for
Fundamentals of Physics, 6/E
by Halliday, Resnick, and Walker
James B. Whitenton
Preface
This booklet includes the solutions relevant to the EXERCISES & PROBLEMS sections of the 6th edition of Fundamentals of Physics, by Halliday, Resnick, and Walker. We also include solutions to problems in the Problems Supplement. We have not included solutions or discussions which pertain to the QUESTIONS sections.
I am very grateful for helpful input from J. Richard Christman, Meighan Dillon, Barbara Moore, and Jearl Walker regarding the development of this document.
Chapter 1
1. The metric prefixes (micro, pico, nano, . . .) are given for ready reference on the inside front cover of the textbook (see also Table 1-2).
(a) Since 1 km = 1× 103m and 1 m = 1
× 106µm,
1 km = 103m = (103m)(106µm/m) = 109µm .
The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0× 109µm.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2m, 1 cm = 10−2m = (10−2m)(106µm/m) = 104µm . We conclude that the fraction of one centimeter equal to 1.0 µm is 1.0× 10−4. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
1.0 yd = (0.91 m)(106µm/m) = 9.1× 105µm .
2. The customer expects 20× 7056 in3 and receives 20× 5826 in3, the difference being 24600 cubic inches, or 24600 in3 2.54 cm 1 inch 3 1 L 1000 cm3 = 403 L where Appendix D has been used (see also Sample Problem 1-2).
3. Using the given conversion factors, we find (a) the distance d in rods to be
d = 4.0 furlongs = (4.0 furlongs)(201.168 m/furlong)
5.0292 m/rod = 160 rods , (b) and that distance in chains to be
d =(4.0 furlongs)(201.168 m/furlong)
20.117 m/chain = 40 chains . 4. (a) Recalling that 2.54 cm equals 1 inch (exactly), we obtain
(0.80 cm) 1 inch 2.54 cm 6 picas 1 inch 12 points 1 pica ≈ 23 points , (b) and (0.80 cm) 1 inch 2.54 cm 6 picas 1 inch ≈ 1.9 picas . 1
5. Various geometric formulas are given in Appendix E. (a) Substituting
R = 6.37× 106m
10−3km/m = 6.37 × 103km into circumference = 2πR, we obtain 4.00× 104km.
(b) The surface area of Earth is
4πR2= 4π 6.37
× 103km2
= 5.10× 108km2 . (c) The volume of Earth is
4π 3 R 3= 4π 3 6.37× 10 3km3 = 1.08× 1012 km3 . 6. (a) Using the fact that the area A of a rectangle is width×length, we find
Atotal = (3.00 acre) + (25.0 perch)(4.00 perch) = (3.00 acre) (40 perch)(4 perch)
1 acre
+ 100 perch2 = 580 perch2 .
We multiply this by the perch2→ rood conversion factor (1 rood/40 perch2) to obtain the answer: Atotal= 14.5 roods.
(b) We convert our intermediate result in part (a): Atotal= (580 perch2)
16.5 ft 1 perch
2
= 1.58× 105 ft2 . Now, we use the feet→ meters conversion given in Appendix D to obtain
Atotal= 1.58× 105ft2 1 m 3.281 ft 2 = 1.47× 104 m2 .
7. The volume of ice is given by the product of the semicircular surface area and the thickness. The semicircle area is A = πr2/2, where r is the radius. Therefore, the volume is
V = π 2 r
2z
where z is the ice thickness. Since there are 103m in 1 km and 102 cm in 1 m, we have r = (2000 km) 10 3m 1 km 102cm 1 m = 2000× 105cm . In these units, the thickness becomes
z = (3000 m) 10 2cm 1 m = 3000× 102cm . Therefore, V = π 2 2000× 10 5cm2 3000× 102cm = 1.9 × 1022cm3 .
8. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area A = 20× 12 = 240 m2) in addition to a rectangular box (height h′= 6.0 m and same base). Therefore,
V = 1 2hA + h ′A = h 2 + h ′ A = 1800 m3 .
(a) Each dimension is reduced by a factor of 1/12, and we find Vdoll= 1800 m3 1 12 3 ≈ 1.0 m3 .
(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore, Vminiature= 1800 m3
1 144
3
≈ 6.0 × 10−4 m3 . 9. We use the conversion factors found in Appendix D.
1 acre· ft = (43, 560 ft2)· ft = 43, 560 ft3. Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V = (26 km2)(1/6 ft) = (26 km2)(3281 ft/km)2(1/6 ft) = 4.66× 107ft3. Thus, V = 4.66× 10 7ft3 4.3560× 104ft3/acre · ft = 1.1× 10 3acre · ft .
10. The metric prefixes (micro (µ), pico, nano, . . .) are given for ready reference on the inside front cover of the textbook (also, Table 1-2).
1 µcentury = 10−6century 100 y 1 century 365 day 1 y 24 h 1 day 60 min 1 h = 52.6 min . The percent difference is therefore
52.6 min− 50 min
50 min = 5.2% .
11. We use the conversion factors given in Appendix D and the definitions of the SI prefixes given in Table 1-2 (also listed on the inside front cover of the textbook). Here, “ns” represents the nanosecond unit, “ps” represents the picosecond unit, and so on.
(a) 1 m = 3.281 ft and 1 s = 109ns. Thus,
3.0× 108m/s = 3.0× 108m s 3.281 ft m s 109ns = 0.98 ft/ns . (b) Using 1 m = 103mm and 1 s = 1012ps, we find
3.0× 108m/s = 3.0× 108m s 103mm m s 1012ps = 0.30 mm/ps .
12. The number of seconds in a year is 3.156×107. This is listed in Appendix D and results from the product (365.25 day/y)(24 h/day)(60 min/h)(60 s/min) .
(a) The number of shakes in a second is 108; therefore, there are indeed more shakes per second than there are seconds per year.
(b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans have existed is given by
106
1010 = 10−4u-day , which we may also express as
10−4 u-day 86400 u-sec 1 u-day
= 8.6 u-sec .
13. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to another, it cannot be corrected since it would impossible to tell what the correction should be. The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning.
CLOCK Sun. Mon. Tues. Wed. Thurs. Fri. -Mon. -Tues. -Wed. -Thurs. -Fri. -Sat
A −16 −16 −15 −17 −15 −15
B −3 +5 −10 +5 +6 −7
C −58 −58 −58 −58 −58 −58
D +67 +67 +67 +67 +67 +67
E +70 +55 +2 +20 +10 +10
Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from−5 s to +10 s, for clock E it is in the range from −70 s to −2 s. After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range. From best the worst, the ranking of the clocks is C, D, A, B, E.
14. The time on any of these clocks is a straight-line function of that on another, with slopes6= 1 and y-intercepts6= 0. From the data in the figure we deduce
tC = 2 7tB+ 594 7 tB = 33 40tA− 662 5 . These are used in obtaining the following results.
(a) We find t′B− tB= 33 40(t ′ A− tA) = 495 s when t′ A− tA= 600 s. (b) We obtain t′C− tC= 2 7(t ′ B− tB) = 2 7(495) = 141 s .
(c) Clock B reads tB= (33/40)(400)− (662/5) ≈ 198 s when clock A reads tA= 400 s. (d) From tC= 15 = (2/7)tB+ (594/7), we get tB ≈ −245 s.
15. We convert meters to astronomical units, and seconds to minutes, using 1000 m = 1 km 1 AU = 1.50× 108km 60 s = 1 min . Thus, 3.0× 108m/s becomes 3.0 × 108m s 1 km 1000 m AU 1.50× 108km 60 s min = 0.12 AU/min .
16. Since a change of longitude equal to 360◦ corresponds to a 24 hour change, then one expects to change longitude by 360◦/24 = 15◦ before resetting one’s watch by 1.0 h.
17. The last day of the 20 centuries is longer than the first day by
(20 century)(0.001 s/century) = 0.02 s .
The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulative effect T is
T = (average increase in length of a day)(number of days) = 0.01 s day 365.25 day y (2000 y) = 7305 s
or roughly two hours.
18. We denote the pulsar rotation rate f (for frequency). f = 1 rotation
1.55780644887275× 10−3s
(a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figureconsiderations for a moment), we obtain the number of rotations:
N =
1 rotation
1.55780644887275× 10−3s
(604800 s) = 388238218.4
which should now be rounded to 3.88× 108 rotations since the time-interval was specified in the problem to three significant figures.
