Física para Ingeniería y Ciencias 3ra Edicion Hans Ohanian.pdf

LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN

DESCARGA DIRECTA

SIGUENOS EN:

VISITANOS PARA DESARGALOS GRATIS.

1CHAPTER 1 SPACE, TIME, AND MASS

Select odd-numbered solutions, marked with a dagger (), appear in the Student SolutionsManual, available for purchase. Answers to all solutions below are underscored.

1-1. Assume a height of 5 ft 10 in. Then, 510 = 70 in = 70 in 2.54 cm/in = 178 cm.1-2. There are approximately 300 actual pages in this text (vol. 1). The thickness is 2.5 cm. Therefore

each page = 2.5 cm/300 pg 38.3 10 cm.=

1-3. 100 yd 0.914 m/yd = 91.4 m; 53 1/3 yd 0.914 m/yd 48.7 m=

1-4. 31 step 1000 m steps 1.7 100.60 m 1 km km

N = =

1-5. 1 pica11 in 66 picas.1 in6

L = = 17 1 picain 51 picas12 in6

W = =

1-6. Virus: 2 108 m 10/0.3048 ft/m 12 in/ft 78 10 in=

Similarly: Atom: 1 1010 m 39.4 in/m 94 10 in=

Fe Nucleus: 8 1015 m 39.4 in/m 133 10 in=

Proton: 2 1015 m 39.4 in/m 148 10 in=

1-7. Lets convert 1/2 inch to mm using conversion factors, then use proportional reasoning to do theothers until the number of significant figures becomes large.1 in 25.4 mm/in 12.7 mm2

=

1 12.7 mm in 6.35 mm4 2

= =

1 6.35 mm in 3.175 mm 3.18 mm 8 2

= = = (to three significant figures)

1 3.175 mm in 1.5875 mm 1.59 mm 16 2

= = = (to three significant figures)

The number of digits is becoming large, so lets do direct conversions for the rest of the problems.1 in 25.4 mm/in 0.794 mm32

=

1 in 25.4 mm/in 0.397 mm64

=

1-8.3 2 610 in 2.54 cm 10 m 10 m1 mil 25.4 m.

mil in cm m

=

1000 m 1 mil1 mm 39.4 milmm 25.4 m

=

http://librosysolucionarios.net

CHAPTER 1

2

1-9. (a) Grapefruit diameter 0.1 mRatio of grapefruit/sun = 0.1 m/(1.4 109 m) = 7 1011

Earth diameter 13 106 mComparative size of Earth = 13 106 m (7 1011)= 9 104 m 1 mm.Nearest star distance = 4 1016 mComparative distance = 4 1016 m (7 1011) 62.8 10 m=

(b) Head diameter 0.2 mEarth diameter 13 106 mEarth/head ratio 7 107

Size of atom = 1010 mComparative size of atom = 1010 m (7 107) = 7 103 m = 7 mmSize of red blood cell 7.5 106 mComparative size of cell = 7.5 106 m (7 107) 500 m 1/ 2 km =

1-10. Distance to Q1208 + 1011 = 12.4 109 9.47 1015 = 1.17 1026 mDistance on the diagram (PRELUDE, p. 6)

= 26

520

1.17 10 = 7.8 10 m1.5 10

1-11. Size (diameter) of the sun = 2 6.46 108 = 1.4 109 m distance on the diagram (PRELUDE, p.

6) =9

312

1.4 10 10 m 1 mm1.5 10

=

1-12.9

6 1310 m10 633 nm 6.33 10 m.nm

l

= = According to Table 1-1, the diameter of an

atom is about 1 1010 m, so this is 6.33 103 times the diameter of an atom, or roughly 1/100the diameter of an atom.

1-13. 1 turn = 360, so 5 1 turn/360 = 0.0139 turn. For an English thread,61in 0.0254 m 10 m0.0139 turn = 4.41 m.

80 turns in m For a metric thread,

3 60.5 mm 10 m 10 m0.0139 turn = 6.94 m.turn mm m

1-14. 1 nmi = 1852 m; Circumference of Earth = 4.00 107 mCircumference = (4.00 107 m)/1852 m/nmi 21, 600 nmi=Also 360 60 min/deg 21, 600 min, so 1nmi 1min=


CHAPTER 1

3

1-15. For one of the triangles, (R + 1.75 m)2 = R2 + (4700 m)2. Expandthis to get 2 2 2 22(1.75 m) (1.75 m) (4700 m) .R R R+ + = +We expect R to be much larger than 1.75 m, so we can ignore(1.75 m)2 relative to all the other terms. The R2 terms cancel,leaving(3.50 m)R = (4700 m)2, which gives R = 6.3 106 m.

1-16. 22 yr, 5 mo, 23 days = (8035 + 153 + 23) 8211 days=

(This excludes leap years and assumes average 30.5-day month.)1 day = 1 day 24 h/day 60 min/h 60 s/min = 86,400 s8211 days = 8211 days 86,400 s/day 87.1 10 s=

1-17. 1 yr = 365.25 days. Therefore, 4.5 109 yr= 4.5 109 yr 365.25 day/yr 86,400 s/day 171.4 10 s=

1-18. 12960 min 60s 1 calculation1 h 3.6 10 calculations/h.

h min 10 sN

= =

1-19. 3600s 60s2 h 9 min 21s 2 h 9 min 21s 7761 sh min

= + + =

1-20. 3600s 60s2 h 24 min 51s 2 h 24 min 51s 8692 sh min

= + + =

1-21. 365.25 solar days/year 24 h1 sidereal day 366.25 sidereal days/year solar day

= 23.934 h/sidereal day. Using 1 h = 60

min to convert the 0.934 h to minutes gives 1 sidereal day = 23 h 56 min.

1-22. 7 9s4 ticks 3.2 10 10 years 1.2 10 ticksyear

N = =

1-23. 6 1h 1day10 s 11.6 days3600s 24h

=

1-24. 7 days 24 h1week week day

= 168 h. 53600 s168 h 6.048 10 shr

=

1-25.7

7beats 1 min 3.2 10 s71 ticks 3.8 10 beats/yearmin 60 s yr

N = =

1-26. (a) June 2425: (20 4) s/day = 16 s/24 h 0.67 s/h=

June 2526: (34 20) s/day = 14 s/24 h 0.58 s/h=

June 2627: (51 34) s/day = 17 s/24 h 0.71 s/h=

(b) Average rate = (51 4)/3 day = 47 s/(24 3)h 0.65 s/h=

R

R + 1.75 m

9400


CHAPTER 1

4

(c) 10h30m on June 30 is 70.5 h after noon June 27. By average loss, watch should have lost 70.5h 0.65 s/h = 46 s. Combined with loss of 51 s on June 27 gives total loss of 97 s. Therefore thecorrect WWV time is h m s10 31 37 . With the largest rate, loss is 70.5 h 0.71 s/h = 50 s. Combinedloss is 51 + 50 = 101 s. Then estimated WWV time is h m s10 31 41 . The wristwatch can be trusted toabout 4 s on June 30.

1-27. 1 day = 24 hr = 86,400 s. The earth rotates 360 per day, which corresponds to a rotation rate of360 60 min 0.250 min/s.

86, 400 s degree

= A timing error of 1 s will result in an angular error of 0.250

min. According to Problem 1-14, 1 min = 1852 m, so the corresponding error in position is 0.250min/s 1852 m = 463 m = 0.463 km.

1-28. 140 lb-mass = 140 lb-mass 0.454 kg/lb-mass = 64 kg140 lb-mass = 140 lb-mass 0.454 kg/lb-mass 1/14.6 slug/kg 4.35slug=

140 lb-mass = 140 lb-mass 0.454 kg/lb-mass 1 amu/(1.67 1027 kg) 283.8 10 amu=

1-29. 24 27(0.33 4.9 5.98 0.64 1900 553 87.3 10.8 0.66) 10 kg 2.56 10 kgplanetsm = + + + + + + + + =

(to three significant figures). msun = 1.99 1030 kg, so the total mass is30 27 301.99 10 kg 2.56 10 kg 1.99 10 kgtotal sun planetsm m m= + = + = (to three significant

figures). The fraction of the total mass included in the planets is27

30

2.56 10 100% 100% 0.134%.1.99 10

planets

total

mm

= = The fraction of the mass in the sun is 100%

0.134% = 99.9%.

1-30.26

15

largest length 1 10 msmallest length 2 10 m

= = 5 1040. 17

24

longest time 4 10 sshortest time 1 10 s

= = 4 1041.

53

31

largest mass 1 10 kgsmallest mass 9 10 kg

= = 1 1083. The first two ratios are within an order of magnitude of

each other, and the third is roughly equal to the square of the other two. (Note that the first tworatios will keep increasing because the universe is expanding and aging.)

1-31. From the periodic table in the Appendix, we see that the uranium nucleus contains about 238nucleons, each with the mass of a proton. Table 1.7 gives 1.7 1027 kg for the mass of a proton.Then the total mass of the electrons is (92)( 9.1 1031 kg) = 8.4 1029 kg, and the total mass ofthe nucleus is (238)( 1.7 1027 kg) = 4.0 1025 kg. To two significant figures, the total mass ofthe atom is 4.0 1025 kg. The fraction of the total mass in the electrons is 8.4 1029/4.0 1025

= 2.1 104 = 0.021%. The fraction of mass in the nucleus is 99.98%.1-32. Let m represent the mass of material and M represent the mass of one mole. Then

623 14atoms 0.1 10 g6.02204 10 3 10 atoms.

mol 197 g/molAmN NM

= = =

1-33. 1 lb avoirdupois = 0.453 59 kg = 435.59 g (given in the text)0.453 59 kg1 lb troy 0.822 86 lb avoidupois

lb avoidupois= = 0.373 24 kg = 373.24 g

1-34.21

23 4atoms 10 kg6.02204 10 1.1 10 atomsmol 0.055 85 kg/molA

mN NM

= = =


CHAPTER 1

5

1-35. (a) Since the density of water is 1 g/cm3, 250 cm3 of water has a mass of 250 g, which is 250g/(18 g/mol) = 14 mol. Therefore, the number of molecules is 14 mol 6.02 1023

molecules/mol 248.4 10 molecules.=

(b) Mass of sea-water (density = 1030 kg/m3)is 1.3 1018 m3 1030 kg/m3 = 1.3 1021 kg = 1.3 1024 gNumber of molecules of sea-water is1.3 1024 g 1 mol/18 g 6.02 1023 molecules/mol 46 4.3 10 molecules.=

(c) Ratio of molecules of water from cup to molecules from sea is8.4 1024/4.3 1046 222.0 10 .=

The probability of a single molecule drawn from the ocean being originally from the cup istherefore 2.0 1022. Drawing a cup full then should result in 2.0 1022 8.4 1024

1680 molecules.=

1-36. The mass of H2 and He, respectively, is 0.70 1.99 1030 kg = 1.39 1030 kg and 0.30 1.99 1030 kg = 0.60 1030 kg. The atomic mass of hydrogen and helium, respectively, is 1 g/mol and 4g/mol. Therefore, the number of hydrogen atoms isNH = 1.39 1033 g 1 mol/g 6.02 1023 molecules/mol 56 8.38 10 molecules.=

The number of helium atoms is NHe = 0.60 1033 g 1 mol/4 g 6.02 1023 molecules/mol55

9.03 10 atoms.=

Total number 569.28 10=

1-37. Molecular mass of N2 = 28 g/molMolecular mass of O2 = 32 g/molMolecular mass of Ar = 40 g/molTherefore, 1000 g of air will contain:755 g N2 = 755 g/(28 g/mol) = 27.0 mol232 g O2 = (232/32) mol = 7.25 mol13 g Ar = (13/40) mol = 0.325 molThe percentage by number of molecules of these substances is:N2: 27.0/(27.0 + 7.25 + 0.325) 100% = 27.0/34.575 100% = 78.1%O2: (7.25/34.575) 100% = 21%Ar: (0.325/34.575) 100% = 0.9%Therefore, the molecular mass of air is(0.781 28) + (0.21 32) + (0.009 40) 28.95g/mol.=

1-38. The mass of elements is:Oxygen: 0.65 73 kg = 47.4 kg; atomic mass = 16Carbon: 0.185 73 kg = 13.5 kg; atomic mass = 12Hydrogen: 0.095 73 kg = 6.94 kg; atomic mass = 1.008Nitrogen: 0.033 73 kg = 2.41 kg; atomic mass = 14Calcium: 0.015 73 kg = 1.09 kg; atomic mass = 40.08Phosphorous: 0.01 73 kg = 0.73 kg; atomic mass = 31


CHAPTER 1

6

The number of atoms of these substances is:O: 47.4 103 g 1/16 mol/g 6.02 1023/mol = 1.78 1027

C: 13.5 103 g 1/12 mol/g 6.02 1023/mol = 6.77 1026

H: 6.94 103 g 1 mol/g 6.02 1023/mol = 4.17 1027

N: 2.41 103 g 1/14 mol/g 6.02 1023/mol = 1.04 1026

Ca: 1.09 103 g 1/40 mol/g 6.02 1023/mol = 1.64 1025

P: 0.73 103 g 1/31 mol/g 6.02 1023/mol = 1.42 1025

Total number of atoms: = 6.76 1027

1-39. 0.53 = 0.53/360 2 rad = 9.25 103 radd = 9.25 103 rad 1.5 1011 m

= 1.4 109 m86.9 10 mr =

1-40. 8 15m days h s1 ly 3.00 10 365.25 24 3600 9.47 10 ms year day h

= =

1-41. 1 ly = 9.47 1015 m2.2 106 ly = 2.2 106 ly 9.47 1015 m/ly 222.1 10 m=

1-42. 1 light-s (ls) = 3.00 108 m/s 1 s = 3.00 108 m1 light-min (lm) = 3.00 108 m/s 60 s = 1.80 1010 mE-S distance = 1.50 1011 m 1/(1.80 1010) lm/m 8.3 lm=E-M distance = 3.84 108 m 1/(3.00 108) ls/m 1.28 ls=

1-43. 1 second of arc = 13600

degree = 2 1360 3600

rad = 4.85 106 rad

(a) 61 AU 4.85 10pc

=

1 pc = 1 1/(4.85 106)AU 5 2.06 10 AU=

(b) 1 pc = 2.06 105 AU 1.496 1011 m/AU = 3.08 1016 m1 ly = 9.47 1015 m1 pc = 3.08 1016 m 1 ly/(9.47 1015 m) 3.25 ly=

(c) 1 ly = 3.00 108 m/s 1 yr = 3.00 108 m/s 3.156 107 s 15 9.47 10 m=

1-44.2

2 1 ft1m 0.3048 m

= 10.76 ft2. Note that the conversion factor must be squared.

