192266681 Solution of Physics for Scientists and Engineers 3rd Edition by Douglas C Giancoli

825  Download (0)

Full text

(1)Chapter 1 CHAPTER 1 - Introduction, Measurement, Estimating 1.. (a) Assuming one significant figure, we have 11010 yr. 10 billion yr = 10109 yr = 10 7 17 310 s. (b) (110 yr)(310 s/yr) =. 2.. (a) (b) (c) (d) (e) (f) (g). 4 significant figures. Because the zero is not needed for placement, we have 4 significant figures. 3 significant figures. Because the zeros are for placement only, we have 1 significant figure. Because the zeros are for placement only, we have 2 significant figures. 4 significant figures. 2, 3, or 4 significant figures, depending on the significance of the zeros.. 3.. (a) (b) (c) (d) (e) (f). 1,156 = 1.156103. 21.8 = 2.18101. 0.0068 = 6.810–3. 27.635 = 2.7635101. 0.219 = 2.1910–1. 22 = 2.2101.. 4.. (a) (b) (c) (d) (e). 8.69104 = 7.1103 = 6.610–1 = 8.76102 = 8.6210–5 =. 5.. % uncertainty = [(0.25 m)/(3.26 m)] 100 = 7.7%. Because the uncertainty has 2 significant figures, the % uncertainty has 2 significant figures.. 6.. We assume an uncertainty of 1 in the last place, i. e., 0.01, so we have % uncertainty = [(0.01 m)/(1.28 m)] 100 = 0.8%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.. 7.. We assume an uncertainty of 0.2 s. (a) % uncertainty = [(0.2 s)/(5 s)] 100 = 4%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. (b) % uncertainty = [(0.2 s)/(50 s)] 100 = 0.4%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. (c) % uncertainty = [(0.2 s)/(5 min)(60 s/min)] 100 = 0.07%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.. 8.. For multiplication, the number of significant figures in the result is the least number from the multipliers; in this case 2 from the second value. (2.079102 m)(0.07210–1) = 0.15101 m = 1.5 m.. 9.. To add, we make all of the exponents the same: 9.2103 s + 8.3104 s + 0.008106 s = 0.92104 s + 8.3104 s + 0.8104 s 1.0105 s. = 10.02104 s = Because we are adding, the location of the uncertain figure for the result is the one furthest to the left. In this case, it is fixed by the third value.. 86,900. 7,100. 0.66. 876. 0.000 086 2.. Page 1.

(2) Chapter 1 10. We assume an uncertainty of 0.1104 cm. We compare the area for the specified radius to the area for the extreme radius. A1 = pR12 = p(3.8104 cm)2 = 4.54109 cm2 ; A2 = pR22 = p[(3.8 + 0.1)104 cm]2 = 4.78109 cm2 , so the uncertainty in the area is ?A =A2 – A1 = 0.24109 cm2 = 0.2109 cm2. We write the area as A = (4.5 ± 0.2)109 cm2. We could also treat the change as a differential: dA =2pR dR = 2p(3.8104 cm)(0.1104 cm) = 2108 cm2. 11. We compare the volume with the specified radius to the volume for the extreme radius. V1 = )pR13 = )p(2.86 m)3 = 98.0 m3; V2 = )pR23 = )p(2.86 m + 0.08 m)3 = 106.45 m3, so the uncertainty in the volume is ?V =V2 –V1 = 8.5 m3; and the % uncertainty is % uncertainty = [(8.5 m3)/(98.0 m3)] 100 = 9%. 12. (a) (b) (c) (d) (e). 106 volts = 1 megavolt = 1 Mvolt. 10–6 meters = 1 micrometer = 1 µm. 6103 days = 6 kilodays = 6 kdays. 18102 bucks = 18 hectobucks = 1.8 kbucks. 810–9 pieces = 8 nanopieces = 8 npieces.. 13. (a) (b) (c) (d) (e) (f). 286.6 mm = 286.610–3 m = 0.286 6 m. 85 µV = 8510–6 V = 0.000 085 V. 760 mg = 76010–3 g = 76010–6 g = 0.000 760 kg. This assumes that the last zero is significant. 60.0 picoseconds = 60.010–12 s = 0.000 000 000 060 0 s. 22.5 femtometers = 22.510–15 m = 0.000 000 000 000 022 5 m. 2.50 gigavolts = 2.50109 volts = 2,500,000,000 volts.. 14. 50 hectocars = 50102 cars = 5,000 cars. 1 megabuck/yr = 1106 bucks/yr = 1,000,000 bucks/yr 15. If we assume a height of 5 ft 10 in, we have height = 5 ft 10 in = 70 in = (70 in)[(1 m)/(39.37 in)] =. (millionaire). 1.8 m.. 16. (a) 93 million mi = 93106 mi = (93106 mi)[(1610 m)/(1 mi)] = 150Gm. (b) 1.51011 m = 150109 m =. 1.51011 m.. 17. (a) 1 ft2 = (1 ft2)[(1 yd)/(3 ft)]2 = 0.111 yd2. 2 2 2 10.76 ft2. (b) 1 m = (1 m )[(3.28 ft)/(1 m)] = 18. We find the time from time = distance/speed = [(1.0 mi)(1.61 km/mi)]/[2300 km/h)/(3600 s/h)] =. 2.5 s.. 19. (a) 1.010–10 m = (1.010–10 m)[(1 in)/(2.54 cm)][(100 cm)/(1 m)] = 3.910–9 in. (b) We let the units lead us to the answer: 1.0108 atoms. (1.0 cm)[(1 m)/(100 cm)][(1 atom)/(1.010–10 m)] = 20. To add, we make all of the units the same: 2.00 m + 142.5 cm + 7.24105 µm = 2.00 m + 1.425 m + 0.724 m = 4.149m = 4.15m. Because we are adding, the location of the uncertain figure for the result is the one furthest to the left. In this case, it is fixed by the first value.. Page 2.