(b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form
N = f t 1× 106 = 1 rotation 1.55780644887275× 10−3s t
which yields the result t = 1557.80644887275 s (though students who do this calculation on their calculator might not obtain those last several digits).
(c) Careful reading of the problem shows that the time-uncertainty per revolution is ±3 × 10−17 s. We therefore expect that as a result of one million revolutions, the uncertainty should be (±3 × 10−17)(1× 106) =
19. If MEis the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME= N m or N = ME/m. We convert mass m to kilograms using Appendix D (1 u = 1.661×10−27kg). Thus, N = ME m = 5.98× 1024kg (40 u)(1.661× 10−27kg/u) = 9.0× 10 49 .
20. To organize the calculation, we introduce the notion of density (which the students have probably seen in other courses):
ρ = m V .
(a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density ρ = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be
V = m
ρ = 1.430 cm 3 .
We convert the volume to SI units: 1.430 cm3 1 m 100 cm 3 = 1.430× 10−6 m3 .
And since V = A z where z = 1× 10−6m (metric prefixes can be found in Table 1-2), we obtain A = 1.430× 10−6m
3
1× 10−6m = 1.430 m 2 .
(b) The volume of a cylinder of length ℓ is V = Aℓ where the cross-section area is that of a circle: A = πr2. Therefore, with r = 2.500× 10−6 m and V = 1.430× 10−6m3, we obtain
ℓ = V
πr2 = 7.284× 10 4 m .
21. We introduce the notion of density (which the students have probably seen in other courses): ρ = m
V and convert to SI units: 1 g = 1× 10−3kg.
(a) For volume conversion, we find 1 cm3= (1× 10−2m)3= 1
× 10−6m3. Thus, the density in kg/m3 is 1 g/cm3= 1 g cm3 10−3kg g cm3 10−6m3 = 1× 103kg/m3 . Thus, the mass of a cubic meter of water is 1000 kg.
(b) We divide the mass of the water by the time taken to drain it. The mass is found from M = ρV (the product of the volume of water and its density):
M = (5700 m3)(1× 103kg/m3
) = 5.70× 106 kg . The time is t = (10 h)(3600 s/h) = 3.6× 104s, so the mass flow rate R is
R = M t =
5.70× 106kg
22. The volume of the water that fell is V = (26 km2)(2.0 in.) = (26 km2) 1000 m 1 km 2 (2.0 in.) 0.0254 m 1 in. = (26× 106m2)(0.0508 m) = 1.3× 106 m3 .
We write the mass-per-unit-volume (density) of the water as: ρ =m
V = 1× 10
3kg/m3 . The mass of the water that fell is therefore given by m = ρV :
m = 1× 103kg/m3 1.3× 106m3 = 1.3× 109kg .
23. We introduce the notion of density (which the students have probably seen in other courses): ρ =m
V and convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m.
(a) The density ρ of a sample of iron is therefore ρ =7.87 g/cm3 1 kg 1000 g 100 cm 1 m 3
which yields ρ = 7870 kg/m3. If we ignore the empty spaces between the close-packed spheres, then the density of an individual iron atom will be the same as the density of any iron sample. That is, if M is the mass and V is the volume of an atom, then
V =M ρ =
9.27× 10−26kg
7.87× 103kg/m3 = 1.18× 10
−29 m3 .
(b) We set V = 4πR3/3, where R is the radius of an atom (Appendix E contains several geometry formulas). Solving for R, we find
R = 3V 4π 1/3 = 3(1.18 × 10−29m 3) 4π 1/3 = 1.41× 10−10m . The center-to-center distance between atoms is twice the radius, or 2.82× 10−10m.
24. The metric prefixes (micro (µ), pico, nano, . . .) are given for ready reference on the inside front cover of the textbook (see also Table 1-2). The surface area A of each grain of sand of radius r = 50 µm = 50×10−6m is given by A = 4π(50× 10−6)2= 3.14× 10−8m2(Appendix E contains a variety of geometry formulas). We introduce the notion of density (which the students have probably seen in other courses):
ρ =m V
so that the mass can be found from m = ρV , where ρ = 2600 kg/m3. Thus, using V = 4πr3/3, the mass of each grain is m = 4π 50× 10 −6m3 3 ! 2600 kg m3 = 1.36× 10−9 kg .
We observe that (because a cube has six equal faces) the indicated surface area is 6 m2. The number of spheres (the grains of sand) N which have a total surface area of 6 m2 is given by
N = 6 m
2
3.14× 10−8m2 = 1.91× 10 8.
Therefore, the total mass M is given by
M = N m = 1.91× 108
1.36× 10−9 kg = 0.260 kg .
25. From the Figure we see that, regarding differences in positions ∆x, 212 S is equivalent to 258 W and 180 S is equivalent to 156 Z. Whether or not the origin of the Zelda path coincides with the origins of the other paths is immaterial to consideration of ∆x.
(a) ∆x = (50.0 S) 258 W 212 S = 60.8 W (b) ∆x = (50.0 S) 156 Z 180 S = 43.3 Z
26. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally:
(a) (11 tuffet) 2 peck 1 tuffet = 22 peck (b) (11 tuffet) 0.50 bushel 1 tuffet = 5.5 bushel (c) (5.5 bushel) 36.3687 L 1 bushel ≈ 200 L
27. We make the assumption that the clouds are directly overhead, so that Figure 1-3 (and the calculations that accompany it) apply. Following the steps in Sample Problem 1-4, we have
θ 360◦ =
t 24 h
which, for t = 38 min = 38/60 h yields θ = 9.5◦. We obtain the altitude h from the relation d2= r2tan2θ = 2rh
which is discussed in that Sample Problem, where r = 6.37× 106m is the radius of the earth. Therefore, h = 8.9× 104m.
28. In the simplest approach, we set up a ratio for the total increase in horizontal depth x (where ∆x = 0.05 m is the increase in horizontal depth per step)
x = Nsteps∆x = 4.57
0.19
(0.05) = 1.2 m .
However, we can approach this more carefully by noting that if there are N = 4.57/.19≈ 24 rises then under normal circumstances we would expect N− 1 = 23 runs (horizontal pieces) in that staircase. This would yield (23)(0.05) = 1.15 m, which – to two significant figures – agrees with our first result. 29. Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barn
along with appropriate conversion factors: (25 wp)100 hide 1 wp 110 acre 1 hide 4047 m2 1 acre (11 barn)1×10−28m2 1 barn ≈ 1 × 10 36 .
30. It is straightforward to compute how many seconds in a year (about 3× 107). Now, if we estimate roughly one breath per second (or every two seconds, or three seconds – it won’t affect the result) then to within an order of magnitude, a person takes 107 breaths in a year.
31. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so (3.7 m)(106µm/m)
(14 day)(86400 s/day) = 3.1 µm/s . 32. The mass in kilograms is
(28.9 piculs) 100 gin 1 picul 16 tahil 1 gin 10 chee 1 tahil 10 hoon 1 chee 0.3779 g 1 hoon
which yields 1.747× 106 g or roughly 1750 kg.
33. (a) In atomic mass units, the mass of one molecule is 16 + 1 + 1 = 18 u. Using Eq. 1-9, we find (18 u) 1.6605402 × 10−27kg
1 u
= 3.0× 10−26kg .
(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of water molecules:
1.4× 1021
3.0× 10−26 ≈ 5 × 10 46.
34. (a) We find the volume in cubic centimeters (193 gal) 231 in 3 1 gal 2.54 cm 1 in 3 = 7.31× 105cm3 and subtract this from 1× 106 cm3 to obtain 2.69
× 105cm3. The conversion gal
→ in3 is given in Appendix D (immediately below the table of Volume conversions).
(b) The volume found in part (a) is converted (by dividing by (100 cm/m)3) to 0.731 m3, which corre-sponds to a mass of
1000 kg/m3
0.731 m2 = 731 kg
using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in 731 kg
0.0018 kg/min= 4.06× 10 5min
35. (a) When θ is measured in radians, it is equal to the arclength divided by the radius. For very large radius circles and small values of θ, such as we deal with in this problem,
the arcs may be approximated as straight lines – which for our purposes corre-spond to the di-ameters d and D of the Moon and Sun, respec-tively. Thus, ... ... ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . D d RSun RMoon θ . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... θ = d RMoon = D RSun =⇒ RRSun Moon =D d which yields D/d = 400.