1-45.3

3 1 ft1m 0.3048 m

= 35.31 ft3. Note that the conversion factor must be cubed.

1-46. Use the result from 144: 2

2

1 m(78 ft)(27 ft) 10.76 ft

A z= = 196 m2.

1-47. His height is measured to a precision of 0.1 inch, so we want to see how many significant digitsthis implies. 8 feet = 96 inches (exactly), so to the nearest 0.1 inch his height can be expressed as107.1 inches, which contains four significant figures. Converting 11.1 inches to feet gives hisheight in feet: 8 ft + 11.1 in = 8.925 ft, which also contains four significant figures. Thus his

height in meters should be specified to four figures: m8.925 ft 0.3048 ft

= 2.720 m.


CHAPTER 1

7

1-48.2 2

2

3 ft 1 m100 yd 53.33 yd yd 10.76 ft

A

= = 4.46 103 m2, using the result from 144.

1-49.3

3 3 31 kg 100 cm8.9 g/cm 8.9 10 kg/m .1000 g m

= Note that the cm to m conversion factor

must be cubed! 1 ft = 0.3048 m, 1 lb = 0.454 kg.3

3 3 31 lb 0.3048 m8.9 10 kg/m 555 lb/ft ,0.454 kg ft

= or 5.6 10

2 lb/ft3 to two significant

figures. Again, note that the m to ft conversion factor must be cubed. 1 ft = 12 in, so3

3 31 ft555 lb/ft 0.32 lb/in .12 in

= 1-50. Assume a mass of 73 kg and a density of 1000 kg/m3.

V = massdensity

= 73 kg 1/1000 m3/kg 30.073 m=

1-51.3 6 3

3

cm 3600 s 10 m92 24 h s h cm

= 7.9 m3/day.

1-52. 1 liter = 103 cm3, so 5.2 liter = 5.2 103 cm3. 3 3

3

5.2 10 cm 57 s.92 cm /s

t = =

1-53.

4 22

2

212

10 m1 cm cm

m10transistor

N

= = 108 transistors. If theyre stacked, N = the number of transistors per

layer (108) number of layers. If the cube is 1 cm high and each layer is 107 m, or 105 cm,thick, then the cube holds 105 layers and the cube can hold 108 105 = 1013 transistors!

1-54. 1 gal = 3.785 liter. 3

3 33

g kg cm 1 lb literDensity 1.00 10 10 3.875cm g liter 0.4536 kg gal

= = 8.34

lb/gal1-55. (a) (3.6 104) (2.049 102) = (3.6)(2.049) 1042 = 7.4 102

(b) (2.581 102) (7.264 101) = (2.581 0.7264) 102 = 1.855 102

(c) 0.079832 9.43 =2

0

7.9832 109.43 10

= 0.847 1020 = 8.47 103

1-56.30 3

3 6 33 8 3 3

3

3 3(2.0 10 kg) g mDensity 10 10 1.5 g/cm4 4 4 (7.0 10 m) kg cm3

m m mV RR

= = = = =

1-57.30 3

63 3 3 3 3

3

3 3(2.0 10 kg) 1 metric ton mDensity 104 4 4 (20 10 m) 10 kg cm3

m m mV RR

= = = =

= 6.0 107 metric tons/cm3

1-58. Oceans of the earth have 1.3 1018 m3 of water. The mass of the oceans is 1.3 1018 m3 1030kg/m3 = 1.3 1021 kg. Mass of the earth is 5.98 1024 kg, so that the percentage of the mass ofthe earth that is water is:(1.3 1021 kg/5.98 1024 kg) 100% 0.02%.=


CHAPTER 1

8

1-59. From Table 1.10, 1 liter = 103 m3. According to data given in the Conversion of Units sectionof the chapter, the density of water is 1000 kg/m3, so

3 33 310 m 1 min300 liters/min 5.00 10 m /s,

liter 60 s

= and

3 3 35.00 10 m /s 1000 kg/m 5.00 kg/s. =

1-60. 1 in = 1 in 2.54 cm/in 1/100 m/cm = 0.0254 m. Therefore volume on 1 m2 is V = 0.0254 m 1 m2 3 20.0254 m /m .=

Mass of this much water is 0.0254 m3/m2 1000 kg/m3 = 25.4 kg/m2.

1-61. This can be solved using proportional reasoning. 33

3Density ,4 43

m m mV RR

= = = which means

3 3 ,copperlead

lead copper

mmR R

= from which we get 1/ 3 1/ 3

15 3.5(4.8 10 m)1.06

leadlead copper

copper

mR Rm

= =

= 7.1 1015 m. Likewise we get 1/ 3 1/ 3

15 0.27(4.8 10 m)1.06

oxygenoxygen copper

copper

mR R

m

= =

= 3.0 1015 m. Note that it was not necessary to include the factor ( 1025) when expressing themasses because it cancels in the ratio.

1-62. Density is calculated as massvolume

where V = 43r3.

Planet Mass (kg) Vol (m3) Density (kg/m3)Pluto 6.6 1023 ? 1.13 1020 5800 ?Mercury 3.3 1023 5.94 1019 5600Earth 5.98 1024 1.09 1021 5500Venus 4.9 1024 9.51 1020 5100Mars 6.40 1023 1.62 1020 4000Neptune 1.03 1026 4.58 1022 2250Uranus 8.73 1025 5.65 1022 1550Jupiter 1.90 1027 1.52 1024 1250Saturn 5.53 1026 9.23 1023 600

1-63. One dimension of each roof segment must be divided by cos 45, or multiplied by 2. Then totalroof area = floor area 2 = 250 m2 2 = 354 m2.

1-64. Use the radius of the earth from Table 1-1.

6

15 m6.4 10 m

sR

= = = 2.3 106 radian. In

degrees, this is (2.3 106 radian) (180/radian) = 1.3 104 degree.

1-65. The slope is the tangent of the required angle. For a slope of 1:5, 1 1tan 11 .5

= = For a slope of

1:10, 1 1tan 5.7 .10

= = For a slope angle of 0.1, tan = 1.7 103, so the slope is

1:( 1.7 103)1 = 1: 5.7 102, so the rise is about 1 atom per 570 atoms.

R

s = 15 m


CHAPTER 1

9

1-66. tan ,hD

= where D = 75 m. Both D and are

specified to two significant figures, so the valuecalculated for h must be specified to two figures.Thus tan (75 m)( tan 78 )h D = = = 3.5 102 m.

1-67. In this calculation, assume that the calendar year is exactly 365 days long and a circle containsexactly 360, so these are not interpreted as numbers with only three significant figures. The

angle the earth moves through in one calendar year is 365 360365.24

= = 359.76. In four

years, including one leap year with exactly 366 days, the total number of days is 3 365 + 366 =

1461 days. The angle the earth moves through in four years is 1461 360365.24

= = 1440.0,

which is exactly four complete circles to five significant figures. (The angle is actually a littlelarger than 1440.0, which is why there are some four-year intervals that do not include an extraday.)

1-68. The sun will set at Marchena n-minutes after 8 PM where

n = minutes in a day distance between islandscircumference of the earth

= 3

7

24 60 60 104 10

= 2.15 min

Therefore, the sun sets at Marchena at 2.15 min after 8 PM.1-69. The diameter, d of the tree trunk is related to its length, L by

d = AL3/2, where A is a constant7.6 = A(81)3/2

if L = 90 m, then

d = A(90)3/2 = 3 / 290 7.6 = 8.9 m.

81

The mass, m of the tree trunk is related to d and L by

m = 2

2d L, where is the mass density.

6100 = (7.6)4

2 81

for L = 90 m,

m = 2 (8.9)4

90 = 6100 28.9

7.6

90 = 9295 tons81

1-70. Distance from pole to equator = 1/4 circumference. Therefore

d = 1 (24

r) = 2r =

2 6.37 106 m 7 1.00 10 m.=

Straight line distance is, using the Pythagorean theorem:d = 2 2 2 6 6 2 2 6.37 10 m 9.0 10 m.r r r+ = = =

h

D


CHAPTER 1

10

1-71.23

235 g/molatoms6.02204 10mol

atomA

MmN

= = = 3.902 1022 g = 3.902 1025 kg. In atomic mass units,

this is 25

27

3.902 10 kgkg1.66054 10u

atomm

= = 235.0 u.

1-72. The molar mass M of water (H2O) is 18.0 g/mol. The mass m of 1 liter (1000 cm3) of water is

1000 g. 231000 g molecules= = 6.02204 1018.0 g/mol molA

mN NM

= 3.35 1025 molecules. Each

molecule contains one oxygen atom and two hydrogen atoms, so there are 3.35 1025 oxygenatoms and 6.70 1025 hydrogen atoms.

1-73. Molecular mass of N2 is 2 14 g/mol = 28 g/mol. The density of air is 1.3 kg/m3. The number ofmolecules in 1 cm3 of air is then number of molecules = 106 m3 1.3 kg/m3 1000 g/kg 1mol/28 g 6.02 1023 molecules/mol 192.8 10 molecules.=

1-74.3 3

6 33

cells mm m5.1 10 1000 10 5.2 litersmm m liter

N = = 2.7 1013 cells.

1-75. Volume of paint = area thickness.3

3 2volume mThickness (1 liter) 10 (8 m )area liter

= =

= 1.25 104 m (0.125 mm)

1-76. The area is 2

6 2 3 12 2m9.4 10 km 10 9.4 10 m .km

=

6 3 12 2 4 2mass kgArea density 8 10 metric tons 10 (9.4 10 m ) 9 10 kg/marea metric ton

= = =

1-77. 55 mi/h = 55 mi/h 1.609 km/mi 88.5 km/h== 55 mi/h 5280 ft/mi 1/3600 h/s 80.7 ft/s== 88.5 km/h 1000 m/km 1/3600 h/s 24.6 m/s=

1-78.26

3 15 33

3 3(9.5 10 kg)Density 4 4 4 (4.6 10 m)3

m m mV RR

= = = = = 2.33 1017 kg/m3. In metric tons per

cm3 this is 2

173 3 2

kg 1 metric ton 1 m2.33 10 m 10 kg 10 cm

= 2.3 108 metric tons/cm3.

1-79. 120 yr 365 days/yr + 237 days = 44067 days. 9h 3600 s44067 days 24 3.81 10 sday h

=

1-80. 1slope 300 m 300 m = 33.3 m.9

h = =

2 2(33.3 m) (300 m) 302 m.d = + =h d

300 m


CHAPTER 1

11

1-81. The distance traveled by the plane gliding:x = 5 103/tan 15= 18.7 103 m 18.7 km.=

Therefore the pilot can reach San Francisco.

1-82. = cos1 RRth

= cos1 6

6 3

6.4 106.4 10 2.3 10

+

= 1 = 0.027 rad = R = 6.4 106 0.027= 1.71 105 m= 171 km

1-83. The arc AB has a length of 3900 km, so the half angle is 1950 km/RE, where RE is the radius of the earth.Using the value from Table 1-1 gives = (1950 km)/(6.4 103 km) = 0.305 radian = 17.5.(a)The linear distance d from A to B is

32 sin 2(6.4 10 km)(sin17.5 )Ed R = = = 3840km.(b)The depth h at the midpoint is

cos (1 cos ),E E Eh R R R = = which gives3(6.4 10 km)(1 cos17.5 )h = = 296 km.

(c) Horizontal means parallel to the surface of the earth, or tangent to the surface of the earth atthat location. The tangent line is shown in the diagram, and the slope angle between the tangentand the line AB is shown. Since the radius OA is perpendicular to the tangent line and the radiusOC is perpendicular to the line AB, the angle must be the same as . Thus = 17.5 and theslope is the tangent of that angle: slope = tan 17.5 = 0.315. Using the ratio form of slope givesslope = 1:(0.315)1 = 1:3.2. (This is a quite steep slope!) Note that to observers standing at eachend, the tunnel appears to be sloping down into the earth.