(3) Chapter 1 21. (a) 1 km/h = (1 km/h)[(0.621 mi)/(1 km)] = 0.621 mi/h. (b) 1 m/s = (1 m/s)[(1 ft)/(0.305 m)] = 3.28 ft/s. (c) 1 km/h = (1 km/h)[(1000 m)/(1 km)][(1 h)/(3600 s)] = 0.278 m/s. A useful alternative is 1 km/h = (1 km/h)[(1 h)/(3.600 ks)] = 0.278 m/s. 22. For the length of a one-mile race in m, we have 1 mi = (1 mi)[(1610 m)/(1 mi)] = 1610 m; so the difference is 110 m. The % difference is % difference = [(110 m)/(1500 m)] 100 = 7.3%. = (2.998108 m/s)(1.00 yr)[(365.25 days)/(1 yr)][(24 h)/(1 day)][(3600 s)/(1 h)] = 9.461015 m. 6.31104 AU. (b) 1.00 ly = (9.461015 m)[(1 AU)/(1.50108 km)][(1 km)/(1000 m)] = 8 8 (c) speed of light = (2.99810 m/s)[(1 AU)/(1.5010 km)][(1 km)/(1000 m)][(3600 s)/(1 h)] = 7.20 AU/h.. 23. (a) 1.00 ly. 24. For the surface area of a sphere, we have AMoon = 4prMoon2 = 4p[!(3.48106 m)]2 = 3.801013 m2. We compare this to the area of the earth by finding the ratio: AMoon /AEarth = 4prMoon2 /4prEarth2 = (rMoon/rEarth)2 = [(1.74103 km)/(6.38103 km)]2 = 7.44 x 10–2. Thus we have AMoon = 7.44 x 10–2 AEarth. 25. (a) (b) (c) (d). 2800 = 2.8103 ˜ 1103 = 103. 86.30102 ˜ 100102 = 104. 0.0076 = 0.7610–2 ˜ 10–2. 15.0108 = 1.50109 ˜ 109.. 26. We assume that a good runner can run 6 mi/h (equivalent to a 10-min mile) for 5 h/day. Using 3000 mi for the distance across the United States, we have time = (3000 mi)/(6 mi/h)(5 h/day) ˜ 100 days. 27. We assume a rectangular house 40 ft30 ft, 8 ft high; so the total wall area is Atotal = [2(40 ft) + 2(30 ft)](8 ft) ˜ 1000 ft2. If we assume there are 12 windows with dimensions 3 ft5 ft, the window area is Awindow = 12(3 ft)(5 ft) ˜ 200 ft2. Thus we have % window area = [Awindow/Atotal ](100) = [(200 ft2)/(1000 ft2)](100) ˜ 20%. 28. If we take an average lifetime to be 70 years and the average pulse to be 60 beats/min, we have N = (60 beats/min)(70 yr)(365 day/yr)(24 h/day)(60 min/h) ˜ 2109 beats. 29. If we approximate the body as a box with dimensions 6 ft, 1 ft, and 8 in, we have volume = (72 in)(12 in)(8 in)(2.54 cm/in)3 ˜ 1105 cm3.. Page 3.