(b) Various geometric formulas are given in Appendix E. Using rsand rmfor the radius of the Sun and Moon, respectively (noting that their ratio is the same as D/d), then the Sun’s volume divided by that of the Moon is
4 3πr3s 4 3πr3m = rs rm 3 = 4003= 6.4× 107.
(c) The angle should turn out to be roughly 0.009 rad (or about half a degree). Putting this into the equation above, we get
d = θRMoon= (0.009) 3.8× 105 ≈ 3.4 × 103 km . 36. (a) For the minimum (43 cm) case, 9 cubit converts as follows:
(9 cubit) 0.43 m 1 cubit
= 3.9 m . And for the maximum (43 cm) case we obtain
(9 cubit) 0.53 m 1 cubit
= 4.8 m .
(b) Similarly, with 0.43 m → 430 mm and 0.53 m → 530 mm, we find 3.9 × 103mm and 4.8× 103mm, respectively.
(c) We can convert length and diameter first and then compute the volume, or first compute the volume and then convert. We proceed using the latter approach (where d is diameter and ℓ is length).
Vcylinder, min = π 4 ℓ d 2= 28 cubit3 = 28 cubit3 0.43 m 1 cubit 3 = 2.2 m3 .
Similarly, with 0.43 m replaced by 0.53 m, we obtain Vcylinder, max= 4.2 m3. 37. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain
1 ken2 1 m2 =
1.972m2
(b) Similarly, we find
1 ken3 1 m3 =
1.973m3
1 m3 = 7.65 .
(c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus, πr2h = π(3.00)2(5.50) = 155.5 ken3.
(d) If we multiply this by the result of part (b), we determine the volume in cubic meters: (155.5)(7.65) = 1.19× 103m3.
38. Although we can look up the distance from Cleveland to Los Angeles, we can just as well (for an order of magnitude calculation) assume it’s some relatively small fraction of the circumference of Earth – which suggests that (again, for an order of magnitude calculation) we can estimate the distance to be roughly r, where r ≈ 6 × 106 m is the radius of Earth. If we take each toilet paper sheet to be roughly 10 cm (0.1 m) then the number of sheets needed is roughly 6× 106/0.1 = 6× 107≈ 108.
39. Using the (exact) conversion 2.54 cm = 1 in. we find that 1 ft = (12)(2.54)/100 = 0.3048 m (which also can be found in Appendix D). The volume of a cord of wood is 8× 4 × 4 = 128 ft3, which we convert (multiplying by 0.30483) to 3.6 m3. Therefore, one cubic meter of wood corresponds to 1/3.6≈ 0.3 cord. 40. (a) When θ is measured in radians, it is equal to the arclength s divided by the radius R. For a very large radius circle and small value of θ, such as we deal with in Fig. 1-9, the arc may be approximated as the straight line-segment of length 1 AU. First, we convert θ = 1 arcsecond to radians: (1 arcsecond) 1 arcminute 60 arcsecond 1◦ 60 arcminute 2π radian 360◦
which yields θ = 4.85× 10−6 rad. Therefore, one parsec is Ro= s θ = 1 AU 4.85× 10−6 = 2.06 × 10 5AU .
Now we use this to convert R = 1 AU to parsecs: R = (1 AU)
1 pc
2.06× 105AU
= 4.9× 10−6pc .
(b) Also, since it is straightforward to figure the number of seconds in a year (about 3.16× 107 s), and (for constant speeds) distance = speed×time, we have
1 ly = (186, 000 mi/s) 3.16× 107s 5.9 × 1012mi
which we convert to AU by dividing by 92.6× 106 (given in the problem statement), obtaining 6.3× 104 AU. Inverting, the result is 1 AU = 1/6.3
× 104= 1.6
× 10−5 ly. (c) As found in the previous part, 1 ly = 5.9× 1012mi.
(d) We now know what one parsec is in AU (denoted above as Ro), and we also know how many miles are in an AU. Thus, one parsec is equivalent to
92.9× 106mi/AU
2.06 × 105AU = 1.9 × 1013mi . 41. We reduce the stock amount to British teaspoons:
1 breakfastcup = 2× 8 × 2 × 2 = 64 teaspoons 1 teacup = 8× 2 × 2 = 32 teaspoons 6 tablespoons = 6× 2 × 2 = 24 teaspoons 1 dessertspoon = 2 teaspoons
which totals to 122 teaspoons – which corresponds (since liquid measure is being used) to 122 U.S. teaspoons. Since each U.S cup is 48 teaspoons, then upon dividing 122/48≈ 2.54, we find this amount corresponds to two-and-a-half U.S. cups plus a remainder of precisely 2 teaspoons. For the nettle tops, one-half quart is still one-half quart. For the rice, one British tablespoon is 4 British teaspoons which (since dry-goods measure is being used) corresponds to 2 U.S. teaspoons. Finally, a British saltspoon is 1
2 British teaspoon which corresponds (since dry-goods measure is again being used) to 1 U.S. teaspoon. 42. (a) Megaphone.
(b) microphone.
(c) dekacard (“deck of cards”). (d) Gigalow (“gigalo”).
(e) terabull (“terrible”). (f) decimate.
(g) centipede.
(h) nanonannette. (“No No Nanette”). (i) picoboo (“peek-a-boo”).
(j) attoboy (“at-a-boy”).
(k) Two hectowithit (“to heck with it”).
(l) Two kilomockingbird (“to kill a mockingbird”). 43. The volume removed in one year is
V = 75× 104m2 (26 m) ≈ 2 × 107 m3 which we convert to cubic kilometers:
V = 2× 107m3 1 km 1000 m 3 = 0.020 km3 .
44. (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.3, and 1 in.3= 1.639× 10−2L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460ft2/gal becomes
460 ft2 gal 1 m 3.28 ft 2 1 gal 3.79 L = 11.3 m2/L . (b) Also, since 1 m3is equivalent to 1000 L, our result from part (a) becomes
11.3 m2 L 1000 L 1 m3 = 1.13× 104m−1 . (c) The inverse of the original quantity is (460 ft2/gal)−1 = 2.17× 10−3gal/ft2
, which is the volume of the paint (in gallons) needed to cover a square foot of area. From this, we could also figure the paint thickness (it turns out to be about a tenth of a millimeter, as one sees by taking the reciprocal of the answer in part (b)).
Chapter 2
1. Assuming the horizontal velocity of the ball is constant, the horizontal displacement is ∆x = v∆t
where ∆x is the horizontal distance traveled, ∆t is the time, and v is the (horizontal) velocity. Converting v to meters per second, we have 160 km/h = 44.4 m/s. Thus
∆t =∆x v =
18.4 m
44.4 m/s = 0.414 s .
The velocity-unit conversion implemented above can be figured “from basics” (1000 m = 1 km, 3600 s = 1 h) or found in Appendix D.
2. Converting to SI units, we use Eq. 2-3 with d for distance.
savg = d t (110.6 km/h) 1000 m/km 3600 s/h = 200.0 m t
which yields t = 6.510 s. We converted the speed km/h→ m/s by converting each unit (km → m, h → s) individually. But we mention that the “one-step” conversion can be found in Appendix D (1 km/h = 0.2778 m/s).
3. We use Eq. 2-2 and Eq. 2-3. During a time tc when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with ∆x = v tc.
(a) During the first part of the motion, the displacement is ∆x1= 40 km and the time interval is t1=
(40 km)
(30 km/h) = 1.33 h .
During the second part the displacement is ∆x2= 40 km and the time interval is t2= (40 km)
(60 km/h) = 0.67 h .
Both displacements are in the same direction, so the total displacement is ∆x = ∆x1+ ∆x2 = 40 km + 40 km = 80 km. The total time for the trip is t = t1+ t2 = 2.00 h. Consequently, the average velocity is
vavg=
(80 km)
(2.0 h) = 40 km/h .
(b) In this example, the numerical result for the average speed is the same as the average velocity 40 km/h.
(c) In the interest of saving space, we briefly describe the graph (with kilometers and hours understood): two contiguous line segments, the first having a slope of 30 and connecting the origin to (t1, x1) = (1.33, 40) and the second having a slope of 60 and connecting (t1, x1) to (t, x) = (2.00, 80). The average velocity, from the graphical point of view, is the slope of a line drawn from the origin to (t, x).