RE

REd/2

A

B

h

O

C


12

CHAPTER 2 MOTION ALONG A STRAIGHT LINE


2-1. Time required, t = distancespeed

= 30 0.3 s100

=

2-2. Avg speed = distancetime

= 100 yd9.0 s

1 mi1760 yd

3600s 23 mi/h1h

=

2-3. 1 year = 3.156 107 sec, so 20 m/year = 7720 m 6.34 10

1 year 3.156 10 s/year

= m/s (6.3

107 m/s to two significant figures). 1 day = 24 hr = 86,400 s. In cm/day the rate is7 46.34 10 m/s 8.64 10 s/day 100 cm/m = 5.4 cm/day.

2-4. Assume the butterflys speed is 0.5 m/sec. Then the travel time is

t = 33500 10 m 1 81 days.

0.5 m/s 24 hr/day 3600 s/hrdv

=

2-5. 6 days 12 hrs = 156 hrs.dist. 5068 32.5 km/htime 156

= = =

2-6.9

7 15 7

1.4 10 ly2.16 10 m/s 9.47 10 m/ly 1/(3.16 10 )yr/s

dt

= =

10= 1.9 10 yrt

2-7. Estimated distance (by sea) between Java and England is 20,000 km.20,000 km 600 km/h

32 hdt

= =

2-8. (a) 4000 nmiAverage speed 8.3 nmi/hr.20 days 24 hr/day

= = (b) He must cover the remaining 1720

nmi in 7 days, which requires an average speed of 1720 nmi 10.2 nmi/hr.7 days 24 hr/day

= This is

about the same as his maximum possible speed. Since its unlikely that he can maintain thehighest possible speed for the entire 7 days, he should probably conclude that he will not be ableto complete the trip within the 20-day limit.

2-9. 35 kmAverage speed 14 km/hr2.5 hr

dt

= = =

2-10.3110 km 110 10 mAverage speed 1.27 m/s.

1 day 24 h 3600 sec= = = Burst speed = 32 km/hour =

332 10 m 8.9 m/s3600 s

=

2-11. 41 21 km 1000 m 2.5 10 years.

4 cm/year 4 10 m/yeardt tv

= = = = Moving 1000 km will take1000 times as long, or t2 = 2.5 107 years.


CHAPTER 2

13

2-12. 402 mAverage speed 16.9 m/s23.8 s

dt

= = =

2-13. 500 mAverage speed 12.8 m/s,39.10 s

= = or 12.8 103 km/m 3600 s/hr = 46.0 km/hr

2-14. 23.8 m 0.326 s(263 km/hour 1000 m/km / 3600 s/hr)

dtv

= = =

2-15. Use the formula: .dtv

= 5280 km900 km/hrair

t = = 5.87 hr. 5280 km35 km/hourshiip

t = = 151 hr.

2-16. 1100 m 9.49 m/s.10.54 s

v = = 2200 m 9.37 m/s21.34 s

v = =

2-17. 16 m/s (s)100

d v t= = =i 0.06 m2-18. Time taken for the arrow to reach the deer is

50 m 10 s 0.77 s.65 m/s 13

dtv

= = = =

In this time the deer traveled from 40 m to 50 m, i.e., 10 m. Thus10 m 13 m/s.

(10/13s)dvt

= = =

2-19. (a) Take x = 0 to be the cheetahs starting position. Then the cheetahs position is given by.c cx v t= The antelopes starting position is 50 m from the cheetahs starting position, so the

position of the antelope is given by 50.a ax v t= + When the cheetah catches the antelope, theirpositions are the same, and we get 50.c av t v t= + The speeds are c = 101 km/h = 28.1 m/s and

va = 88 km/h = 24.4 m/s. Solving the equation for t gives 50 m 50 m

28.1 m/s 24.4 m/sc at

v v= =

=

13.8 s, or 14 s to two significant figures. During this time, the cheetah travels (28.1 m/s)(13.8 s)= 380 m.(b) The cheetah must catch the antelope within 20 s. Call the antelopes initial position x0.We use the same equation that says the cheetah catches the antelope, 0 ,c av t v t x= + but now weset t = 20 s and calculate what head start x0 the antelope needs. We get

0 ( ) (28.1 m/s 24.4 m/s) (20 s) c ax v v t= = = 72 m. If the antelope is farther away than 72 m,the cheetah will not be able to catch it.

2-20. 100 mAverage speed 10.1 m/s9.86 s

dt

= = =

2-21. 3 426 mi 1.6 10 m/mi 385 yd 0.9144 m/yd 4.195 10 md = + =2 hr 24 min 52 s 2 hr 3600 s/hr 24 min 60 s / min 52 s 8692 st = = + + =

44.195 10 maverage speed8692 s

dt

= = = 4.83 m/s

2-22. 2 0.9 cm 0.094 cm/s60 s

secondsecond

second

dvt

= = =

32 0.9 cm 1.6 10 cm/s60 min 60 s/min

minuteminute

minute

dvt

= = =


CHAPTER 2

14

52 0.5 cm 7.3 10 cm/s12 hr 60 min/hr 60 s/min

hourhour

hour

dvt

= = =

2-23. 24.0 0.50 .x t t= To find the maximum value of x, differentiate with respect to t and set the

derivative equal to zero: 4.0 0.dx tdt

= = The result is t = 4.0 s. (This is the point at which the

runner turns around and moves back toward the starting line.) The distance traveled at this time is24.0(4.0) 0.50(4.0) 8.0 m.x = = At t = 8 seconds, x = 24.0(8.0) 0.5(8.0) 0; = that is when he

comes back to the starting line. The total distance traveled is 16 m. Then average speed =distance 16 m

time 8 s= = 2.0 m/s.

2-24. Distance = 100 km time = 50 km 50 km 1.4660 km/hour 80 km/hr

+ = hr average speed

distance 100 km 69 km/hr.time 1.46 hour

= = = The average speed is not exactly 70 km/hr because the car

moves at 80 km/hr for a shorter period of time than it does at 60 km/hr.

2-25. Planet Orbit circumference (km) Period (s) Speed (km/s) log speed log radiusMercury 3.64 108 7.61 106 47.8 1.68 8.56Venus 6.79 108 1.94 107 35.0 1.54 8.83Earth 9.42 108 3.16 107 29.8 1.47 8.97Mars 1.43 109 5.93 107 24.1 1.38 9.16Jupiter 4.89 109 3.76 108 13.0 1.11 9.69Saturn 8.98 109 9.31 108 9.65 0.985 9.95Uranus 1.80 1010 2.65 109 6.79 0.832 10.26Neptune 2.83 1010 5.21 109 5.43 0.735 10.45Pluto 3.71 1010 7.83 109 4.74 0.676 10.57

The slope of the line through the nine points is 1 .2


CHAPTER 2

15

This means that log speed 12

= log radius + log C.

Therefore log speed = log[C(radius)1/2

]. Thus1/2Speed = C(radius) where Cis some constant.

2-26. Avg speed = distancetime

= 8 8 6.27 m/s2.55

+=

Avg velocity = displacement 0 m/stime

=

2-27. Avg speed (for t = 0 to t = 10 s) = 200 20 m/s10

=

Avg speed (for t = 10 to t = 14.3 s) = 270 200 16.3 m/s14.3 10

=

2-28. Distance = (8 floors + 4 floors + 7 floors) 4 m/floor = 76 m.

Average speed = distance 76 m 1.5 m/s.time 50 s

= = The total change in position is

(12 floors 1 floors) 4 m/floorx = = 44 m. The average velocity is44 m 0.88 m/s.50 s

xvt

= = =

2-29. Distance = (12 blocks + 6 blocks + 3 blocks) 81 m/block = 1701 m.

The elapsed time = 14 min 5 s + 6 min 28 s + 3 min 40 s = 23 min 73 s = 1453 s. Then average

speed = distance 1701 m 1.17 m/s.time 1453 s

= = The total displacement is x = (12 blocks 6 blocks +

3 blocks) 81 m/block = 729 m. The average velocity is 729 m 0.502 m/s.1453 s

xvt

= = =

2-30. 2 2( 0) 0; ( 8) 4 8 0.5 8 0; ( 10) 4 10 0.5 10 10x t x t x t= = = = = = = = m.

Average velocity between t = 0 to t = 8.0 s is: ( 8) ( 0) 08 0

x t x t= ==

;

Average velocity between t = 8.0 s to t = 10.0 s is: ( 10) ( 8) 10 m 0 5.0 m/s.10 8 2 s

x t x ts s

= =

= =

2-31. Total distance = 3 0.25 mile 1609 m/mile 1207 m.= Displacement = 0 m because the horsereturns to the starting point. Total time = 1 min 40 s = 100 s. Then average speed =distance 1207 m 12.1 m/s.

time 100 s= = The average velocity is 0.xv

t

= =

2-32. Total distance = 35 m + 22 m = 57 m. The total time is = 4.5 s + 3.6 s = 8.1 s. The average speed

is distance 57 m 7.0 m/s.time 8.1 s

= = The total displacement is x = 35 m 22 m = 13 m. Then

average velocity is x 13 m 1.6 m/s.t 8.1 s

v = = =


CHAPTER 2

16

2-33. From the graph, the total distance traveled by the squirrel is 6 m + 6 m + 2 m + 6 m = 20 m. The

total elapsed time is 30 s. Then average speed distance 20 m 0.67 m/s.time 30 s

= = = The total

displacement is x = 16 m, so the average velocity is 13 m 0.53 m/s.8.1 s

xvt

= = =

2-34. For 0 2 s,t displacement = 25 m. displacement 25 mtime 2s

v = = = 12.5 m/s. For

2.0 4.0 s,t displacement = 40 m 25 m = 15 m. (This requires an estimate for the position

at 4.0 s. Your value may be slightly different.) displacement 15 m 7.5 m/s.time 2s

v = = =

To find the instantaneous velocity at any time, draw a tangent to the position vs time curve at thattime and determine the slope of the line. Your numbers may be slightly different from the onesgiven here. At 1.0 s, we get a tangent passing through points with coordinates (0.3 s, 0 m) and(2.5 s, 25 m). This gives a slope of 11 m/s. At 3.0 s, the position vs time graph is a straight line,so the instantaneous velocity will be the same as the average velocity between 2.0 s and 4.0 s, or7.5 m/s. Again, your value may be slightly different if you estimate a different position at 4.0 s.

2-35. a = /t = a t = 282.6 gee (9.807 m/s )

gee 0.04 s32.4 m/sv =

2-36. a = /t = [(96 0) km/h]/2.2 s 1 h/3600 s 1000 m/km 212 m/s=

2-37. 3 22 1 32 1

27 m/s 0 3.4 10 m/s .8.0 10 s

v vat t

= = =

2-38. 80 (km/hr) 22.22 m/s,iv = = 0, 2.8 s.fv t= =

20 22.22 m/s 7.94 m/s2.8 s

f iv vat

= = =

2-39. (a) t(s) 2(m/s )a (in gees)a0 6.1 0.62 Method:

10 1.4 0.14 i) Draw tangent to curve.20 0.83 0.085 ii) Get slope of line by counting squares to find and t.30 0.56 0.057 iii) Convert from km/h to m/s.40 0.49 0.050

(b) t(s) 2(m/s )a (gees)a

0 0.74 0.07510 0.44 0.04520 0.44 0.04530 0.31 0.03240 0.22 0.022

2-40. 2 32.5 3.1 4.5 ;x t t t= + 22.5 6.2 13.5 ;dxv t tdt

= = + 6.2 27.0 .dva tdt

= =

At t = 0 s, instantaneous velocity = 20| 2.5 6.2 0 13.5 0 2.5 m/s.tdxdt =

= + = Instantaneous

acceleration = 6.2 m/s2. At t = 2. 0 s, instantaneous velocity =


CHAPTER 2

17

22| 2.5 6.2 2 13.5 2t

dxdt =

= + = 39.1 m/s (39 m/s to two significant figures).

Instantaneous acceleration = 26.2 27.0(2) 48 m/s . =

For 0 2 s,t 2 3( 2 ) ( 0) 2.5(2) 3.1(2) 4.5(2) 9.3 m/s.

2 0 2x t s x tv = = + = = =

( 2) ( 0) 39.1 m/s 2.5 m/s2 0 2 s

v t v ta = = = = =

21 m/s2.

2-41. 2 33.6 2.4 ,x t t= 27.2 7.2 .dxv t tdt

= = 0v = if27.2 7.2t t = 0 t = 0 s or t = 1.0 s. At t = 0, x = 0.

At t = 1.0 s, x = 3.6 2.4 = 1.2 m. To make a sketch,

consider that x = 0 when t = 0 and t = 3.6 1.52.4

= s.

Also, dx/dt = 0 when t = 0 and t = 1s.

