.
Chapter 3 Solutions *3.1 x = r cos = (5.50 m) cos 240 = (5.50
m)(0.5) = 2.75 m y = r sin = (5.50 m) sin 240 = (5.50 m)(0.866) =
4.76 m 3.2 (a) d = (x2 x1)2 + (y2 y1)2 = (2.00 [3.00]2) + (4.00
3.00)2 d = 25.0 + 49.0 = 8.60 m (b) r1 = (2.00)2 + (4.00)2 = 20.0 =
4.47 m 1 = tan1 4.00 2.00 = 63.4 r2 = (3.00)2 + (3.00)2 = 18.0 =
4.24 m 2 = 135 measured from + x axis. 3.3 We have 2.00 = r cos
30.0 r = 2.00 cos 30.0 = 2.31 and y = r sin 30.0 = 2.31 sin 30.0 =
1.15 3.4 (a) x = r cos and y = r sin , therefore x1 = (2.50 m) cos
30.0, y1 = (2.50 m) sin 30.0, and (x1, y1) = (2.17, 1.25) m x2 =
(3.80 m) cos 120, y2 = (3.80 m) sin 120, and (x2, y2) = (1.90,
3.29) m (b) d = (x)2 + (y)2 = 16.6 + 4.16 = 4.55 m
2 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 3.5 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 = x2 + y2 = (2.00 m)2 + (1.00 m)2 = 5.00 m2 = 2.24 m
(b) = Arctan 1 2 = 26.6; r = 2.24 m, 26.6 3.6 We have r = x2 + y2
and = Arctan y x (a) The radius for this new point is (x) 2 + y 2 =
x 2 + y 2 = r and its angle is Arctan y (x) = 180 (b) (2x)2 + (2y)2
= 2r This point is in the third quadrant if (x, y) is in the first
quadrant or in the fourth quadrant if (x, y) is in the second
quadrant. It is at angle 180 + . (c) (3x)2 + (3y)2 = 3r This point
is in the fourth quadrant if (x, y) is in the first quadrant or in
the third quadrant if (x, y) is in the second quadrant. It is at
angle . 3.7 (a) The distance d from A to C is d = x2 + y2 where x =
(200) + (300 cos 30.0) = 460 km and y = 0 + (300 sin 30.0) = 150 km
d = (460)2 + (150)2 = 484 km (b) tan = y x = 150 460 = 0.326 =
tan-1(0.326) = 18.1 N of W 3.8 R 14 km = 65 N of E d 300 km C B 200
km A 3030 R 13 km 6 km
Chapter 3 Solutions 3 2000 by Harcourt College Publishers. All
rights reserved. 3.9 tan 35.0 = x 100 m x = (100 m)(tan 35.0) =
70.0 m 35.035.0 100 m x 3.10 R = 310 km at 57 S of W base R B A 200
km100 km0 E 3.11 (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 a t 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
4 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 3.12 Find the resultant F1 + F2 graphically by
placing the tail of F2 at the head of F1. The resultant force
vector F1 + F2 is of magnitude 9.5 N and at an angle of 57 above
the x-axis . 0 1 2 3 N x y F1 F2 F1 + F2 3.13 (a) d = 10.0i = 10.0
m since the displacement is a straight line from point A to point
B. (b) The actual distance walked is not equal to the straight-line
displacement. The distance follows the curved path of the
semi-circle (ACB). s = 1 2 (2 r) = 5 = 15.7 m (c) If the circle is
complete, d begins and ends at point A. Hence, d = 0 . 3.14 Your
sketch should be drawn to scale, and should look somewhat like that
pictured below. 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 . E N W S 30.030.0 R
8.20 meters 8.20 meters 15.0 meters 3.50 meters 3.50 meters 15.0
meters B d A C 5.00 m5.00 m
Chapter 3 Solutions 5 2000 by Harcourt College Publishers. All
rights reserved. 3.15 To find these vector expressions graphically,
we draw each set of vectors. Measurements of the results are taken
using a ruler and protractor. (a) A + B = 5.2 m at 60 (b) A B = 3.0
m at 330 B A + B 4 m2 m a A 0 B A B A 4 m2 m b 0 (c) B A = 3.0 m at
150 (d) A 2B = 5.2 m at 300 B B A A 4 m2 m c 0 2B A 2B A 4 m2 m d 0
*3.16 (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 ~ 105(0.03 m) + 102(1 m) ~103 m
upward .