4. If the plane (with velocity v) maintains its present course, and if the terrain continues its upward slope of 4.3◦, then the plane will strike the ground after traveling
∆x = h tan θ =
35 m
tan 4.3◦ = 465.5 m≈ 0.465 km .
This corresponds to a time of flight found from Eq. 2-2 (with v = vavg since it is constant) t = ∆x
v =
0.465 km
1300 km/h= 0.000358 h≈ 1.3 s . This, then, estimates the time available to the pilot to make his correction.
5. (a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, the average speed is
savg 1=D T =
(55 km/h)T2 + (90 km/h)T2
T = 72.5 km/h
which should be rounded to 73 km/h.
(b) Using the fact that time = distance/speed while the speed is constant, we find savg 2= D T = D D/2 55 km/h+ D/2 90 km/h = 68.3 km/h
which should be rounded to 68 km/h.
(c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain for the two-way trip
savg= 2D D 72.5 km/h+ D 68.3 km/h = 70 km/h .
(d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero. (e) In asking for a sketch, the problem is allowing the student to arbitrarily set the distance D (the intent
is not to make the student go to an Atlas to look it up); the student can just as easily arbitrarily set T instead of D, as will be clear in the following discussion. In the interest of saving space, we briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line segments, the first having a slope of 55 and connecting the origin to (t1, x1) = (T /2, 55T /2) and the second having a slope of 90 and connecting (t1, x1) to (T, D) where D = (55 + 90)T /2. The average velocity, from the graphical point of view, is the slope of a line drawn from the origin to (T, D).
6. (a) Using the fact that time = distance/velocity while the velocity is constant, we find vavg= 73.2 m + 73.2 m 73.2 m 1.22 m/s+ 73.2 m 3.05 m = 1.74 m/s .
(b) Using the fact that distance = vt while the velocity v is constant, we find vavg=(1.22 m/s)(60 s) + (3.05 m/s)(60 s)
(c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before – the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer.
... . ... . ... ... ... . ... . ... ... ... . ... . ... ... x t 60 84 73 146 . . ... . ... . . ... ... . ... ... . ... . . ... ... . ... ... . ... . . ... ... . . ... ... . ... . . ... . . ... x t 60 120 73 256
7. Using x = 3t− 4t2+ t3with SI units understood is efficient (and is the approach we will use), but if we wished to make the units explicit we would write x = (3 m/s)t− (4 m/s2)t2+ (1 m/s3
)t3. We will quote our answers to one or two significant figures, and not try to follow the significant figure rules rigorously. (a) Plugging in t = 1 s yields x = 0. With t = 2 s we get x =−2 m. Similarly, t = 3 s yields x = 0 and
t = 4 s yields x = 12 m. For later reference, we also note that the position at t = 0 is x = 0. (b) The position at t = 0 is subtracted from the position at t = 4 s to find the displacement ∆x = 12 m.
(c) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement ∆x = 14 m. Eq. 2-2, then, leads to
vavg= ∆x ∆t =
14
2 = 7 m/s . (d) The horizontal axis is 0≤ t ≤ 4 with SI units understood.
Not shown is a straight line drawn from the point at (t, x) = (2,−2) to the highest point shown (at t = 4 s) which would represent the answer for part (c).
t 0
10
x
8. Recognizing that the gap between the trains is closing at a constant rate of 60 km/h, the total time which elapses before they crash is t = (60 km)/(60 km/h) = 1.0 h. During this time, the bird travels a distance of x = vt = (60 km/h)(1.0 h) = 60 km.
9. Converting to seconds, the running times are t1= 147.95 s and t2= 148.15 s, respectively. If the runners were equally fast, then
savg 1= savg 2 =⇒ L1 t1
=L2 t2 . From this we obtain
L2− L1=
148.15 147.95− 1
L1≈ 1.35 m
where we set L1≈ 1000 m in the last step. Thus, if L1and L2 are no different than about 1.35 m, then runner 1 is indeed faster than runner 2. However, if L1 is shorter than L2 than 1.4 m then runner 2 is actually the faster.
10. The velocity (both magnitude and sign) is determined by the slope of the x versus t curve, in accordance with Eq. 2-4.
(a) The armadillo is to the left of the coordinate origin on the axis between t = 2.0 s and t = 4.0 s. (b) The velocity is negative between t = 0 and t = 3.0 s.
(c) The velocity is positive between t = 3.0 s and t = 7.0 s. (d) The velocity is zero at the graph minimum (at t = 3.0 s). 11. We use Eq. 2-4.
(a) The velocity of the particle is v = dx
dt = d
dt 4− 12t + 3t
2 = −12 + 6t .
Thus, at t = 1 s, the velocity is v = (−12 + (6)(1)) = −6 m/s. (b) Since v < 0, it is moving in the negative x direction at t = 1 s.
(c) At t = 1 s, the speed is|v| = 6 m/s.
(d) For 0 < t < 2 s,|v| decreases until it vanishes. For 2 < t < 3 s, |v| increases from zero to the value it had in part (c). Then, |v| is larger than that value for t > 3 s.
(e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t→ +∞, we have v → +∞). One can check that v = 0 when t = 2 s.
(f) No. In fact, from v =−12 + 6t, we know that v > 0 for t > 2 s.
12. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and work with distances in centimeters and times in seconds.
(a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x2= 21.75 cm and x3= 50.25 cm, respectively. The average velocity during the time interval 2.00≤ t ≤ 3.00 s is
vavg= ∆x ∆t =
50.25 cm− 21.75 cm 3.00 s− 2.00 s which yields vavg= 28.5 cm/s.
(b) The instantaneous velocity is v = dxdt = 4.5t2, which yields v = (4.5)(2.00)2 = 18.0 cm/s at time t = 2.00 s.
(c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)2= 40.5 cm/s. (d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)2= 28.1 cm/s.
(e) Let tm stand for the moment when the particle is midway between x2 and x3 (that is, when the particle is at xm= (x2+ x3)/2 = 36 cm). Therefore,
xm= 9.75 + 1.5t3m =⇒ tm= 2.596
(f) The answer to part (a) is given by the slope of the straight line between t = 2 and
t = 3 in this x-vs-t plot. The an-swers to parts (b), (c), (d) and (e) correspond to the slopes of tangent lines (not shown but easily imag-ined) to the curve at the appropriate points. (cm) x (a) 20 40 60 2 t 3
13. Since v = dxdt (Eq. 2-4), then ∆x =R v dt, which corresponds to the area under the v vs t graph. Dividing the total area A into rectangular (base×height) and triangular (1
2base×height) areas, we have A = A0<t<2+ A2<t<10+ A10<t<12+ A12<t<16 = 1 2(2)(8) + (8)(8) + (2)(4) +1 2(2)(4) + (4)(4)
with SI units understood. In this way, we obtain ∆x = 100 m.
14. From Eq. 2-4 and Eq. 2-9, we note that the sign of the velocity is the sign of the slope in an x-vs-t plot, and the sign of the acceleration corresponds to whether such a curve is concave up or concave down. In the interest of saving space, we indicate sample points for parts (a)-(d) in a single figure; this means that all points are not at t = 1 s (which we feel is an acceptable modification of the problem – since the datum t = 1 s is not used).
(c) (a)
(b) (d)
Any change from zero to non-zero values of ~v represents in-creasing |~v| (speed). Also, ~v k ~a implies that the particle is going faster. Thus, points (a), (b) and (d) involve increasing speed.
15. We appeal to Eq. 2-4 and Eq. 2-9.
(a) This is v2– that is, the velocity squared. (b) This is the acceleration a.
(c) The SI units for these quantities are (m/s)2= m2/s2and m/s2, respectively. 16. Eq. 2-9 indicates that acceleration is the slope of the v-vs-t graph.
Based on this, we show here a sketch of the acceleration (in m/s2) as a function of time. The values along the acceleration axis should not be
taken too seriously. –20 –10 0 10 a 5 t
17. We represent its initial direction of motion as the +x direction, so that v0= +18 m/s and v =−30 m/s (when t = 2.4 s). Using Eq. 2-7 (or Eq. 2-11, suitably interpreted) we find
aavg=
(−30) − (+18)
2.4 =−20 m/s 2
which indicates that the average acceleration has magnitude 20 m/s2and is in the opposite direction to the particle’s initial velocity.