2-42. 2.v Bt Ct= V = 0 if t = 0 or t =2

3

6.0 m/s 3.0 s.2.0 m/s

BC

= = 2 .dva B Ctdt

= = a = 0 if

1.5 s.2BtC

= = To make a sketch: consider the

above information, plus when t = 1.5 s, =2 3 2(6.0 m/s ) (1.5 s) (2.0 m/s ) (1.5 s) = 4.5 m/s.

2-43. For 0 5.0 s,t ( 5 s) ( 0) 5 m/s 05 s 5 s

v t v ta = = = = 1 m/s2. (Your value may be slightly

different depending on how you read the values of at 0 and 5 s.) For 5.0 10.0 s,t 2( 10 s) ( 5 s) 9.5 m/s 5.0 m/s 0.9 m/s .

5 s 5 sv t v ta = = = = (Again, your value may be slightly

different depending on how you estimate the values of at 5 and 10 s.) To find the instantaneousacceleration at 3 s, draw a tangent to the curve at that time. Your estimate may be slightlydifferent from ours. We get a tangent line that passes through the points (1 s, 0 m/s) and (5 s, 5

m/s). The slope of this line is the instantaneous acceleration 25 m/s 0 1.3 m/s .5 s 1 s

a = =

2-44. 0/ 2.5 / 2.50[ ( ) ] .(2.5 s)ft s t s

f f

v vdv da v v v e edt dt

= = + = At t = 0,

00

m/s 200 km/hr 18 km/hr|2.5 s 2.5 s

ft

v vdvadt =

= = =

2182 km/hr 1000 m/km 20 m/s .2.5 s 3600 s/hr

= =


CHAPTER 2

18

2-45. 20 0 02 2 2 2 22( ) .

(1 ) (1 ) (1 )v v Av tdv d da At

dt dt At At dt At

= = = = + + + At t = 0, a = 0.

at t = 2 s, 2

222 2

2(25 m/s)(2 s )(2 s) 2.5 m/s .1 (2 s )(2 s)

a

= = + As ,t 02 4

2 , so 0.Av ta aA t

2-46. (a) Estimated average velocity (by mid-point method) is about 30 km/h. Because v = x/t, we

have x = v t 30 km/h 5 s 5 m/s 42 m.18 km/h

(b) Time interval (s) (km/h)v x(m) 5 m / sby km / h 5s 18 km / h

v

510 70 1001015 90 1201520 110 1502025 130 1802530 140 1903035 150 2103540 160 2204045 170 240

(c) Total distance traveled is the sum of the last column plus the 42 m traveled in the first5 s: 1450 m 1.45 km.d = =

2-47. (a)

(b) Time interval (s) Avg speed (m/s) Distance traveled (m)00.3 647.5 194

0.30.6 628.5 1890.60.9 611.5 1830.91.2 596.0 1791.21.5 579.5 1741.51.8 564.0 1691.82.1 549.5 1652.12.4 535.0 1612.42.7 521.0 1562.73.0 508.0 152

Total distance traveled = 1722 m


CHAPTER 2

19

(c) Counting the number of squares under the versus t curve gives the same answer withinabout 2 m.

2-48. = 655.9 61.14t + 3.26t2

acceleration, a = dvdt

= 61.14 + 6.52t

a |t = 0 s 261.14 m/s=

a |t = 1.5 s = 61.14 + (6.52) (1.5) 251.36 m/s=

a |t = 3.0 s = 61.14 + (6.52) (3.0) 241.58 m/s=

2-49. (a)

(b) Calculus method:If x = 0, then cos t = 0, so t = /2 or 3/2 s. Particle crosses x = 0 at 1.6 st = and 4.7 s.

= dx/dt = 2.0 sin t (/2 s) = 2.0 sin(/2) = 2.0 m/s

(3/2 s) = 2.0 sin(3/s) = 2.0 m/s

a = dv/dt = 2.0 cos ta(/2) = 2.0 cos(/2) = 20 m / s

a(3/2) = 2.0 cos(3/2) = 20 m / s(c) Maximum distance achieved when cos t = 1, i.e., when t = 0, , 2, or 0s,3.1s and 6.3s.t =

v = 2.0 sin tv(0) = 2.0 sin(0) = 0 m/ sv() = 2.0 sin() = 0 m/ sv(2) = 2.0 sin(2) = 0 m/ s

a = dv/dt = 2.0 cos ta(0) = 2.0 cos(0) = 22.0 m / s

a() = 2.0 cos() = 22.0 m/ s

a(2) = 2.0 cos(2) = 22.0 m / s

2-50. x = uext + uex(1/b t) ln(1 bt )(a) Instantaneous velocity, v = dx/dtdxdt

= uex + uex ln(1 ) (1/ )1bbt b tbt

+

= uex[1 ln(1 bt) 1]

ln(1 )exv u bt=

(b) a = 2

2

d xdt

= uex[bt(1 bt)] ex1u b

bt=


CHAPTER 2

20

(c) dxdt

= 3.0 103 ln(1 7.5 103t) m/s

(t = 0) = 3.0 103 ln 1 = 0 m/s

(t = 120) = 3.0 103 ln(1 7.5 103 120) m/s(120) = 3.0 103 ln(1 0.9) m/s = 3.0 103 (2.80) m/s

3(120) 6.9 10 m/sv =

(d) a = ddt =

2

2

d xdt

= 3 3

3

3.0 10 7.5 101 7.5 10 t

m/s2 = 322.5

1 7.5 10 tm/s2

a(0) = 22.5 m/s2

a(120) = 322.5

1 7.5 10 120m/s2 = 22.5

1 0.9m/s2

2(120) 225m/sa =

2-51. Use Equation (25): a(x x0) = 2 201 ( )2

v v

x x0 = 2100 m; v = 360 km/h 5 m/s

18 km/h = 100 m/s; v0 = 0

a = 12

2 2

0

0

v vx x

= 12

2100

2100m/s2

22.4 m/sa =

2-52. Acceleration, a = 2 2

0

02( )v v

x x

= 2(657)

2 6.63 = 4 23.26 10 m/s

Time to travel the length of the barrel, t = 0v va

t = 4657

3.26 10 = 22.02 10 s

2-53. 4.2 ly = 4.2 ly 9.46 1015 m/ly = 3.97 1016 m

Use x x0 = 0t + 21 ( )2

at with x0 = 0 = 0 and x = 1 (4.2ly)2

t 1/2 = 2xa

= 16

2

3.97 10 m9.807 m / s

= 76.36 10 s

Because the magnitude of the acceleration is the same for both parts of the trip, the time for thesecond half is identical to that of the first half. Thus, the total time for the trip, T, isT = 2t 1/2 = 2 6.36 107 s = 81.3 10 s 4.0 yr

Speed at midpoint: Use a(x x0) = 2 201 ( )2

v v

0 = 0; x x0 = 1.99 1016 m; a = 9.807 m/s2

Then22v = 2a(x x0) = (2 9.807 1.99 10

16) m2/s2

2 = 3.90 1017 m2/s286.2 10 m / s (This exceeds the speed of light!!)v =


CHAPTER 2

21

2-54. 2 20 0 0 02 ( ) 2 ( )a x x v v v a x x = = From the information given in the problem, x x0 =290 m, and a = 10 m/s2. Then 20 2( 10 m/s )(290 m) 76 m/s,v = = which corresponds to

about 270 km/h, or 170 mph.

2-55. Use 2 201 ( )2

v v = a(x x0) with v = 0; 0 = 80 km/h 5 m/s

18 km/h = 22.2 m/s; x x0 = 0.7 m.

Therefore a = 2 2 2

0

20

1 1( ) ( 22.2)2 2 .

0.7m/s

v v

x x

=

2a 350m/s (will probably survive)= 2-56. 2 20 0 0 02 ( ) 2 ( )a x x v v v a x x = = From the information given in the problem, x x0 =

9.6 km = 9.6 103 m, and a = 5 m/s2. Then 2 3 20 2( 5 m/s )(9.6 10 m) 3.1 10 m/s,v = =

which corresponds to about 700 mph. The time to stop is 2

02

0 3.1 10 m/s 62 s.5 m/s

v vta

= = =

2-57. Use 2 201 ( )2

v v = a(x x0) with x x0 = 50 m; v = 0;

v0 = 96 km/h = 26.67 m/s2 2 2

0

20

1 1( ) ( 26.67)2 2

50 m/s

v va

x x

= =

27.1m/sa = Use 0v v at= + with v = 0 m/s, v0 = 96 km/h = 26.7 m/s, and a = 7.1 m/s

2. Then

0 3.8 s.vta

= =

2-58.2 2 3

0

0

(260 10 / 3600)2( ) 2 1500v va

x x

= =

2 = 1.74 m/s (The minus sign denotes deceleration.)a Use 0v v at= + with v = 0 m/s, v0 = 260 km/h = 72.2 m/s, and a = 1.74 m/s

2. Then

0 42.5 s.vta

= =

2-59. 20 1.5 m/s 20 s 30 m/sv v at= + = =2 2 2

01 10 1.5m/s (20 s) 300 m.2 2

x v t at= + = + =

2-60. 2 20 01 11.0 m 5.0 m 3.0 m/s 4.0s (4.0s)2 2

x x v t at a = + = +i i 4 m = 12 m + 8 a 22.0 m/s .a = The velocity is 0v v at= + =

23.0 m/s (2.0 m/s )(4.0s) 5.0 m/s. =

2-61. 2 201 12 2

x v t at at= + = (since v0 = 0). 22 2 150 m

1.2 m/sxt

a= = = 16 s.


CHAPTER 2

22

2-62. 0550 km/hr 1000 m/km550 km/hr

3600 s/hrv = = = 153 m/s. 0v v at= + =

2 2153 m/s (0.60 m/s )(90 s) 2.1 10 m/s.+ =

2-63. The sketch should be based on the following:

For 0 6t s, a = 3.0 m/s2; 0 0 3 ;v v at t= + = + 2

0 0

33 .2

t t

x v dt t dt t= = = At t = 6s,3 6 18 m/s;v = = 23 6 54 m.

2x = =i For 6 10 s,t a = 4.5 m/s2, and

0 ( 6) 18 4.5 ( 6)v v a t t= + = m/s = 45 4.5t m/s.2 2 2

0 01 1( 6) ( 6) 54 18( 6) 4.5( 6) 2.25( 6) 18( 6) 542 2

x x v t a t t t t t= + + = + = + +i

At t = 10 s, 45 4.5 10 0;v = = 22.25 4 18 4 54 90 m.x = + + =i i

2-64.

2

2 2 22

0 01 700 m 0.5 0.05 m/s 30 s22 30s

atxx v t at v

t

= + = = = 22.6 m/s.2

0 22.6 m/s 0.05 m/s 30sv v at= + = + =i 24 m/s.

2-65.

2

2 2 22

0 01 550 m 0.5 0.5 m/s 15 s2 32.92 15s

atxx v t at v

t

= + = = = m/s.2

0 32.9 m/s 0.5 m/s 15 sv v at= + = + =i 40.4 m/s.2-66. Speed at the end of the 440-yard mark,

v = 250.69 1760 360 60

= 367.7 ft/s


CHAPTER 2

23

(a) Average acceleration, a = vt

= 2367.7 65.23ft/s5.637

=

(b) For a constant acceleration, the distance traveled would be = avg speed time =367.7 (5.637)

2 = 1036.36 ft 345.45 yd.=Therefore the acceleration was not constant.

(c) Assuming constant acceleration, distance = 440 3 = (5.637)2tv

vt = 2 440 3

5.637 = 463 ft/s 319 mi/h.=

2-67. The distance traveled by the elevator is2 21 2 3 3

1 12 2

x at vt vt at = + + where v = constant speed reached at at1; t1, t2 and t3 are the times spent in the following,respectively: accelerating, traveling at constant speed, and decelerating.

21 2.5 m = 12

a(5)2 + (a5)7 + (a5)5 12

a52

52.5 m = 60 a20.875m/sa =

The maximum speed of the elevator is vmax = at1vmax = 0.875 m/s2 5 s

max 4.4 m/sv =

2-68. (a) Average speed = 40055

= 7.3 m/s

(b) For minimum values of acceleration and deceleration, the elevator should travel half thedistance in half the time. Therefore

21 552002 2

a = 20.53m/sa = where a is the acceleration and a is the deceleration.

Maximum speed is given by vmax = at = 0.53 55 14.6 m/s2

=

(This is twice the average speed. What is its significance?)

2-69. v0 (km/h) v0 (m/s) 0v t (m) 202va

(m)Total stoppingdistance (m)

15 4.17 8.3 1.1 9.430 8.33 16.7 4.3 21.045 12.5 25.0 10 35.060 26.7 33.3 18 51.375 20.8 41.7 27 68.790 25.0 50.0 39 89.0


CHAPTER 2

24

2-70. Table on Page 47:

0v t = 20

2va

20 2 2 8 m/s 0.75 sv at = = = 12 m/s.