6 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 3.17 The scale drawing for the graphical solution
should be similar to the figure at 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 ft
and 3 3.18 x y 0 km 3.00 km 1.41 1.41 4.00 0 2.12 2.12 4.71 2.29 R
= |x|2 + |y|2 = 5.24 km = tan1 y x = 154 or = 25.9 N of W 3.19 Call
his first direction the x direction. R = 10.0 m i + 5.00 m(j) +
7.00 m(i) = 3.00 m i 5.00 m j = (3.00)2 + (5.00)2 m at Arctan 5 3
to the right R = 5.83 m at 59.0 to the right from his original
motion 3.20 Coordinates of super-hero are: x = (100 m) cos (30.0) =
86.6 m y = (100 m) sin (30.0) = 50.0 m 135 ft 200 ft d 135 ft y x
40.040.0 30.030.0 E N 3.00 km 2.00 km 4.00 km 45.045.0 45.045.0
3.00 km3.00 km 3.00 km 2.00 km 4.00 km R t
Chapter 3 Solutions 7 2000 by Harcourt College Publishers. All
rights reserved. 100 m x y 30.0
8 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 3.21 The person would have to walk 3.10 sin(25.0)
= 1.31 km north , and 3.10 cos(25.0) = 2.81 km east . 3.22 + x
East, + y North x = 250 + 125 cos 30 = 358 m y = 75 + 125 sin 30
150 = 12.5 m d = (x)2 + (y)2 = (358)2 + (12.5)2 = 358 m tan = (y)
(x) = 12.5 358 = 0.0349 = 2.00 d = 358 m at 2.00 S of E *3.23 Let
the positive x-direction be eastward, positive y-direction be
vertically upward, and the positive z-direction be southward. The
total displacement is then d = (4.80 cm i + 4.80 cm j) + (3.70 cm j
3.70 cm k) or d = 4.80 cm i + 8.50 cm j 3.70 cm k (a) The magnitude
is d = (4.80)2 + (8.50)2 + (3.70)2 cm = 10.4 cm (b) Its angle with
the y-axis follows from cos = 8.50 10.4 , giving = 35.5 . 3.24 B =
Bxi + Byj + B2k B = 4.00i + 6.00j + 3.00k |B|= (4.00)2 + (6.00)2 +
(3.00)2 = 7.81 = cos1 4.00 7.81 = 59.2 = cos1 6.00 7.81 = 39.8 =
cos1 3.00 7.81 = 67.4
Chapter 3 Solutions 9 2000 by Harcourt College Publishers. All
rights reserved. 3.25 Ax = 25.0 Ay = 40.0 A = A 2 x + A 2 y =
(25.0)2 + (40.0)2 = 47.2 units From the triangle, we find that =
58.0, so that = 122 Goal Solution A vector has an x component of
25.0 units and a y component of 40.0 units. Find the magnitude and
direction of this vector. r x y 3025201510 5 0 5 10 40 30 20 10 G:
First we should visualize the vector either in our mind or with a
sketch. Since the hypotenuse of the right triangle must be greater
than either the x or y components that form the legs, we can
estimate the magnitude of the vector to be about 50 units. The
direction of the vector appears to be about 120 from the +x axis.