18. We use Eq. 2-2 (average velocity) and Eq. 2-7 (average acceleration). Regarding our coordinate choices, the initial position of the man is taken as the origin and his direction of motion during 5 min≤ t ≤ 10 min is taken to be the positive x direction. We also use the fact that ∆x = v∆t′when the velocity is constant during a time interval ∆t′.
(a) Here, the entire interval considered is ∆t = 8− 2 = 6 min which is equivalent to 360 s, whereas the sub-interval in which he is moving is only ∆t′ = 8− 5 = 3 min = 180 s. His position at t = 2 min is x = 0 and his position at t = 8 min is x = v∆t′= (2.2)(180) = 396 m. Therefore,
vavg=
396 m− 0
360 s = 1.10 m/s .
(b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, keeping the answer to 3 significant figures,
aavg=
2.2 m/s− 0
360 s = 0.00611 m/s 2
.
(c) Now, the entire interval considered is ∆t = 9− 3 = 6 min (360 s again), whereas the sub-interval in which he is moving is ∆t′ = 9− 5 = 4 min = 240 s). His position at t = 3 min is x = 0 and his position at t = 9 min is x = v∆t′ = (2.2)(240) = 528 m. Therefore,
vavg= 528 m− 0
360 s = 1.47 m/s .
(d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequently, aavg= 2.2/360 = 0.00611 m/s2just as in part (b).
the man stand-ing at x = 0 for 0 ≤ t < 300 s and the linearly rising line for 300 ≤ t ≤ 600 s represents his constant-velocity motion. The dotted lines represent the answers to part (a) and (c) in the sense that their slopes yield those results.
(c)
(a)
0 500x
0t
500The graph of v-vs-t is not shown here, but would consist of two horizontal “steps” (one at v = 0 for 0 ≤ t < 300 s and the next at v = 2.2 m/s for 300 ≤ t ≤ 600 s). The indications of the average accelerations found in parts (b) and (d) would be dotted lines connected the “steps” at the appropriate t values (the slopes of the dotted lines representing the values of aavg).
19. In this solution, we make use of the notation x(t) for the value of x at a particular t. Thus, x(t) = 50t + 10t2 with SI units (meters and seconds) understood.
(a) The average velocity during the first 3 s is given by vavg= x(3)− x(0) ∆t = (50)(3) + (10)(3)2 − 0 3 = 80 m/s .
(b) The instantaneous velocity at time t is given by v = dx/dt = 50 + 20t, in SI units. At t = 3.0 s, v = 50 + (20)(3.0) = 110 m/s.
(c) The instantaneous acceleration at time t is given by a = dv/dt = 20 m/s2. It is constant, so the acceleration at any time is 20 m/s2.
(d) and (e) The graphs below show the coordinate x and velocity v as functions of time, with SI units understood. The dotted line marked (a) in the first graph runs from t = 0, x = 0 to t = 3.0 s, x = 240 m. Its slope is the average velocity during the first 3 s of motion. The dotted line marked (b) is tangent to the x curve at t = 3.0 s. Its slope is the instantaneous velocity at t = 3.0 s.
... . ... . ... ... . ... . ... ... . ... . ... ... . ... . ... ... . ... ... ... ... ... ... ... . . ... . . ... . . ... . . ... 1.0 2.0 3.0 4.0 0 100 200 300 400 (a) (b) t x ... 1.0 2.0 3.0 4.0 50 0 100 150 200 t v
20. Using the general property d
dxexp(bx) = b exp(bx), we write v = dx dt = d (19t) dt · e−t+ (19t)· d e−t dt .
If a concern develops about the appearance of an argument of the exponential (−t) apparently having units, then an explicit factor of 1/T where T = 1 second can be inserted and carried through the computation (which does not change our answer). The result of this differentiation is
v = 16(1− t)e−t
with t and v in SI units (s and m/s, respectively). We see that this function is zero when t = 1 s. Now that we know when it stops, we find out where it stops by plugging our result t = 1 into the given function x = 16te−twith x in meters. Therefore, we find x = 5.9 m.
21. In this solution, we make use of the notation x(t) for the value of x at a particular t. The notations v(t) and a(t) have similar meanings.
(a) Since the unit of ct2 is that of length, the unit of c must be that of length/time2, or m/s2 in the SI system. Since bt3has a unit of length, b must have a unit of length/time3, or m/s3.
(b) When the particle reaches its maximum (or its minimum) coordinate its velocity is zero. Since the velocity is given by v = dx/dt = 2ct− 3bt2, v = 0 occurs for t = 0 and for
t = 2c 3b =
2(3.0 m/s2)
3(2.0 m/s3) = 1.0 s .
For t = 0, x = x0= 0 and for t = 1.0 s, x = 1.0 m > x0. Since we seek the maximum, we reject the first root (t = 0) and accept the second (t = 1 s).
(c) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and goes back to x(4 s) = (3.0 m/s2)(4.0 s)2− (2.0 m/s3)(4.0 s)3=−80 m .
The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m.
(d) Its displacement is given by ∆x = x2− x1, where x1= 0 and x2=−80 m. Thus, ∆x = −80 m. (e) The velocity is given by v = 2ct− 3bt2= (6.0 m/s2
)t− (6.0 m/s3)t2. Thus v(1 s) = (6.0 m/s2)(1.0 s)− (6.0 m/s3)(1.0 s)2= 0 v(2 s) = (6.0 m/s2)(2.0 s)− (6.0 m/s3)(2.0 s)2=−12 m/s v(3 s) = (6.0 m/s2)(3.0 s)− (6.0 m/s3)(3.0 s)2= −36.0 m/s v(4 s) = (6.0 m/s2)(4.0 s)− (6.0 m/s3)(4.0 s)2= −72 m/s . (f) The acceleration is given by a = dv/dt = 2c− 6b = 6.0 m/s2− (12.0 m/s3)t. Thus
a(1 s) = 6.0 m/s2− (12.0 m/s3)(1.0 s) =−6.0 m/s2 a(2 s) = 6.0 m/s2− (12.0 m/s3)(2.0 s) =−18 m/s2 a(3 s) = 6.0 m/s2− (12.0 m/s3)(3.0 s) =−30 m/s2 a(4 s) = 6.0 m/s2− (12.0 m/s3)(4.0 s) =−42 m/s2 . 22. For the automobile ∆v = 55− 25 = 30 km/h, which we convert to SI units:
a =∆v ∆t = (30 km/h)1000 m/km3600 s/h (0.50 min)(60 s/min) = 0.28 m/s 2 .
The change of velocity for the bicycle, for the same time, is identical to that of the car, so its acceleration is also 0.28 m/s2.
(a) Setting v = 0 and x0= 0 in v2= v02+ 2a(x− x0), we find x =−12v 2 0 a =− 1 2 5.00× 106 −1.25 × 1014 = 0.100 m .
Since the muon is slowing, the initial velocity and the acceleration must have opposite signs. (b) Below are the time-plots of the position x and velocity v of the muon from the moment it enters
the field to the time it stops. The computation in part (a) made no reference to t, so that other equations from Table 2-1 (such as v = v0+ at and x = v0t + 12at2) are used in making these plots.
... 10 20 30 40 t (ns) 0 2.5 5.0 7.5 10 x (cm) . ... 10 20 30 40 t (ns) 0 2.0 4.0 6.0 8.0 v (Mm/s)
24. The time required is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7). First, we convert the velocity change to SI units:
∆v = (100 km/h) 1000 m/km 3600 s/h
= 27.8 m/s . Thus, ∆t = ∆v/a = 27.8/50 = 0.556 s.
25. We use v = v0+ at, with t = 0 as the instant when the velocity equals +9.6 m/s.
(a) Since we wish to calculate the velocity for a time before t = 0, we set t =−2.5 s. Thus, Eq. 2-11 gives
v = (9.6 m/s) +3.2 m/s2(−2.5 s) = 1.6 m/s . (b) Now, t = +2.5 s and we find
v = (9.6 m/s) +3.2 m/s2(2.5 s) = 18 m/s .
26. The bullet starts at rest (v0= 0) and after traveling the length of the barrel (∆x = 1.2 m) emerges with the given velocity (v = 640 m/s), where the direction of motion is the positive direction. Turning to the constant acceleration equations in Table 2-1, we use
∆x =1
2(v0+ v) t . Thus, we find t = 0.00375 s (about 3.8 ms).