2-71. 50 km/h = 50 518

m/s = 13.9 m/s; 2.0 ft = (2 0.3048) m = 0.610 m. In the inertial frame that is

traveling at constant velocity with the car, all velocities are zero. In this frame, a = 200 m/s2 also.Therefore the speed with which the dashboard hits the passenger isv = 02 ( )a x x =

22(200 m/s )(0.610 m) 15.6 m/s.=

2-72. Consider the position of the car in the reference frame of the truckx0c = 12 17 = 29 m

The final position of the car is xc = 17 m, therefore,

xc = x0c + v0c t + 12

at2

17 = 29 + 0 + 12

at2

2146 .2

at= (i)

The final speed of the car relative to the truck is 24 km/h

= 324 10

60 60 = 6.67 m/s.

at = 6.67 m/s (ii)Divide equation (i) by (ii) to get 13.8t s= and

a = 26.67 0.48 m/s .13.8

=

2-73. v = g + get/

(a) acceleration, /tdv gedt

=

(b)Lim

t e t/ = 0, therefore

Limt

v = g + Lim

t e t/ gt=

(c) v = ( )2 / 2 0xtd g t gt e gdt + += g + g

2

e t/

/tg g e = +

(d) for t

CHAPTER 2

25

2-74. x x0 = v0t 12

gt2 with x x0 = 380 m; v0 = 0; g = 9.8 m/s2

380 m = 21 (9.8 m/s )2

t2

t2 = 77.6 s2

8.8st =

For the impact velocity use v = v0 gtv = 0 9.8 m/s2 8.8 s 86 m/s=

2-75. 130 km/h = 36.1 m/s. The acceleration of the freely falling falcon is g, so use2 2 2

2 2 00 0 0 2

(36.1 m/s) 02 ( ) 66.5 m.2 2(9.81 m/s )

v vv v g x x x xg

= = = =

2-76. 2 20 2 .v v ax= + Here v0 = 0, a = g. 22 2(9.81m s )(8.7 m) 13 m/s.v gh= = =

2-77. 2 20 2 .v v ax= + Here v = 0, a = g because the displacement is upward and the acceleration is

downward. 20 2 2( 9.81m s )(1.9 m) 6.1 m/s.v gh= = =

2-78. mgh = 12

mv2, where h is the maximum height reached

h = 21

2vg

=

2(366) 6834 m2 9.8

=

2-79. Neglecting air resistance, we have

x x0 = v0t 12

gt2 with v0 = 0; g = 9.8 m/s2; t = 3.0 s

x x0 = 0 21 (9.8 m/s ) (3.0 s)2

2 44 m=

2-80. Use 2 20v v = 2g(x x0) with v = 0; g = 1.80 m/s2; x x0 = 200,000 m

20v = 2 1.80 m/s

2 2 105 m = 7.2 105 m2/s2

0 849 m/sv =

2-81. v = 45 km/h = 12.5 m/s. 2 2

2

(12.5 m/s) 1 floor8.4 m 2 2(9.81 m/s ) 2.9 mvhg

= = = = 3 floors.

v = 75 km/h = 20.8 m/s. 2 2

2

(20.8 m/s) 1 floor22.1 m 2 2(9.81 m/s ) 2.9 mvhg

= = = = 8 floors.

v = 105 km/h = 29.2 m/s. 2 2

2

(29.2 m/s) 1 floor43.4 m 2 2(9.81 m/s ) 2.9 mvhg

= = = = 15 floors.

2-82. 22 2(10 m)

9.81 m/sht

g= = = 1.43 s. This is 1.43 s/(3 half turns) = 0.48 s per half turn.

2-83. 22 2(9.5 m) 1.39 s.

9.81 m/supht

g= = = Total time = 2(1.39 s) = 2.8 s.

20 (9.81 m/s )(1.4 s) 14 m/s.upv gt= = = This is the initial speed. The initial velocity is 14 m/s

up.


CHAPTER 2

26

2-84. v0 = 1.0 m/s downward, and the ball is initially some distance h above the ground. After fallingthat distance h, the ball will strike the ground with some speed v and will rebound (reverse itsdirection of motion) at the same speed if the collision with the ground is elastic. (This conceptwill be introduced in a later chapter.) Since it starts moving back up with the same speed it hadjust before it hit the ground, the time required to return to its starting point a distance h above thefloor will be the same as the time required for it to reach the floor in the first place, and it willarrive at the distance h above the ground with same speed with which it was initially thrown.Thus the downward and upward travel times are each equal to half the total travel time. The speed

of the ball just before striking the ground is 0 ,2tv v g= + where t is the total travel time. Then

the height h can be found using 2 20 2 ,v v gh= + which gives 2 2

0 .2

v vhg

= Substituting, we get

2 0.75 s1 m/s (9.81 m/s ) 4.68 m/s.2

v = + = Then 2 2

2

(4.68 m/s) (1 m/s)2(9.81 m/s )

h = = 1.1 m.

Comment: This requires conservation of momentum! The ball will collide with the ground andrebound with a speed equal to its speed just before it hit, a topic that obviously isnt covered inthis chapter.

2-85. Take the coordinate axis to point up. Then the final displacement is 9.2 m, a = g, t = 2.5 s.2

2 2

0 09.2 m ( 9.81 m/s )(2.5 s)2 8.57 m/s,

2 2 2.5 s 2

atdat d atd v t vt t

= + = = = = or 8.6m/s (to two significant figures). The height the penny reaches above its launch point is

2 2 2 20 0 0

2

(8.57 m/s) 3.7 m.2 2( ) 2 2(9.81 (m/s )v v vha g g

= = = = =

2-86. v0 = 0, a = g, x = h 20 2 2 .v v ax gh= + = 02v vv += for constant acceleration, so

2.

2gh

v =

2-87. Assume that the collision of the ball with the floor simply reverses the direction of the ballsvelocity. Then the time for the ball to reach the floor is the same as the time for the ball to returnto its starting point, and its speed upon returning will be the same as the speed with which it

started. Thus 0.90 s 0.45 s.2down

t = = Since the ball begins by moving down, choose the

direction of the axis for the motion to point down. Then the acceleration and initial velocity are

represented by positive numbers. 2012

h v t gt= +2 2 2

02 2(1.5 m) (9.81 m/s )(0.45 s)

2 2(0.45 s)h gtv

t = = = 1.13 m/s, or 1.1 m/s to two significant

figures. The velocity just before hitting the floor is 0v v gt= + =21.13 m/s (9.81 m/s )(0.45 s)+ = 5.5 m/s.


CHAPTER 2

27

2-88. 21 .2

x gt= At t = 0,1,2,3 s the distance fallen is 0, 4.91 m, 19.6 m, 44.1 m, 78.5 m. The distance

traveled from 0 1 s is 4.91 m; from 1 s to 2 s the distance traveled is 19.6 m 4.91 m = 14.7 m.Likewise the distance traveled from 2 s to 3 s is 24.5 m. Dividing each of these by 4.91 m givesthe ratios 1:3:5 and so on.

2-89.2

0 .2ath v t= + v0 = 0 2

2 .hat

=

2

2 hat =

10 cm4 fingers 4 fingers 100% 0.22%.46 cm100 cubits

cubit

a ha h

= = =

2-90. Time taken to fall through a distance of 45 m,

y2 = y1 + v1t + 12

at2

0 = 45 + 0 1 (9.8)2

t2

t = 3.03 s

Since 12 ,ytg

= to measure y1 within 10%, the time must be known to within 5% or within 0.15 s.

Therefore a stopwatch must be used. An ordinary wristwatch will have an uncertainty of 1 s.

2-91. Velocity of ball on impact 212 2(9.8 m/s ) (1.5 m)gx= = = 5.42 m/s

Velocity of ball after impact = 22gx = 22(9.8m/s ) (1.1m) = 4.64 m/s

vat

=

= 4 24

4.64 m / s ( 5.42m/s) 1.6 10 m/s6.2 10 s

=

2-92. (a) Impact speed, v = 2gh = 2 9.8 96 = 43.4 m/s

(b) a = 2 22

22( )v v

x x

= 2

2(43.4) 254.5m / s2 3.7

=

(The minus sign denotes deceleration.)2-93. The muzzle speed is given by 2 3 30 2 2(9.81 m/s )(180 10 m) 1.88 10 m/s.v gh= = = To

find out how long the projectile remains above 100 km, we can use the fact that the time for theprojectile to climb from 100 km to 180 km is the same as the time for it to fall from 180 km backto 100 km. So we can just calculate the time to fall a distance of 80 km from rest and double that

value. The time to fall can be calculated from 2 3

2

2 2(80 10 m) 128s,2 9.81 m/s

gt yy tg

= = = =so the total time above 100 km is 2t = 256 s.

2-94. Given,(a) g = 978.0318 cm/s2 (1 + 53.024 104 sin2 5.9 106 sin2 2 ) for = 45g = 978.0318 cm/s2 (1 + (53.024 104)/2 5.9 106) 2980.6190 cm/s=

At the pole = 90, g = 978.0318 cm/s2 (1 + 53.024 104) 2983.2177 cm/s=


CHAPTER 2

28

(b) Let g = A(1 + B sin2 C sin2 2 )where A = 978.0318 cm/s2, B = 53.024 104, C = 5.9 106. The condition

( sin 2 4 sin 2 cos 2 )dg A B Cd

=

= A (B 4C cos 2 )sin 2 = 0 gives extrema at = 0, 2

To distinguish between extrema at = 0, ,2 evaluate

2

2

d gd

at = 0, 2

2

02 0(2 cos 2 8 cos 4 sin 4 )d g A B C

d = ==

= 2AB > 0 (Thus at = 0, g has a minimum.)Similarly

2

2 / 2

2 0d g ABd

=

= > 2 s), find its value at 2 s:

2 s 2 32 s 2

0 0 02 20 0

8(1 ) ( ) | 20 m/s (2 ) s 26.7 m/s4.0 s 12.0 s 12

v t tv adt a dt a t= = = = = The distance traveled is

2 s 2 s 3 2 4 2 42 s 2

0 0 0 02 2 20 0

(2 s) (2 s)( ) ( ) | (20 m/s )12.0 s 2 48 s 2 48 s

t t tx x vdt a t dt a

= = = = = 33.3 m.

2-102.2 s 2 3

2 2 s 2 2 2 4 3 30

0 0

1 1( ) | 15 m/s 2 s 25 m/s 2 s2 3 2 3

v At Btv adt At Bt dt= = + = + = + i i i i= 96.7 m/s


CHAPTER 2

31

Distance traveled 2 s 2 s 2 3 3 4

2 s1 s

1 s 1 s

( ) |2 3 6 12

At Bt At Btx vdt dt = = + = + =

3 3 3 4 4 415 m/s (2 1) s 25 m/s (2 1) s6 12

+ = 48.75 m

2-103. dva g Avdt

= = dv dtg Av

=

0

v

v

dv tg Av

=

0

0 0

1 ln ( )At Atg Av g Avt e g Av g Av eA g Av g Av

= = =

0 01 [ ( ) ] (1 )At At Atgv g g Av e e v eA A

= = +

After a long time, 1/ ,t A>> 0,Ate .gvA

2-104. 2dva Bvdt

= = 0

0

2

1 |v

vv

v

dv B t B tv v

= = 4 1

0

4 1

1 1 1 1 1 16.1 10 m 90 km/hr 120 km/hr

1 1 1 3600 s/hr6.1 10 m 1000 m/km 90 km/hr 120 km/hr

tB v v

= =

= i= 16.4 seconds

2-105. Crams speed is 1 mile 1 mile .3 min 46.32 s 226.32 sC

v = = The time difference for the two runners

getting to the finish line is 3min 46.32sec 3min 44.39sect = = 1.93 s. So Cram is behind

by a distance of 3C1 mile 1609 m 1.93 s 8.53 10 mile 13.7 m.

226.32 s milev t = = =

2-106. distance 1500 1500 m 14.43 m/stime 1min 43.95sec 103.95 s

mv = = = =

2-107. Average speed = distance (100 100) m 1.3 m/stime (10 60 80) s

+= =

+ +

Average velocity = 0, because the woman returns to her starting point.

2-108. (a) 1135 km/hr 30 min 35 km/hr 30 min 17.5 km;

60 min/hrd = = =

2185 km/hr 30 min 85 km/hr 30 min 42.5 km;

60 min/hrd = = =

(b) Average speed 1 2 (17.5 42.5) kmtime 60 min

d d+ += = = 1.0 km/min, or 60 km/hr

2-109. Define: t = time from when the sailfish spots the mackerel to when it catches the mackerel. Then:distance for the sailfish = 109 km/hr ,t distance for the mackerel = 33 km/hr .t Theseparation between the fish is 20 m, so the time for the sailfish to catch the mackerel is given by109 km/hr 33 km/hr 20 mt t =


CHAPTER 2

32

20 m 20 m 3600 s/hr76 km/hr 76 km/hr 1000 m/km

t = = i i = 0.95 s. During this time, the sailfish travels a

distance 109 km/hr 1000 m/km 0.95 s109 km/hr 0.95 s 28.8 m.3600 s/hr

d = = =

2-110. Distance traveled by the first plane = d1 = 720 km/hr ,ti where t is measured beginning at 10:00.Distance traveled by the second plane = d2 = 640 km/hr ( 1 hr),t i since that plane left one hourlater. 1 2 1286d d+ = km 720 640( 1) 1286 kmt t + = (1286 640) km 1.42 hr1360 km/hrt

+ = == 1 hr 25 min. The distance traveled by the first plane is 1 (720 km/hr)(1.42 hr) 1022 km,d = =so the planes meet 1022 km north of San Francisco. Since the total elapsed time is 1 hr 25 afterdeparture of the first plane, they meet at 11:25 AM.