O: The graphical analysis and visual estimates above may be
sufficient for some situations, but we can use trigonometry to
obtain a more precise result. A: The magnitude can be found by the
Pythagorean theorem: r = x2 + y2 r = (25.0 units)2 + (40 units)2 =
47.2 units We observe that tan = y x (if we consider x and y to
both be positive) . = tan1 y x = tan1 40.0 25.0 = tan1 (1.60) =
58.0 The angle from the +x axis can be found by subtracting from
180. = 180 58 = 122 L: Our calculated results agree with our
graphical estimates. We should always remember to check that our
answers are reasonable and make sense, especially for problems like
this where it is easy to mistakenly calculate the wrong angle by
confusing coordinates or overlooking a minus sign. Quite often the
direction angle of a vector can be specified in more than one way,
and we must choose a notation that makes the most sense for the
given problem. If compass directions were stated in this question,
we could have reported the vector angle to be 32.0 west of north or
a compass heading of 328. t 40.040.0 25.025.0 A x y
10 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. *3.26 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: dDCeast
= dDAeast + dACeast = 730 cos 5.00 560 sin 21.0 = 527 miles.
dDCnorth = dDAnorth + dACnorth = 730 sin 5.00 + 560 cos 21.0 = 586
miles. By the Pythagorean theorem, d = (dDCeast)2 + (dDCnorth)2 =
788 mi Then tan = dDCnorth dDCeast = 1.11 and = 48.0. Thus, Chicago
is 788 miles at 48.0 north east of Dallas . 3.27 x = d cos = (50.0
m)cos(120) = 25.0 m y = d sin = (50.0 m)sin(120) = 43.3 m d = (25.0
m)i + (43.3 m)j 3.28 (a) A + B A B B B A B A + B A B B B A B (b) C
= A + B = 2.00i + 6.00j + 3.00i 2.00j = 5.00i + 4.00j C = 25.0 +
16.0 at Arctan 4 5 C = 6.40 at 38.7 D = A B = 2.00i + 6.00j 3.00i +
2.00j = 1.00i + 8.00j D = (1.00)2 + (8.00)2 at Arctan 8.00
(1.00)
Chapter 3 Solutions 11 2000 by Harcourt College Publishers. All
rights reserved. D = 8.06 at (180 82.9) = 8.06 at 97.2
12 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 3.29 d = (x1 + x2 + x3)2 + (y1 + y2 + y3)2 = (3.00
5.00 + 6.00)2 + (2.00 + 3.00 + 1.00)2 = 52.0 = 7.21 m = tan-1 6.00
4.00 = 56.3 3.30 A = 8.70i + 15.0j B = 13.2i 6.60j A B + 3C = 0 3C
= B A = 21.9i 21.6j C = 7.30i 7.20j or Cx = 7.30 cm Cy = 7.20 cm
3.31 (a) (A + B) = (3i 2j) + (i 4j) = 2i 6j (b) (A B) = (3i 2j) (i
4j) = 4i + 2j (c) A + B = 22 + 62 = 6.32 (d) A B = 42 + 22 = 4.47
(e) A + B = tan1 6 2 = 71.6 = 288 A B = tan1 2 4 = 26.6 3.32 Let i
= east and j = north. R = 3.00b j + 4.00b cos 45 i + 4.00b sin 45 j
5.00b i R = 2.17b i + 5.83b j R = 2.172 + 5.832 b at Arctan 5.83
2.17 N of W = 6.22 blocks at 110 counterclockwise from east
Chapter 3 Solutions 13 2000 by Harcourt College Publishers. All
rights reserved. 3.33 x = r cos and y = r sin , therefore: (a) x =
12.8 cos 150, y = 12.8 sin 150, and (x, y) = (11.1i + 6.40j) m (b)
x = 3.30 cos 60.0, y = 3.30 sin 60.0, and (x, y) = (1.65i + 2.86j)
cm (c) x = 22.0 cos 215, y = 22.0 sin 215, and (x, y) = (18.0i
12.6j) in 3.34 (a) D = A + B + C = 2i + 4j D = 22 + 42 = 4.47 m at
= 63.4 (b) E = A B + C = 6i + 6j E = 62 + 62 = 8.49 m at = 135 3.35
d1 = (3.50j) m d2 = 8.20 cos 45.0i + 8.20 sin 45.0j = (5.80i +