27. The constant acceleration stated in the problem permits the use of the equations in Table 2-1. (a) We solve v = v0+ at for the time:
t = v− v0 a = 1 10 3.0× 10 8m/s 9.8 m/s2 = 3.1× 10 6 s
(b) We evaluate x = x0+ v0t +12at2, with x0= 0. The result is x = 1 2 9.8 m/s2 3.1× 106s2 = 4.7× 1013m . 28. From Table 2-1, v2− v2
0= 2a∆x is used to solve for a. Its minimum value is amin= v2 − v2 0 2∆xmax = (360 km/h) 2 2(1.80 km) = 36000 km/h 2 which converts to 2.78 m/s2.
29. Assuming constant acceleration permits the use of the equations in Table 2-1. We solve v2= v2
0+ 2a(x− x0) with x0= 0 and x = 0.010 m. Thus,
a = v 2 − v2 0 2x = 5.7× 1052 − 1.5 × 1052 2(0.01) = 1.62× 10 15 m/s2 . 30. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7).
a = ∆v ∆t = (1020 km/h)1000 m/km3600 s/h 1.4 s = 202.4 m/s 2 .
In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s2 as follows: a =202.4
9.8 g = 21g .
31. We choose the positive direction to be that of the initial velocity of the car (implying that a < 0 since it is slowing down). We assume the acceleration is constant and use Table 2-1.
(a) Substituting v0= 137 km/h = 38.1 m/s, v = 90 km/h = 25 m/s, and a =−5.2 m/s2into v = v0+at, we obtain
t = 25 m/s− 38 m/s
−5.2 m/s2 = 2.5 s . (b) We take the car to be at x = 0 when the brakes are applied
(at time t = 0). Thus, the coordinate of the car as a function of time is given by
x = (38)t +1 2(−5.2)t
2
in SI units. This function is plotted from t = 0 to t = 2.5 s on the graph to the right. We have not shown the v-vs-t graph here; it is a descending straight line from v0 to v. ... 0 20 40 60 80 0.5 1.0 1.5 2.0 2.5 x (m) t (s)
32. From the figure, we see that x0 =−2.0 m. From Table 2-1, we can apply x − x0 = v0t + 12at2 with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, v0 and a. SI units are understood.
0.0− (−2.0) = v0(1.0) + 1 2a(1.0) 2 6.0− (−2.0) = v0(2.0) + 1 2a(2.0) 2 .
Solving these simultaneous equations yields the results v0 = 0.0 and a = 4.0 m/s2. The fact that the answer is positive tells us that the acceleration vector points in the +x direction.
33. The problem statement (see part (a)) indicates that a = constant, which allows us to use Table 2-1. (a) We take x0 = 0, and solve x = v0t +12at2 (Eq. 2-15) for the acceleration: a = 2(x− v0t)/t2.
Substituting x = 24.0 m, v0= 56.0 km/h = 15.55 m/s and t = 2.00 s, we find a = 2 (24.0 m− (15.55 m/s)(2.00 s))
(2.00 s)2 =−3.56 m/s 2
.
The negative sign indicates that the acceleration is opposite to the direction of motion of the car. The car is slowing down.
(b) We evaluate v = v0+ at as follows:
v = 15.55 m/s−3.56 m/s2(2.00 s) = 8.43 m/s which is equivalent to 30.3 km/h.
34. We take the moment of applying brakes to be t = 0. The deceleration is constant so that Table 2-1 can be used. Our primed variables (such as v′
o= 72 km/h = 20 m/s) refer to one train (moving in the +x direction and located at the origin when t = 0) and unprimed variables refer to the other (moving in the−x direction and located at x0= +950 m when t = 0). We note that the acceleration vector of the unprimed train points in the positive direction, even though the train is slowing down; its initial velocity is vo=−144 km/h = −40 m/s. Since the primed train has the lower initial speed, it should stop sooner than the other train would (were it not for the collision). Using Eq 2-16, it should stop (meaning v′= 0) at x′= (v′) 2 − (v′ o) 2 2a′ = 0− 202 −2 = 200 m . The speed of the other train, when it reaches that location, is
v =pv2
o+ 2a∆x =p(−40)2+ 2(1.0)(200− 950) = √
100 = 10 m/s
using Eq 2-16 again. Specifically, its velocity at that moment would be−10 m/s since it is still traveling in the−x direction when it crashes. If the computation of v had failed (meaning that a negative number would have been inside the square root) then we would have looked at the possibility that there was no collision and examined how far apart they finally were. A concern that can be brought up is whether the primed train collides before it comes to rest; this can be studied by computing the time it stops (Eq. 2-11 yields t = 20 s) and seeing where the unprimed train is at that moment (Eq. 2-18 yields x = 350 m, still a good distance away from contact).
35. The acceleration is constant and we may use the equations in Table 2-1.
(a) Taking the first point as coordinate origin and time to be zero when the car is there, we apply Eq. 2-17 (with SI units understood):
x = 1
2(v + v0) t = 1
2(15 + v0) (6) .
With x = 60.0 m (which takes the direction of motion as the +x direction) we solve for the initial velocity: v0= 5.00 m/s.
(b) Substituting v = 15 m/s, v0 = 5 m/s and t = 6 s into a = (v− v0)/t (Eq. 2-11), we find a = 1.67 m/s2.
(c) Substituting v = 0 in v2= v2
0+ 2ax and solving for x, we obtain x =−v 2 0 2a =− 52 2(1.67) =−7.50 m .
(d) The graphs require computing the time when v = 0, in which case, we use v = v0+ at′= 0. Thus, t′= −v0
a = −5
1.67=−3.0 s indicates the moment the car was at rest. SI units are assumed.
... −5 5 10 t −20 20 40 60 x ... −5 0 5 10 t 5 10 15 v
36. We denote the required time as t, assuming the light turns green when the clock reads zero. By this time, the distances traveled by the two vehicles must be the same.
(a) Denoting the acceleration of the automobile as a and the (constant) speed of the truck as v then ∆x = 1 2at 2 car = (vt)truck which leads to t = 2v a = 2(9.5) 2.2 = 8.6 s . Therefore, ∆x = vt = (9.5)(8.6) = 82 m . (b) The speed of the car at that moment is
vcar= at = (2.2)(8.6) = 19 m/s .
37. We denote tr as the reaction time and tb as the braking time. The motion during tr is of the constant-velocity (call it v0) type. Then the position of the car is given by
x = v0tr+ v0tb+ 1 2at
2 b
where v0is the initial velocity and a is the acceleration (which we expect to be negative-valued since we are taking the velocity in the positive direction and we know the car is decelerating). After the brakes are applied the velocity of the car is given by v = v0+ atb. Using this equation, with v = 0, we eliminate tb from the first equation and obtain
x = v0tr− v2 0 a + 1 2 v2 0 a = v0tr− 1 2 v2 0 a . We write this equation for each of the initial velocities:
x1= v01tr− 1 2 v2 01 a
and x2= v02tr−1 2 v2 02 a . Solving these equations simultaneously for trand a we get
tr= v2 02x1− v201x2 v01v02(v02− v01) and a =−1 2 v02v012 − v01v022 v02x1− v01x2 . Substituting x1= 56.7 m, v01= 80.5 km/h = 22.4 m/s, x2 = 24.4 m and v02 = 48.3 km/h = 13.4 m/s, we find tr= 13.42(56.7) − 22.42(24.4) (22.4)(13.4)(13.4− 22.4) = 0.74 s and a =−12 (13.4)22.4 2− (22.4)13.42 (13.4)(56.7)− (22.4)(24.4)=−6.2 m/s 2 .
The magnitude of the deceleration is therefore 6.2 m/s2. Although rounded off values are displayed in the above substitutions, what we have input into our calculators are the “exact” values (such as v02= 16112 m/s).