2-111. Distance that my car travels = 80 km/hr ti Distance that the other car travels = 50 km/hr .ti To gofrom 10 m behind the slower car to 10 m ahead of it requires traveling a total relative distance of10 m + 10 m + 4 m, because of the length of the car. Thus 80 km/hr ti 50 km/hr ti = (10 + 10 +4) m 24 m 24 m 3600 s/hr

30 km/hr 30 km/hr 1000 m/kmt = = =i i 2.9 s.

2-112.2 2 2 2 2

3 20 (105 0) km /hr 1.72 10 km/hr2 2(3.2 km)

v vax

= = =

03 2

105 km/hr 0 0.0610 hr,1.72 10 km/hr

v vta

= = = or 219 seconds.

2-113. (a) 22.0 6.0 3.0 .x t t= + At t = 0.50 s, 22.0 6.0 0.50 3.0 (0.50)x = + i = 4.3 m.(b) 6.0 6.0 .dxv t

dt= = At t = 5.0 s, v = 6.0 6.0 0.50 =i 3.0 m/s

(c) (6.0 6.0 )dv da tdt dt

= = = 6.0 m/s2 at all times.

2-114. (a) 100 km/hr = 27.8 m/s. The position of the speeder after 8.0 s is,1 1 27.8 m/s 8 ssx vt= = = 222 m from the starting point.

(b) The speed of the police cruiser goes from 0 to 120 km/hr (33.3 m/s) in 10 s, so itsacceleration is 3.33 m/s2. The position of the police cruiser after reaching its final speed is

2 2 2,1

,1(3.33 m/s )(10 s) 168 m

2 2p

p

atx = = = from the starting position. At this time the position of

the speeder is ,2 ,2 27.8 m/s 18 s 500 m,s sx vt= = = so the speeder is 332 m ahead of the

police cruiser.(c) Let t be the time from when the cruiser reaches its final speed of 120 km/hr until it catches upto the speeder. When the cruiser catches up to the speeder, both vehicles have traveled the samedistance from the point where the cruiser reached 120 km/hr. Mathematically this means(33.3 m/s) (27.8 m/s) 332 m,t t= + which gives t = 60 s. So the total time that has elapsed sincethe cruiser began pursuit is 70 s. During this time the cruiser traveled a total distance of

3168 m (33.3 m/s)(60 s) 2.18 10 m,+ = or 2.18 km.2-115. The initial speed of the car is v0 = 90 km/hr = 25.0 m/s. The distance traveled during the reaction

time t1 = 0.75 s is 1 0 1 (25 m/s)(0.75 s) 18.8 m.d v t= = = The remaining distance to the cow is d2= 30 m 18.8 m = 11.2 m. The cars acceleration as it travels this distance is a = 8.0 m/s2. Its


CHAPTER 2

33

final speed when it hits the cow is given by2 2 2 2

0 2 0 22 2 (25 m/s) 2( 8.0 m/s)(11.2 m) 21.1 m/s,v v ad v v ad = = + = + = or76 km/hr.

2-116. Use v2 v 20 = 2g(x x0) with v = 0; g = 9.8 m/s2; v0 = 26 m/s

x x0 = 2

2

(26 m/s) 34m2 9.8 m/s

=

The total time of flight for any particle of water is:

t = 02vg

= 22 26 m/s9.8 m/s

= 5.3 s = 8.84 102 min

The discharge rate is 280 l/min so the total amount of water in the air after 5.3 s is vol = 280 l/min 8.84 102 minvol 25 l=

2-117. Speed upon impact, v = 2gh = 2 9.8 56 33.1m/s=

Average deceleration, a = vt

= 233.1 2209 m/s0.015

=

2-118. The final speed of the part with acceleration g is v = 200 km/hr = 55.6 m/s.

The time with acceleration is 01 255.6 m/s 5.66 s.9.81 m/s

v vtg

= = = The distance she falls during this

time period 2 2 2 2

01 2

55.6 (m/s)2 2 9.81 m/s

v vhg

= = = 158 m. The rest of the height is h2 = 1000 158 m

= 842 m. She falls the distance h2 with a constant speed of 55.6 m/s. The time for this part of the

fall is 22842 m

55.6 m/shtv

= = = 15.2 s total time = (5.66 + 15.2) s = 20.9 s.

2-119. (a) 1 s.t = The distance the first ball falls in that interval is 2112

y gt=

2 21 (9.81 m/s )(1 s) 4.90 m,2

= = so the first ball is 13 m 4.9 m = 8.1 m above the ground when

the second ball is released. The time for the first ball to fall the total distance of 13 m is

1 2

2 2 13 m 1.63 s.9.81 m/s

htg

= = = When the first ball hits the ground, the second ball has been

falling for t2 = 0.63 s. The distance second ball has fallen during this time is2 2 2

2 21 1 (9.81 m/s )(0.63 s)2 2

y gt= = = 1.95 m, so the second ball is 13 m 1.95 m = 11.1 m

above the ground when the first ball lands.(b) v (1st) = gt1 = 29.81 m/s 1.63 s 16.0 m/s=iv(2nd) = gt2 = 29.81 m/s 0.63 s 6.18 m/s=i instantaneous velocity of the first ball relative to the second just before the first hits the groundis: 16.0 m/s 6.18 m/s = 9.8 m/s down.(c) Both balls have same acceleration, (9.81 m/s2 down,) so the relative acceleration is zero.


34

CHAPTER 3 VECTORS


3-1. RGP = RGA + RAP

= 10.2 x + 5.9 y

| RGP | = 2 2(10.2) (5.9)+

11.8 km=

= tan1 5.9 30N of E10.2

=

3-2. The two displacements areOA = 240(sin 29 x + cos 29 y )

= 116.4 x + 209.9 y

AB = 560(cos 29 x + sin 29 y )= 489.8 x + 271.5 y

The resultant displacementvector is:OB = OA + AB = 373.4 x + 481.4 y

| OB | = 2 2(373.4) (481.4)+= 609.2 m

= tan1 481.4373.4

= 52.2 N of W

3-3. A reduced copy of the diagram is shown. In the actualdiagram, 1 cm = 1 km. From the diagram, the linerepresenting the resultant R is 11.2 cm long, so the lengthof R is 11.2 km. The angle is measured to be 27.5.Graphically we find R = 11.2 km @ 27.5 S of E.To do the problem trigonometrically, take x to point eastand y to point north. Then


CHAPTER 3

35

1

2

3

18.0sin 60 18.0cos 60 15.6 9.0 km9.5cos 60 9.5sin 60 4.75 8.23 km

12.0sin 60 12.0 cos 60 10.4 6.0 km

= + = +

= =

= =

r i j i jr i j i jr i j i j

Adding gives 9.95 5.23 km.= R i j The magnitude is 2 29.95 5.23 11.2 km,R = + = and the

direction is 1 5.23tan = 27.7.9.95

= In terms of compass directions, this is R = 11.2 km @

27.7 S of E.3-4. We can combine the two vectors at 45 E of N to make a vector

5.8 km at 45 E of N (see diagram). ThenR2 = 5.82 + 4.52 2(5.8)(4.5) cos 85

R = 249 m = 7.0 km

By the law of sines:sin4.5

= sin 857.0

= sin1 4.5 sin 85 = 407.0

Therefore, the angle between R and N is (45 40) = 5.Thus, the resultant is magnitude 7.0 km, 5 E of N.

3-5.


CHAPTER 3

36

The graph is drawn with a scale of 1 cm = 20 m. Measuring the distance from South Bay toMosquito Rock, we get: r 0.6 cm i 3.0 cm j on the graph = 120 m i 600 m j.

The distance between them is 2 2( 120) ( 600) = 612 + m at an angle 1 120tan600

= =

11.3 west of south.3-6. r1 = 2.5 km j,

r2 = (1.5 km) sin 30 i + (1.5 km) cos 30 j = 0.75 km i + 1.30 km jr = r1 + r2 = 0.75 km i + 3.80 km j.|r| = 2 20.75 3.80+ = 3.87 km

The angle of this vector with the x axis: 1 3.80tan0.75

= =

78.8 north of east.

3-7. A reduced copy of the graphical solution is shown inthe diagram. In the actual graph, the scale is 1.0 cm =100 km, so the line for the first displacement is 4.8cm long and the line for the second displacement is3.7 cm long. The line for the total displacement ismeasured to be 4.4 cm long, corresponding to amagnitude of 440 km. The direction is measured tobe 7.5 W of N. The total displacement is D = 440km @ 7.5 W of N.Check analytically. Let r1 = 480 sin (40) km i + 480km cos (40) j = 308.5 km i + 367.7 km jand r2 = 370 cos (10) km i + 370 km cos (10) j =364.4 km i + 64.2 km j, where E corresponds to xand N corresponds to y. Then D = r1 + r2 = 55.9 kmi + 431.9 km j.|D| = 2 2( 55.9) 431.9 km + = 436 km. To get the direction relative to the x direction, note thatthe resultant vector is in the third quadrant (negative x component, positive y component). Thestandard method of finding the direction will give an angle measured from E, the x axis.Using a calculator is used to find the inverse tangent will give a negative angle, so the correctangle (in the second quadrant) is found by adding 180 to the calculator result:

1 431.9180 tan 97.4 ,55.9

= = which measured N of E. This is 7.4 W of N. The analytical

result is D = 436 km @ 7.4 W of N, which agrees almost exactly with the graphical result.3-8. Distance east = 14 cos 60 = 7 km

Distance south = 14 sin 60 = 12.1 km

1.5 km30

2.5 km


CHAPTER 3

37

3-9. The resultant displacement vector, R = (5 m). The given displacement vector, A = (2.2 m) sin35 i + (2.2 m) cos 35 j = (1.26 m) i + (1.8 m)j. Let the other displacement vector be B = Bxi +Byj. Therefore, R = A + B = (1.26 m + Bx) i + (1.8 m + By) j = 5 mj. Comparing i and jcomponents gives1.26 m + Bx = 0 or Bx = 1.26 m1.8 m + By = 5 or By = 3.2 mThus, ( 1.26 m) (3.2 m) .= +B i j

3-10. Graphs were drawn at actual size. Reduced copies are shown.

A + B + C = 13.7 cm @ 8.0 E of NA + B C = 7.7 cm @ 14.0 E of N

3-11. The easiest method is to write the vectors in component form, with j as N and i as E.Then R = 1.2 j + (sin 38 i cos 38 j )6.1 + (sin 59 i cos 59 j )2.9 +

(sin 89 i + cos 89 j ) 4.0 + (sin 31 i cos 31 j )6.5= 1.2 j + 3.76 i 4.81 j + 2.49 i 1.49 j + 4.00 i + 0.07 j + 3.35 i + 5.57 j= 13.6 i + 0.54 j

This translates into a vector of length13.6 nmi at 88 E of N.

3-12. (a) (b)

(b) (cont.) The radius of the earth is 1.49 1011 m, so using the Pythagorean theorem, we get forthe displacement vector

11 2 11 2 11 (1.49 10 ) + (1.49 10 ) = 2.1 10 md =


CHAPTER 3

38

3-13. The straight-line distance is

d2 = R2 + R2 2R2 cos 178= R2 + R2 + 1.999 R2

= 3.999 R2

d = 2.000 = 12,750 kmR

Distance around the equator is

S = 2360

178 R

19,810 kmS =

3-14. By completely filling in all of the angles inthe triangle, we find that it is an isoscelestriangle, so that .P P AP =

3-15. Position vector at 11h10m is 4.2 km at 33 E ofN. Displacement vector from position at10h30m to 11h10m isA2 = 9.52 + 4.22 2(9.5)(4.2) cos 27

= 36.8 km2

A = 6.07 km.sin /4.2 = sin 27/6.07 = sin1 (4.2/6.07 sin 27)The displacement vector then isA = 6.07 mi, 78.3 W of S.

In another 20 min, it would have traveled adistance of

6.07 6040

= 9.1 km.

Therefore, its position at11h30m should beB 2 = 9.52 + 9.102 2(9.5)(9.1) cos 18.3

= 8.90 km2

B = 3.0 kmsin = 9.10/3.0 sin 18.3 = 0.952 = 72.3Thus, the position vector is3.0 km, 12.3 W of N


CHAPTER 3

39

3-16. If the bearing of each ship remains constantwith respect to the other, the angles 1 and2 are constant in the diagram where thetwo ships have been denoted by A and B.A, B shows their positions at a laterinstant. For 1, 2 constant, the separationbetween the ships will eventually go tozero, otherwise they will miss each other,assuming they are point masses.

3-17. N: 12.0 cos 40 9.19 km=W: 12.0 sin 40 7.71 km=

3-18. Ax = 5.0 cos 30 4.3 m=Ay = 5.0 sin 30 2.5 m=

3-19. Place the x axis along the sloping line as shown. Then: xcomponent = (4.0 m) sin(25) = 1.7 m.