5.80j) m d3 = (15.0i) m R = d1 + d2 + d3 = (15.0 + 5.80)i + (5.80
3.50)j = (9.20i + 2.30j) m (or 9.20 m west and 2.30 m north) The
magnitude of the resultant displacement is |R|= R 2 x + R 2 y =
(9.20)2 + (2.30)2 = 9.48 m The direction is = Arctan 2.30 9.20 =
166 3.36 Refer to the sketch R = A + B + C = 10.0i 15.0j + 50.0i =
40.0i 15.0j R = [(40.0)2 + (15.0)2]1/2 = 42.7 yards |A| = 10.0 |B|
= 15.0 |C| = 50.0 R
14 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 3.37 (a) F = F1 + F2 F = 120 cos (60.0)i + 120 sin
(60.0)j 80.0 cos (75.0)i + 80.0 sin (75.0)j F = 60.0i + 104j 20.7i
+ 77.3j = (39.3i + 181j) N F = (39.3)2 + (181)2 = 185 N ; = tan1
181 39.3 = 77.8 (b) F3 = F = (39.3i 181j) N Goal Solution The
helicopter view in Figure P3.37 shows two people pulling on a
stubborn mule. Find (a) the single force that is equivalent to the
two forces shown and (b) the force that a third person would have
to exert on the mule to make the resultant force equal to zero. The
forces are measured in units of newtons. G: The resultant force
will be larger than either of the two individual forces, and since
the two people are not pulling in exactly the same direction, the
magnitude of the resultant should be less than the sum of the
magnitudes of the two forces. Therefore, we should expect 120 N
< R < 200 N. The angle of the resultant force appears to be
straight ahead and perhaps slightly to the right. If the stubborn
mule remains at rest, the ground must be exerting on the animal a
force equal to the resultant R but in the opposite direction. 75 60
80 N 120 N O: We can find R by adding the components of the two
force vectors. A: F1 = (120 cos 60)i N + (120 sin 60)j N = 60.0i N
+ 103.9j N F2 = (80 cos 75)i N + (80 sin 75)j N = 20.7i N + 77.3j N
R = F1 + F2 = 39.3i N + 181.2j N R = |R| = (39.3)2 + (181.2)2 = 185
N The angle can be found from the arctan of the resultant
components. = tan1 y x = tan1 181.2 39.3 = tan1 (4.61) = 77.8
counterclockwise from the +x axis The opposing force that the
either the ground or a third person must exert on the mule, in
order for the overall resultant to be zero, is 185 N at 258
counterclockwise from +x. L: The resulting force is indeed between
120 N and 200 N as we expected, and the angle seems reasonable as
well. The process applied to solve this problem can be used for
other statics problems encountered in physics and engineering. If
another force is added to act on a system that is already in
equilibrium (sum of the forces is equal to zero), then the system
may accelerate. Such a system is now a dynamic one and will be the
topic of Chapter 5.
Chapter 3 Solutions 15 2000 by Harcourt College Publishers. All
rights reserved. 3.38 East North x y 0m 4.00 m 1.41 1.41 0.500
0.866 +0.914 4.55 R = x 2 + y 2 = 4.64 m at 78.6 N of E 3.39 A =
3.00 m, A = 30.0, B = 3.00 m, B = 90.0 Ax = A cos A = 3.00 cos 30.0
= 2.60 m, Ay = A sin A = 3.00 sin 30.0 = 1.50 m so, A = Axi + Ayj =
(2.60i + 1.50j) m Bx = 0, By = 3.00 m so B = 3.00j m A + B = (2.60i
+ 1.50j) + 3.00j = (2.60i + 4.50j) m *3.40 The y coordinate of the
airplane is constant and equal to 7.60 103 m whereas the x
coordinate is given by x = vit 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 = (268