38. In this solution we elect to wait until the last step to convert to SI units. Constant acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and denote the train’s initial velocity as vt and the locomotive’s velocity as vℓ (which is also the final velocity of the train, if the rear-end collision is barely avoided). We note that the distance ∆x consists of the original gap between them D as well as the forward distance traveled during this time by the locomotive vℓt. Therefore,
vt+ vℓ 2 = ∆x t = D + vℓt t = D t + vℓ. We now use Eq. 2-11 to eliminate time from the equation. Thus,
vt+ vℓ 2 = D (vℓ− vt) /a+ vℓ leads to a = vt+ vℓ 2 − vℓ vℓ− vt D =−2D1 (vℓ− vt)2 . Hence, a =−2(0.676 km)1 29km h − 161 km h 2 =−12888 km/h2 which we convert as follows:
a =−12888 km/h2 1000 m 1 km 1 h 3600 s 2 =−0.994 m/s2
so that its magnitude is 0.994 m/s2. A graph is shown below for the case where a collision is just avoided (x along the vertical axis is in meters and t along the horizontal axis is in seconds). The top (straight) line shows the motion of the locomotive and the bottom curve shows the motion of the passenger train.
The other case (where the colli-sion is not quite avoided) would be similar except that the slope of the bottom curve would be greater than that of the top line at the point where they meet. 0 200 400 600 800
x
10 20 30t
39. We assume the periods of acceleration (duration t1) and deceleration (duration t2) are periods of constant a so that Table 2-1 can be used. Taking the direction of motion to be +x then a1 = +1.22 m/s2 and a2=−1.22 m/s2. We use SI units so the velocity at t = t1is v = 305/60 = 5.08 m/s.
(a) We denote ∆x as the distance moved during t1, and use Eq. 2-16: v2= v20+ 2a1∆x =⇒ ∆x =
5.082 2(1.22) which yields ∆x = 10.59≈ 10.6 m.
(b) Using Eq. 2-11, we have
t1= v− v0
a1 = 5.08
1.22 = 4.17 s .
The deceleration time t2 turns out to be the same so that t1+ t2= 8.33 s. The distances traveled during t1 and t2 are the same so that they total to 2(10.59) = 21.18 m. This implies that for a distance of 190− 21.18 = 168.82 m, the elevator is traveling at constant velocity. This time of constant velocity motion is
t3=
168.82 m
5.08 m/s = 33.21 s . Therefore, the total time is 8.33 + 33.21≈ 41.5 s.
40. Neglect of air resistance justifies setting a =−g = −9.8 m/s2 (where down is our
−y direction) for the duration of the fall. This is constant acceleration motion, and we may use Table 2-1 (with ∆y replacing ∆x).
(a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward), we have v =−
q v2
0− 2g∆y = −p0 − 2(9.8)(−1700) = −183 in SI units. Its magnitude is therefore 183 m/s.
(b) No, but it is hard to make a convincing case without more analysis. We estimate the mass of a raindrop to be about a gram or less, so that its mass and speed (from part (a)) would be less than that of a typical bullet, which is good news. But the fact that one is dealing with many raindrops leads us to suspect that this scenario poses an unhealthy situation. If we factor in air resistance, the final speed is smaller, of course, and we return to the relatively healthy situation with which we are familiar.
41. We neglect air resistance, which justifies setting a =−g = −9.8 m/s2(taking down as the
−y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1 (with ∆y replacing ∆x).
(a) Starting the clock at the moment the wrench is dropped (vo= 0), then v2= vo2− 2g∆y leads to ∆y =−(−24)
2
2(9.8) =−29.4 m so that it fell through a height of 29.4 m.
(b) Solving v = v0− gt for time, we find: t = v0− v
g =
0− (−24)
9.8 = 2.45 s .
(c) SI units are used in the graphs, and the initial position is taken as the coordinate origin. In the interest of saving space, we do not show the acceleration graph, which is a horizontal line at −9.8 m/s2. ... 1 2 3 t 0 −10 −20 −30 y . . ... 1 2 3 t 0 −10 −20 −30 v
42. We neglect air resistance, which justifies setting a =−g = −9.8 m/s2(taking down as the
−y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1 (with ∆y replacing ∆x).
(a) Noting that ∆y = y− y0=−30 m, we apply Eq. 2-15 and the quadratic formula (Appendix E) to compute t: ∆y = v0t−1 2gt 2 = ⇒ t = v0±pv 2 0− 2g∆y g
which (with v0 =−12 m/s since it is downward) leads, upon choosing the positive root (so that t > 0), to the result:
t = −12 +p(−12)
2− 2(9.8)(−30)
9.8 = 1.54 s .
(b) Enough information is now known that any of the equations in Table 2-1 can be used to obtain v; however, the one equation that does not use our result from part (a) is Eq. 2-16:
v =qv2
0− 2g∆y = 27.1 m/s
where the positive root has been chosen in order to give speed (which is the magnitude of the velocity vector).
43. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a = −g = −9.8 m/s2 (we take downward to be the
−y direction). We use the equations in Table 2-1 (with ∆y replacing ∆x) because this is a = constant motion.
(a) At the highest point the velocity of the ball vanishes. Taking y0= 0, we set v = 0 in v2= v20− 2gy, and solve for the initial velocity: v0=√2gy. Since y = 50 m we find v0= 31 m/s.
(b) It will be in the air from the time it leaves the ground until the time it returns to the ground (y = 0). Applying Eq. 2-15 to the entire motion (the rise and the fall, of total time t > 0) we have
y = v0t−1 2gt
2 =
⇒ t = 2vg0
which (using our result from part (a)) produces t = 6.4 s. It is possible to obtain this without using part (a)’s result; one can find the time just for the rise (from ground to highest point) from Eq. 2-16 and then double it.
(c) SI units are understood in the x and v graphs shown. In the interest of saving space, we do not show the graph of a, which is a horizontal line at −9.8 m/s2.
. ... 2 4 6 8 t 0 20 40 60 y ... 2 4 6 8 t 0 −20 20 40 v
44. There is no air resistance, which makes it quite accurate to set a =−g = −9.8 m/s2 (where downward is the−y direction) for the duration of the fall. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is constant acceleration motion; in fact, when the acceleration changes (during the process of catching the ball) we will again assume constant acceleration conditions; in this case, we have a2= +25g = 245 m/s2.
(a) The time of fall is given by Eq. 2-15 with v0= 0 and y = 0. Thus, t =r 2y0
g = r
2(145)
9.8 = 5.44 s .
(b) The final velocity for its free-fall (which becomes the initial velocity during the catching process) is found from Eq. 2-16 (other equations can be used but they would use the result from part (a)).
v =− q
v2
0− 2g (y − y0) =−p2gy0=−53.3 m/s where the negative root is chosen since this is a downward velocity.
(c) For the catching process, the answer to part (b) plays the role of an initial velocity (v0=−53.3 m/s) and the final velocity must become zero. Using Eq. 2-16, we find
∆y2= v2− v2 0 2a2 =−(−53.3) 2 2(245) =−5.80 m
where the negative value of ∆y2 signifies that the distance traveled while arresting its motion is downward.
45. Taking the +y direction downward and y0 = 0, we have y = v0t + 12gt2 which (with v0 = 0) yields t =p2y/g.
(a) For this part of the motion, y = 50 m so that t =
r 2(50)
9.8 = 3.2 s .
(b) For this next part of the motion, we note that the total displacement is y = 100 m. Therefore, the total time is
t = r
2(100)
9.8 = 4.5 s .
The difference between this and the answer to part (a) is the time required to fall through that second 50 m distance: 4.5− 3.2 = 1.3 s.
46. We neglect air resistance, which justifies setting a =−g = −9.8 m/s2(taking down as the−y direction) for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is constant acceleration motion. The ground level is taken to correspond to y = 0.
(a) With y0= h and v0 replaced with−v0, Eq. 2-16 leads to v = q (−v0)2− 2g (y − y0) = q v2 0+ 2gh .
The positive root is taken because the problem asks for the speed (the magnitude of the velocity). (b) We use the quadratic formula to solve Eq. 2-15 for t, with v0 replaced with−v0,
∆y =−v0t−1 2gt 2 = ⇒ t = −v0+ q (−v0)2− 2g∆y g
where the positive root is chosen to yield t > 0. With y = 0 and y0= h, this becomes t = pv
2
0+ 2gh− v0
g .
(c) If it were thrown upward with that speed from height h then (in the absence of air friction) it would return to height h with that same downward speed and would therefore yield the same final speed (before hitting the ground) as in part (a). An important perspective related to this is treated later in the book (in the context of energy conservation) .