3-20. By the Pythagorean theorem, the distancealong the ground is:

2 215 0.5 km 15 kmD =

x : 15.0 km cos 56 = 8.4 km.y : 15.0 km cos 35 = 12.3 km.z : 0.5 km = 500 m

25

254 m

x

y


CHAPTER 3

40

3-21. (a) (b) A = 3i + 2j cmB = 1i + 3j cm

A + B = 3i + 2j 1i + 3j cm= (2 + 5 cm)i j

3-22. Ax = 8.0 cos 52 4.9 units=Ay = 8.0 sin 52 6.3 units=Az = 8.0 cos 90 0 units=

3-23. x : 6.0 cos 45 4.2 units=y : 6.0 cos 85 0.5 units=

Since the magnitude of the vector is 6.0 units, we know thatA = 2 2 2x y zA A A+ + = 6.0

2. Thus Az = 2 2 2 6.0 4.2 0.5 = 4.2 units

z can be either 4.2 units, so it is not uniquely determined.A

3-24. Magnitude = 2 2 23 2 1+ + = 14 units 3.7 units=

3-25. (a) A + B = (5i 3j + k) + (2i + j 3k) 3 2 2= i j k

(b) A B = (5i 3j + k) (2i + j 3k)7 4 + 4= i j k

(c) 2A 3B = 2(5i 3j + k) 3(2i + j 3k)= 16 9 + 11i j k

3-26. r = 6 i 8 j m|r| 2 26 8= + = 10 m.

The angle between r and +x is: 1 8tan 53.16

= =

3-27. = 14 cos(135) = 9.9 mx = 14 sin(135 ) = 9.9 my

3-28. E = (2.5 + 1.0 + 1.5 + 0) i + (3.5 + 4.5 + 2.0 + 3.0) j + (0 + 2.5 + 3.0 + 1.5) k= 5.0 i + 13.0 j + 7.0 k

F = (2.5 1.0 + 1.5 0) i + (3.5 4.5 + 2.0 3.0) j + (0 2.5 + 3.0 1.5) k= 3.0 i 2.0 j 1.0 k

|E| = 2 2 25 13 7+ + = 15.6

|F| = 2 2 23 ( 2) ( 1)+ + = 3.74

3-29.2 2 2

2 4 4| | 2 4 4

+ += = =

+ +

A i j krA

13

i + 23

j + 23

k


CHAPTER 3

41

3-30. A = 2.0 i + 3.0 j + 1.0 k,B = 1.0 i + 2.0 j + Bz k,

A + B = (2.0 1.0) i + (3.0 + 2.0) j + (1.0 + Bz) k|A + B| 2 2 21 5 (1 ) 6.0zB= + + + =

226 (1 ) 36zB + + =1 10zB + = 10 1,zB = or, 10 1zB =

3-31. c1A + c2B = c1(2.0 i + 3.0 j) + c2(1.0 i + 5.0 j)= (2.0 c1 + 1.0c2)i + (3.0 c1 + 5.0c2)j

c1A + c2B = C = 1.0 i + 3.0 j 2.0c1 + 1.0c2 = 1.0 (1)

3. 0c1 + 5.0c2 = 3.0 (2)

Solve for c1, c2: (1) 3 (2) 2 : 2 297 9 ,7

c c = = 1 87c = 3-32. Let the planes be denoted by A and B and i east and j west.

A = 120 103(cos 20 i sin 20 j ) + 2500 kB = 110 103(cos 25 i sin 25 j ) + 3500 kA = 112.8 103 i 41.0 103 j + 2500 kB = 99.7 103 i 46.5 103 j + 3500 k

The position of B relative to A isB A = 13.1 103 i 5.5 103 j + 103 k

The displacement vector is: horizontal distance = ( )2 2(13.1) (5.5)+ 103 m 14.2 km=The bearing is

= tan1 5.5 = 23 W of S13.1

Altitude difference = 1000 m 1 km=3-33. Let the directions of i, j and k be as shown at the

right, where GM is the Greenwich meridian. NewYork will have a position vector:


CHAPTER 3

42

NY: R(0.728 i + 0.209 j + 0.653 k ) where R = radius of the earth.

And for Cape Wrath:

CW: R(0.068i + 0.779j + 0.624k)The displacement vector is= [(0.068 0.728) i + (0.779 0.209) j + (0.624 0.653)k]R= (0.66 i + 0.570 j 0.029 k )RThe magnitude is= (0.6582 + 0.5702 + 0.0292)1/2R= 0.873R = 5550 km

3-34. Magnitude = 2 2 25.0 3.0 1.0+ + = 35 = 5.9 units

) x-axis : cos x = xAA = 5.05.9

x = cos1 5.0 = 325.9

) y-axis : cos y = yAA

= 3.05.9

y = cos1 3.0 = 605.9

) z-axis : cos z = zAA = 1.05.9

z = cos1 1.0 = 805.93-35. 3i 6j+ 2k has magnitude 2 2 23 6 2+ + = 49 = 7 units

The unit vector in this direction is ( )1 3 6 27

+i j k

Therefore the vector with magnitude 2 is 2 (unit vector)6 12 47 7 7

= +i j k


CHAPTER 3

43

3-36. 3A 2C = 4B. Then 3A 4B = 2C

C = 32

A 2B

C = ( ) ( )3 6 2 2 4 3 8 = 17 3 162

+ +i j i j k i j k

3-37. ( ) ( )5 2 2+ + i j k i ki = (5 2) + (2 0) + (1 [1]) = 93-38. The vectors A and B in Example 1 are perpendicular to each other, so the angle between them is

90. Because cos 90 = 0, A B = 0, regardless of their magnitudes.3-39. A = 2 2 22 1 2+ + = 3

B = 2 2 23 6 2+ + = 7Dot product A B = (2)(3) + (1)(6) + (2)(2) 8= A B = AB cos

= cos1 A B

A Bi= cos1 8 112

7 3=

3-40. ( ) ( )x y z x y zA A A B B B= + + + +A B i j k i j ki i = x x y y z zA B A B A B+ +

3-41. 12 2 2

3 1 4 0 2 0cos 0.557 cos 0.557 56.1| | 3 4 2

+ += = = = = + +

A iAi i i i

3-42. 4 6 2 4= + + =A Bi 1 6 11= +A B i j k

3-43. | | cos=A B A || Bi | | sin =A B A || B

If A B = A B, then cosA B = |A| |B| sin cos sin =which gives: = 45

3-44. A = 50 sin 30 i + 50 cos 30 j = 25 i + 43.3 jB = 35 sin 70 i + 35 cos 70 j = 32.9 i + 12 jA B = (25)(12) i j (43.3)(32.9) j i

= 1725 2 (magnitude 1725 m ; direction along the -axis)zkB A = 1725 (the same magnitude as ;direction along the axis)z k A B

3-45. Take north to be j and east to be i. Then2 6 2 6 2 (2180 ) ( 1790 ) km 3.90 10 ( ) km 3.90 10 km= = = A B i j i j k

3-46. See Eqs. (3.31) through (3.33). (2 5 3 ) ( 0 2 ) + + =i j k i j k[( 5)( 2) (3)(0)] [(3)(1) (2)( 2)] [(2)(0) ( 5)(1)] 10 5 3 + + = +i j k i j k

3-47. cos sin t t = +A i j

sin cos d t tdt

= +A i j

A ddtA

= sin t cos t + sin t cos t = 0

Since both A and ddtA are nonzero, they must be perpendicular to each other.


CHAPTER 3

44

3-48.

(a) (b)Define B to be the y axis as shown in (a). The x axis is shown perpendicular to B. The anglebetween A and B is 130 measured from the y axis, so = 30sin130 + 30cos130A i j= 23.0 19.3 .i j The component of A along B is 19.3 m.Next, define A to be the x axis as shown in (b). The y axis is shown perpendicular to A. Now theangle between A and B is 130 measured from the x axis, so = 40cos130 + 40sin130B i j= 25.7 + 30.6 .i j The component of B along A is 25.7 m.

3-49. ( ) ( ) ( )y z z y z x x y x y y xA B A B A B A B A B A B= + + A B i j k (5.0 2.0 3.0 ) ( 3.0 )x zB B= + + + =A B i j k i j k ( 2 9) + (3 5 ) + (15 + 2 )z x z xB B B B i j k 2.0 .zC= = +A B C j k )x y z z y xA B A B C= =(A B 2 9 0 4.5z zB B = =

( ) y z x x z yA B A B C= =A B 13 5 2 (2 5 )3x z x zB B B B = = +1 [2 5 ( 4.5)] 6.83.3

= + =

( )z x y y x zA B A B C= =A B 15 2 15 2( 6.83) 1.34x z zB C C+ = = + =3-50. A = ,x yA A+i j B = x yB B+i j

= | || | cos x x y yA B A B = +A B A Bi (1) = | || | sin x y y xA B A B = A B A B (2)

(2) (1) tan x y y xx x y y

A B A BA B A B

=+

3-51. A = 2i j 4kA = 2 2 22 1 4+ + = 21

unit vector is ( )1 2 4 0.44 0.22 0.8721

= i j k i j k

3-52. Let A = j + 2k and B = 3i j + k. Let the unit vector along A be 25A

+=

i ju and the unit vector

along B be 3 .5B

+=

i j ku

20

20

B

20

A

y

E

N

B20

2020

A

x

E

N


CHAPTER 3

45

A vector C pointing from 0 to the midpoint of thedifference vector A Bu u will bisect the angle betweenA and B,

C = ( )1 212B + u i i = ( )1 212

+u u

= 1 3 1 1 2 12 11 5 11 5 11

+ + + i j k

Dividing C by its magnitude (|C| = 1/ 2 ) gives therequired unit vector

2C =u C ( )0.64 0.10 0.81= + +i j k3-53. The displacement vector pointing from A to B is B A. The vector that points halfway between

A and B is (B A)/2 from the tip of A, which makes its position vector at A + (B A)/2 =

(A + B)/2. Here, ( )12

+A B = ( )1 4 2 3 22

+ + +i j i j k = ( )1 3 5 2 .2

+ +i j k Its magnitude

is 2 2 21 3 5 22

+ + = 38 .2

The unit vector that points in this direction is

( )1 3 5 2 0.49 0.81 0.32 .38

+ + = + +i j k i j k

3-54. Edges of the cube are parallel to the x-y-z axes.The diagonal of the unit cube is thenAC = i + j + kwhose magnitude is

2 2 21 1 1+ + = 3.

Let AB be the unit vector in thex-direction. Then,

cos = | || |AC AB

AC ABi = ( )

1 3+ +i i j kii =

13

1cos 54.73

= =

3-55. Since |A| 0, |B| 0, yet A Bi = 0, we can conclude the angle between A and B is 90.So | 24 | = | || | sin 90 4 6 = =A B A B

3-56. (a) Right-hand-rule: fingers wrap from the first vector toward the second vector, the thumbpoints at the direction of the product.(b) ( ) ( )x y z x y zA A A B B B= + + + +A B i j k i j k

( ) ( ) ( )

x x y x z x x y

y y z y x z y z z z

y z z y z x x z x y y x

A B A B A B A BA B A B A B A B A B

A B A B A B A B A B A B

= + + +

+ + + + +

= + +

i i j i k i i jj j k j i k j k k k

i j k

0 CuB

uA

uA -0 CuB

uA

uA uB


CHAPTER 3

46

(c) If A || B, , ,x x x xz zy y y y z z

A B A BA BA B A B A B

= = =

0x y y xA B A B =0x y y xA B A B =0z x x zA B A B =

0 =A B3-57.

+ y z x yx zx y zy z x z x y

x y z

i j k A A A AA AA A A

B B B B B BB B B

= i j k

= (AyBz AzBy) i + (AzBx AxBz) j + (AxBy AyBx) k3-58. [( ) ] ( ) = = = i j i i k i i j i k

[( ) ] = ( ) = = i j j j k j j i j k3-59. 2 3 2 .= +A i j k 3 4 .= +B i k (2 3 2 ) ( 3 0 4 )= + + +A B i j k i j k

( ) ( ) ( )y z z y z x x z x y y xA B A B A B A B A B A B= + + i j k[( 3)(4) (2)(0)] [(2)( 3) (2)(4)] [(2)(0) ( 3)( 3)] 12 14 9 .= + + = i j k i j k

3-60. (a) A (B + C) = (3 2 2 ) +i j k (2 3 4 ) +i j k

(b) A (B + C) = (3 2 2 ) +i j k (2 3 4 ) +i j k

= 3 2 22 3 4

i j k = (8 + 6)i (12 4)j + (9 + 4) = 2 8 5 k i j k

(c) A (B C) = x y z

x y z

x y z

A A AB B BC C C

= 3 2 20 0 42 3 0

= 3(12) + 2(8) + 2(0) 20=

(d) B C = 0 0 42 3 0

i j k = 12 i + 8 j

A (B C) = 3 2 2 16 24 4812 8 0

= + +

i j ki j k

3-61. The cross product of the two vectors will give a vector perpendicular to both.