m/s)t i + (7.60 103 m)j. At t = 45.0 s, P = [1.21 104 i + 7.60 103
j] m. The magnitude is P = (1.21 104)2 + (7.60 103)2 m = 1.43 104 m
and the direction is = Arctan 7.60 103 1.21 104 = 32.2 above the
horizontal 3.41 We have B = R A Ax = 150 cos 120 = 75.0 cm Ay = 150
sin 120 = 130 cm Rx = 140 cos 35.0 = 115 cm Ry = 140 sin 35.0 =
80.3 cm Therefore, B = [115 (75)]i + [80.3 130]j = (190i 49.7j) cm
A B x y 120.0 35.035.0 120.0 R = A + B
16 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. B = [1902 + (49.7)2]1/2 = 196 cm , = tan1 49.7 190
= 14.7
Chapter 3 Solutions 17 2000 by Harcourt College Publishers. All
rights reserved. *3.42 Since A + B = 6.00j, we have (Ax + Bx)i +
(Ay + By)j = 0i + 6.00 j giving Ax + Bx = 0, or Ax = Bx (1) and Ay
+ By = 6.00 (2) Since both vectors have a magnitude of 5.00, we
also have: A 2 x + A 2 y = B 2 x + B 2 y = (5.00)2 From Ax = Bx, it
is seen that A 2 x = B 2 x . Therefore A 2 x + A 2 y = B 2 x + B 2
y gives A 2 y = B 2 y . Then Ay = By, and Equation (2) gives Ay =
By = 3.00. Defining as the angle between either A or B and the y
axis, it is seen that cos = Ay A = By B = 3.00 5.00 = 0.600 and =
53.1 The angle between A and B is then = 2 = 106 . 3.43 (a) A =
8.00i + 12.0j 4.00 k (b) B = A/4 = 2.00i + 3.00j 1.00k (c) C = 3A =
24.0i 36.0j + 12.0k 3.44 R = 75.0 cos 240i + 75.0 sin 240j + 125
cos 135i + 125 sin 135j + 100 cos 160i + 100 sin 160j R = 37.5i
65.0j 88.4i + 88.4j 94.0i + 34.2j R = 220i + 57.6j R = (220)2 +
57.62 at Arctan 57.6 220 above the x-axis R = 227 paces at 165 3.45
(a) C = A + B = (5.00i 1.00j 3.00k) m |C|= (5.00)2 + (1.00)2 +
(3.00)2 m = 5.92 m (b) D = 2A B = (4.00i 11.0j + 15.0k) m tt A + B
AB x y = 2t = 2
18 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. |D|= (4.00)2 + (11.0)2 + (15.0)2 m = 19.0 m
Chapter 3 Solutions 19 2000 by Harcourt College Publishers. All
rights reserved. *3.46 The displacement from radar station to ship
is S = (17.3 sin 136i + 17.3 cos 136j) km = (12.0i 12.4j) km From
station to plane, the displacement is P = (19.6 sin 153i + 19.6 cos
153j + 2.20k) km, or P = (8.90i 17.5j + 2.20k) km. (a) From plane
to ship the displacement is D = S P = (3.12i + 5.02j 2.20k) km (b)
The distance the plane must travel is D= |D| = (3.12)2 + (5.02)2 +
(2.20)2 km = 6.31 km 3.47 The hurricane's first displacement is
41.0 km h (3.00 h) at 60.0 N of W, and its second displacement is
25.0 km h (1.50 h) due North. With i representing east and j
representing north, its total displacement is: 41.0 km h cos 60.0
(3.00 h)(i) + 41.0 km h sin 60.0 (3.00 h) j + 25.0 km h (1.50 h) j
= 61.5 km (i) + 144 km j with magnitude (61.5 km)2 + (144 km)2 =
157 km *3.48 (a) E = (17.0 cm) cos 27.0i + (17.0 cm) sin 27.0j E =
(15.1i + 7.72j) cm (b) F = (17.0 cm) sin 27.0i + (17.0 cm) cos
27.0j F = (7.72i + 15.1j) cm (c) G = +(17.0 cm) sin 27.0i + (17.0
cm) cos 27.0j G= (+7.72i + 15.1j) cm F G E x y 27.0 27.0 27.0 27.0
27.0 27.0
20 Chapter 3 Solutions 2000 by Harcourt College Publishers. All
rights reserved. 3.49 Ax = 3.00, Ay = 2.00 (a) A = Axi + Ayj =
3.00i + 2.00j (b) |A|= A 2 x + A 2 y = (3.00)2 + (2.00)2 =
![Solucionario Serway Fisica Mecanica[1]](https://static.cupdf.com/doc/110x72/577cdc701a28ab9e78aa8c33/solucionario-serway-fisica-mecanica1.jpg)