(d) Having to travel up before it starts its descent certainly requires more time than in part (b). The calculation is quite similar, however, except for now having +v0 in the equation where we had put in−v0 in part (b). The details follow:
∆y = v0t− 1 2gt 2 = ⇒ t = v0+pv 2 0− 2g∆y g
with the positive root again chosen to yield t > 0. With y = 0 and y0= h, we obtain t = pv
2
0+ 2gh + v0
g .
47. We neglect air resistance, which justifies setting a =−g = −9.8 m/s2(taking down as the
−y direction) for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y axis.
(a) Using y = v0t−12gt2, with y = 0.544 m and t = 0.200 s, we find v0=
y +12gt2
t =
0.544 +12(9.8)(0.200)2
(b) The velocity at y = 0.544 m is
v = v0− gt = 3.70 − (9.8)(0.200) = 1.74 m/s . (c) Using v2 = v2
0− 2gy (with different values for y and v than before), we solve for the value of y corresponding to maximum height (where v = 0).
y = v 2 0 2g = 3.72 2(9.8) = 0.698 m . Thus, the armadillo goes 0.698− 0.544 = 0.154 m higher.
48. We neglect air resistance, which justifies setting a =−g = −9.8 m/s2(taking down as the −y direction) for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. The total time of fall can be computed from Eq. 2-15 (using the quadratic formula).
∆y = v0t− 1 2gt 2 = ⇒ t = v0+pv 2 0− 2g∆y g
with the positive root chosen. With y = 0, v0= 0 and y0= h = 60 m, we obtain t = √2gh g = s 2h g = 3.5 s .
Thus, “1.2 s earlier” means we are examining where the rock is at t = 2.3 s: y− h = v0(2.3)−
1 2g(2.3)
2 =
⇒ y = 34 m where we again use the fact that h = 60 m and v0= 0.
49. The speed of the boat is constant, given by vb= d/t. Here, d is the distance of the boat from the bridge when the key is dropped (12 m) and t is the time the key takes in falling. To calculate t, we put the origin of the coordinate system at the point where the key is dropped and take the y axis to be positive in the downward direction. Taking the time to be zero at the instant the key is dropped, we compute the time t when y = 45 m. Since the initial velocity of the key is zero, the coordinate of the key is given by y = 12gt2. Thus t =r 2y g = s 2(45 m) 9.8 m/s2 = 3.03 s . Therefore, the speed of the boat is
vb= 12 m
3.03 s = 4.0 m/s .
50. With+y upward, we have y0= 36.6 m and y = 12.2 m. Therefore, using Eq. 2-18 (the last equation in Table 2-1), we find y− y0= vt + 1 2gt 2 = ⇒ v = −22 m/s
at t = 2.00 s. The term speed refers to the magnitude of the velocity vector, so the answer is |v| = 22.0 m/s.
51. We first find the velocity of the ball just before it hits the ground. During contact with the ground its average acceleration is given by
aavg= ∆v
where ∆v is the change in its velocity during contact with the ground and ∆t = 20.0× 10−3 s is the duration of contact. Now, to find the velocity just before contact, we put the origin at the point where the ball is dropped (and take +y upward) and take t = 0 to be when it is dropped. The ball strikes the ground at y =−15.0 m. Its velocity there is found from Eq. 2-16: v2=
−2gy. Therefore, v =−p−2gy = −p−2(9.8)(−15.0) = −17.1 m/s
where the negative sign is chosen since the ball is traveling downward at the moment of contact. Con-sequently, the average acceleration during contact with the ground is
aavg= 0− (−17.1)
20.0× 10−3 = 857 m/s 2
.
The fact that the result is positive indicates that this acceleration vector points upward. In a later chapter, this will be directly related to the magnitude and direction of the force exerted by the ground on the ball during the collision.
52. The y axis is arranged so that ground level is y = 0 and +y is upward.
(a) At the point where its fuel gets exhausted, the rocket has reached a height of y′= 1
2at
2= (4.00)(6.00)2
2 = 72.0 m .
From Eq. 2-11, the speed of the rocket (which had started at rest) at this instant is v′= at = (4.00)(6.00) = 24.0 m/s .
The additional height ∆y1 the rocket can attain (beyond y′) is given by Eq. 2-16 with vanishing final speed: 0 = v′ 2− 2g∆y1. This gives
∆y1= v′ 2
2g = (24.0)2
2(9.8) = 29.4 m .
Recalling our value for y′, the total height the rocket attains is seen to be 72.0 + 29.4 = 101 m. (b) The time of free-fall flight (from y′ until it returns to y = 0) after the fuel gets exhausted is found
from Eq. 2-15:
−y′= v′t−1 2gt
2 =
⇒ −72.0 = (24.0)t −9.802 t2.
Solving for t (using the quadratic formula) we obtain t = 7.00 s. Recalling the upward acceleration time used in part (a), we see the total time of flight is 7.00 + 6.00 = 13.0 s.
53. The average acceleration during contact with the floor is given by aavg= (v2− v1)/∆t, where v1 is its velocity just before striking the floor, v2 is its velocity just as it leaves the floor, and ∆t is the duration of contact with the floor (12× 10−3s). Taking the y axis to be positively upward and placing the origin at the point where the ball is dropped, we first find the velocity just before striking the floor, using v2
1 = v20− 2gy. With v0= 0 and y =−4.00 m, the result is
v1=−p−2gy = −p−2(9.8)(−4.00) = −8.85 m/s
where the negative root is chosen because the ball is traveling downward. To find the velocity just after hitting the floor (as it ascends without air friction to a height of 2.00 m), we use v2= v2
2−2g(y −y0) with v = 0, y =−2.00 m (it ends up two meters below its initial drop height), and y0=−4.00 m. Therefore,
Consequently, the average acceleration is aavg=v2− v1 ∆t = 6.26 + 8.85 12.0× 10−3 = 1.26× 10 3m/s2 .
The positive nature of the result indicates that the acceleration vector points upward. In a later chapter, this will be directly related to the magnitude and direction of the force exerted by the ground on the ball during the collision.
54. The height reached by the player is y = 0.76 m (where we have taken the origin of the y axis at the floor and +y to be upward).
(a) The initial velocity v0 of the player is
v0=p2gy =p2(9.8)(0.76) = 3.86 m/s .
This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y1= 0.76−0.15 = 0.61 m, his speed v1 satisfies v02− v12= 2gy1, which yields
v1= q
v2
0− 2gy1=p(3.86)2− 2(9.80)(0.61) = 1.71 m/s .
The time t1 that the player spends ascending in the top ∆y1 = 0.15 m of the jump can now be found from Eq. 2-17:
∆y1= 1
2(v1+ v) t1 =⇒ t1=
2(0.15)
1.71 + 0 = 0.175 s
which means that the total time spend in that top 15 cm (both ascending and descending) is 2(0.17) = 0.35 s = 350 ms.
(b) The time t2when the player reaches a height of 0.15 m is found from Eq. 2-15: 0.15 = v0t2−1 2gt 2 2= (3.86)t2−9.8 2 t 2 2 ,
which yields (using the quadratic formula, taking the smaller of the two positive roots) t2= 0.041 s = 41 ms, which implies that the total time spend in that bottom 15 cm (both ascending and descending) is 2(41) = 82 ms.
55. We neglect air resistance, which justifies setting a =−g = −9.8 m/s2(taking down as the
−y direction) for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. The time drop 1 leaves the nozzle is taken as t = 0 and its time of landing on the floor t1 can be computed from Eq. 2-15, with v0= 0 and y1=−2.00 m.
y1=− 1 2gt 2 1 =⇒ t1=r −2y g = r −2(−2.00) 9.8 = 0.639 s .
At that moment,the fourth drop begins to fall, and from the regularity of the dripping we conclude that drop 2 leaves the nozzle at t = 0.639/3 = 0.213 s and drop 3 leaves the nozzle at t = 2(0.213) = 0.426 s. Therefore, the time in free fall (up to the moment drop 1 lands) for drop 2 is t2= t1− 0.213 = 0.426 s and the time in free fall (up to the moment drop 1 lands) for drop 3 is t3= t1− 0.426 = 0.213 s. Their positions at that moment are
y2 = −1 2gt 2 2=− 1 2(9.8)(0.426) 2= −0.889 m y3 = −1 2gt 2 3=− 1 2(9.8)(0.213) 2= −0.222 m ,
respectively. Thus, drop 2 is 89 cm below the nozzle and drop 3 is 22 cm below the nozzle when drop 1 strikes the floor.