( )4 3+i j ( 3 2 ) +i j k = 4 3 01 3 2

i j k = 6i 8j 9k

The magnitude of this vector is 2 2 26 8 9+ + = 181 = 13.45


CHAPTER 3

47

The unit vector perpendicular to the two vectors is then

( )1 6 8 9 = 0.44 0.59 0.6713.45

i j k i j k

3-62. Volume of parallelopiped is equal to base area height= Base area A cos The base area = BC sin .Hence the base area = B C.The volume then is | A B | A cos .But B C points vertically upward and the anglebetween it and A is . Socos = ( )

| | AB C AB C

i

V = |B C | A cos = | B C |( ) = ( )| | A

B C A A B CB C

i i

3-63. Pick B along the x-axis and C along the x-y plane.

xB=B i

x yC C= +C i i

x y zA A A= + +A i i k

Then,

(B C) = 0 00

x

x y

BC C

i j k = BxCy k

Therefore A (B C) = 0 0

x y z

x y

A A AB C

i j k = AyBxCy i AxBxCy j

But B(A C) C(A B) = Bx ( )x x y yA C A C+i (Cxi + Cyj)AxBx

= AxBxCx i + AyBxCy i AxBxCx i AxBxCy j = = ( )y x y x x yA B C A B Ci j A B C

3-64. Let x and y be along magnetic east and north, respectively.(a) Coordinates based on magnetic northE comp: 5.0 sin 56 4.15 km=N comp: 5.0 cos 56 2.80 km=

(b) Coordinates based on geographic northE comp: 5.0 sin 445 3.48 km=N comp: 5.0 cos 445 3.59 km=

Note: We can also use the rotation angle 1155E comp: 4.15 cos (1155) + 2.80 sin (1155) 3.48 km=

N comp: 4.15 sin (1155) + 2.80 cos (1155)3.59 km=


CHAPTER 3

48

3-65. If the axis of the original coordinate system is rotatedso that it points in the direction of this vector, theonly non-zero component will be the x component.This vector makes an angle tan1(3/6) = tan1(1/2)= 26.6 with the x-axis.Thus a rotation of 26.6 will do the job.The vector will have coordinates 2 26 3+ i = 6.71i

3-66. magnitude = 2 290 70 114 km+ =

direction: 1 70tan90

= = 37.9 west of north.3-67. By the law of cosines:

C 2 = A2 + B2 2AB cos 115= 3502 + 1202 + 2(350)(120)(0.4226)= 172,400 m2

415 mC =By the law of sines, the angle betweenA and C is given bysin120

= sin 115415

= sin1 120 sin 115 = 15.2415

Therefore, the angle between C and north is 45 15.2 = 29.8 and the resultant has a magnitude of415 m, at 29.8 W of N.

3-68. ave(5 2 4) (0 1 2) (3 3 1) 7 1

3 3 3 + + + + + +

= = + +i j kr i j k

| 2 2

2ave

7 1 59| 13 3 3

= + + = r

3-69. Horizontal component = 2.0 sin(30 ) = 1.0Vertical component = 2.0 cos(30 ) = 1.7

3-70. Magnitude = 2 cos(30 ) 250 = 433 km. The direction isdue south.

30

30

S


CHAPTER 3

49

3-71. A = 6.2 cos 30i 6.2 sin 30jB = 9.6jTherefore, A + B = 6.2 cos 30i + ( 6.2 sin 30 9.6)j = 5.4 12.7i jand A B = 6.2 cos 30i (6.2 sin 30 9.6)j = 5.4 6.5+i j

3-72. The displacement D = 4I + 5j + 3k m. | D | = 2 2 2(4) (5) (3) m+ + = 7.1 m.

By trial and error, one discovers that the shortest path across the room starting at the origin is

diagonally across the floor (4 m 5 m) and up the wall (3 m). The distance is ( )2 24 5+ + 3 m =9.4 m. If the lizard crawls only along the walls as the problem says, the shortest distance will bediagonally along the wall that is 4 m long and 3 m high then horizontally along the 5 m long wall.

Now the distance is ( )2 24 3+ + 5 m = 10 m.3-73. 2 2 22 1 ( 4)+ + = 4.583-74. (a) A + B = i 6j

(b) A B = 7 i + 2j(c) 3A B = 15i 2j

3-75. A = 50 sin 30 i + 50 cos 30 jB = 35 sin 70 i + 35 cos 70 jA B = (50)(35) {sin 30 sin 70 cos 30 cos 70}= 1750 (0.47 0.30) 2= 304 m

3-76. The angle between A and B is 100.Therefore, the magnitude of A B is| A B | = (6.0)(8.0) sin 100 m 47.3 m=By the right-hand rule, the direction of A B is downward.

3-77. If is the angle between A and B, then the component of A along B is A cos . But cos =(A B)/(AB). Therefore A cos = A

AB

A Bi= .

BA Bi B = 2 2 21 3 2+ + = 14 = 3.74.

A B = (3)(1) + (4)(3) + (0) (2) = 15. Then A cos = 15 4.0.3.74

=

Similarly, the component of B along A is A

A Bi with A = 2 23 4+ = 5. Thus, B cos

= 15 = 3.0.5

3-78. (a) A + B = 5i + j k(b) A B = i + 3j k(c) =iA B 6 2 + 0 = 4(d) [(2)(0) ( 1)( 1)] [( 1)(3) (0)(2)] [(2)( 1) (2)(3)] 3 8= + + = i j k i j kA B


50

CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS


4-1. (a) Total displacement is (3.2 + 2.6) km 45 E of N + 4.5 km 50 W of N = 7.0 km , 5 E of N(see Problem 3, Chapter 3)(b) Avg vel = r/t = 7.0 km/1.25 h, 5 E of N 5.6 km/h, 5 E of N=

(c) total distance 3.2 2.6 4.5 kmAvg speed 8.24 km/htime 1.25 h

+ += = =

4-2.11distance 1.50 10 mavg speed 29.9 km/s

time 1/ 2 yr 1/ 2 (365.25 24 3600) sr

= = = =

| Avg vel | = 11

7

2 2 1.50 10 m 19.0 km/s1/ 2 yr 1/ 2 3.16 10 s

rt

= = =

r

4-3. 171 km/h = 47.5 m/sIf the bullet is fired at an angle , its position vector is given by rbul = 366 sin t i + 366 cos t jFor the bird, rbird = (47.5)t i + 30 jFor the bird to be hit rbul = r bird.Comparing the x-components, we have 366 sin = 47.5 = sin1(47.5/366) = 7.45Then d = 30 tan = 30 tan 7.45 = 3.93 m.Thus the hunter must aim 3.93 m ahead of the bird in order to hit the bird.

4-4. At t = 0, the velocity is 30 km/h N,=v or0 (30 km/h) .= +v i j At t = 10 s,30 km/h @ 45 W of N,= v or

(21 km/h) (21 km/h) .= +v i j At t = 20 s,30 km/h W,=v or (30 km/h) 0 .= +v i j At t = 30 s,30 km/h @ 45 S of W,= v or

(21 km/h) (21 km/h) .= v i j At t = 40 s, the driver is

halfway around and v = 30 km/h S, or(30 km/h) 0 .= +v i j

4-5. r = (4 + 2t)i + (3 + 5t + 4t2)j + (2 2t 3t 2)k(a) v = dr/dt 2 (5 8 ) (2 6 )t t= + + +i j k(b) a = dv/dt 8 6= j k

magnitude, | a | = 2 2 2(8) (6) 10 m/s+ =

direction = tan1 6 37 (37 below the -axis in the - plane)8

y z y =

t = 0

t = 10 s

t = 20 s

t = 30 s

t = 40 s


CHAPTER 4

51

4-6. (a) The components of v are vx = dx/dt and vy = dy/dt. Also, the components of a are ax = dvx/dtand ay = dvy/dt.

sinxv Ab bt= cos yv Ab bt=2cos xa Ab bt=

2sin ya Ab bt=

(b) | v | = 2 2 2 2 2 2 ( sin ) ( cos ) (sin cosAb bt ab bt A b bt bt Ab + = + =

| a | = 2 2 2 2( cos ) ( sin )Ab bt Ab bt + = Ab22 2 2

cos sinbt bt Ab+ =

4-7. At t = 2.0 s, the missile has been falling with acceleration 20 (9.81 m/s )= a i j and horizontalvelocity equal to the velocity of the airplane. This means the missile is still directly below the

airplane. Its displacement relative to the plane is 2 2 2(9.81 m/s )(2.0 s)0 0

2 2gt

= = =r i j i j

0 (19.6 m) ,i j or 19.6 m @ 90 below the direction of travel of the airplane. At t = 3.0 s, whichis 1.0 s after igniting the engine, the acceleration is 2 2(6.0 m/s ) (9.81 m/s ) .= a i j Its

displacement during this 1.0 s interval is 2 2

1 2 2xa t gt

= r i j2 2 2 2 2 2 2(6 m/s ) (6 m/s )(1.0 s) (9.81 m/s )(1.0 s)

2 2 2 2t gt

= = =i j i j (3.0 m) (4.9 m) .i j Now

the missiles total displacement relative to the plane is2 (0 m) (19.6 m) (3.0 m) (4.9 m)= + r i j i j (3.0 m) (24.5 m) .= i j The magnitude is

2 22 (3.0 m) (24.5 m) 24.7 m.r = + = The direction is given by

1 24.5tan 83.0 ,3

= = or

24.7 m @ 83 below the direction of travel of the plane.

4-8. 2 2 3(5 4 ) (3 2 ) 0 m.t t t t= + + + +r i j k 2(5 8 ) (6 6 ) 0 m/s.d t t tdt

= = + + + +rv i j k

22

2 (8) (6 12 ) 0 m/s .d d tdt dt

= = = + + +r va i j k At t = 2.0 s, (21 m/s) (36 m/s) 0 m/s,= + +v i j k

which gives a speed of 2 2(21 m/s) (36 m/s)v = + = 42 m/s.

4-9. ,average t

=

rv where r is the total displacement and t = 1.5 h is the total elapsed time. To find

the total displacement, find the displacement during each part of the trip:1 (300 km/h @ 30 N of E) 0.50 h 150 km @ 30 N of E. = = r Taking the y direction to

point N and the x direction to point E, this is 1 (150 km)( cos 30) (150 km)( sin 30) = +r i j(130 km) (75 km) .= +i j For the second part of the trip,

2 (300 km/h @ 30 W of S) 1.0 h = r 300 km @ 30 W of S.= In terms of x and y, thisis 2 (300 km)( sin 30) (300 km)( cos 30) (150 km) (260 km) . = = r i j i j The total

displacement is 1 2 20 km 185 km . = + = r r r i j The average velocity is average t

=

rv


CHAPTER 4

52

20 km 185 km 13.3 km/h 123 km/h .1.5 h

= =

i j i j To give this as a

speed and heading, use 2 213.3 123 km/h 124 km/h.averagev = + = Since

both velocity components are negative, the vector is located in the thirdquadrant (between W and S), so we can give the direction

as 1 123tan 83.8 S of W.13.3

= = Thus vaverage = 124 km/h @ 83.8 S of

W.

The average acceleration is 2 1 ,average t t

= =

v vva where v1 is the velocity during the 0.5, v2 is

the velocity during the next 1.0 h, and t is the time interval during which the velocity changes.Since no value is given for t, we can calculate v and give the answer symbolically. Using thedefinitions above, 1 (300 km/h) cos 30 (300 km/h) sin 30= + v i j (260 km/h) (150 km/h) .= +i j30 W of S is the same as 240 N of E, so 2 (300 km/h) cos 240 (300 km/h) sin 240= + v i j

(150 km/h) (240 km/h) .= i j Then (410 km/h) (410 km/h) , = v i j

and (410 km/h) (410 km/h) .average t

=

i ja As a magnitude and heading, this

is 410 2 km/h @ 45 S of W,average t=

a or 580 km/h @ 45 S of W.average t

=

a Even though we

dont have a numerical value, it is important to note that the average acceleration is not zero, eventhough the speed of the airplane is constant. The acceleration is caused by the change in directionof the velocity.

4-10. cos .A bt Bt= +r i j sin .d bA bt Bdt

= = +rv i j

22

2 cos 0 .d d b A btdt dt

= = = +r va i j The equation

for the speed is 2 2 2 2( sin ) .x yv v v bA bt B= + = +

4-11. 90 (500 15 )t t= + r i j m. 90 15 m/s.ddt

= =

rv i j The equation

for the speed is2 2 2 2(90 m/s) (15 m/s) 91 m/s.x yv v v= + = + = The direction

of the velocity is 1 1 15tan tan 9.5.90

y

x

vv

= = = This is 9.5 below the x axis.

vaverage

E

N

90 m/s

y

x15 m/s


CHAPTER 4

53

4-12. Measure time from when the ball is thrown. At that instant, thereceiver has traveled (7 m/s)(2.0 s) = 14 m along the x directionrelative to the quarterback. The receiver is at an initialdisplacement 0 (14 m) (20 m)= +r i j relative to the quarterback.Since hes running at a constant velocity of vR = 7.0i m/s, his totaldisplacement is (14 7 ) 20 R t= + +r i j m. Once its thrown, thedisplacement of the ball relative to the quarterback is

, , ( sin ) ( cos ) ,B B x B y B Bv t v t v t v t = + = +r i j i j where vB = 15 m/s

Documents

Física para Ingeniería y Ciencias 3ra Edicion Hans Ohanian.pdf