(4) Chapter 1 30. We assume the distance from Beijing to Paris is 10,000 mi. (a) If we assume that today’s race car can travel for an extended period at an average speed of 40 mi/h, we have time = [(10,000 mi)/(40 mi/h)](1 day/24 h) ˜ 10 days. (b) If we assume that in 1906 a race car could travel for an extended period at an average speed of 5 mi/h, we have time = [(10,000 mi)/(5 mi/h)](1 day/24 h) ˜ 80 days. 31. We assume that each dentist sees 10 patients/day, 5 days/wk for 48 wk/yr, for a total number of visits of Nvisits = (10 visits/day)(5 days/wk)(48 wk/yr) ˜ 2400 visits/yr. We assume that each person sees a dentist 2 times/yr. (a) We assume that the population of San Francisco is 700,000. We let the units lead us to the answer: N = (700,000)(2 visits/yr)/(2400 visits/yr) ˜ 600 dentists. (b) Left to the reader to estimate the population. 32. If we assume that the person can mow at a speed of 1 m/s and the width of the mower cut is 0.5 m, the rate at which the field is mown is (1 m/s)(0.5 m) = 0.5 m2/s. If we take the dimensions of the field to be 110 m by 50 m, we have time = [(110 m)(50 m)/(0.5 m2/s)]/(3600 s/h) ˜ 3 h. 33. We assume an average time of 3 yr for the tire to wear d = 1 cm and the tire has a radius r = 30 cm and a width w = 10 cm. Thus the volume of rubber lost by a tire in 3 yr is V = wd2pr. If we assume there are 100 million vehicles, each with 4 tires, we have m = (100106 vehicles)(4 tires/vehicle)(0.1 m)(0.01 m)2p(0.3 m/tire)(1200 kg/m3)/(3 yr) ˜ 3108 kg/yr. 34. We assume that one-third of the floor space can be used for shelves and there are five shelves, each with a depth of 25 cm, in a stack. If the average book has a width of 4.0 cm, we have N = [@(1500 m2)(5 shelves)/(4.0 cm/book)(25 cm/shelf)](100 cm/m)2 = 2.5105 books. 35. For the right triangle shown on the diagram we have r2 + d2 = (r + h)2; d2 = 2rh + h2 = 2(6.4106 m)(200 m) + (200 m)2, which gives d = 5.1104 m = 51 km.. 36. Because every term must have the same dimensions, we have v = At3 – Bt; [L/T] = [A][T]3 – [B][T]. Thus we get [A] = [L/T]/[T3] = [L/T4], and [B] = [L/T]/[T] = [L/T2]. 37. For the units we have A = [L/T4] = m/s4,. and. B = [L/T2] = m/s2.. Page 4.

(5) Chapter 1. 38. We test to see if each term has the same dimensions. (a) x = vt2 + 2at; [L] =? [L/T][T]2 + [L/T2][T]; [L] ? [LT] + [L/T]; therefore, this is not correct. 2 (b) x = v0t + !at ; [L] =? [L/T][T] + [L/T2][T]2; [L] = [L] + [L]; therefore, this may be correct. (c) x = v0t + 2at2; [L] =? [L/T][T] + [L/T2][T]2; [L] = [L] + [L]; therefore, this may be correct. 39. (a) (b) (c) (d). 1.0 Å = (1.010–10 m)/(10–9 m/nm) = 0.10 nm. 1.0 Å = (1.010–10 m)/(10–15 m/fm) = 1.0105 fm. 1.0 m = (1.0 m)/(10–10 m/Å) = 1.01010 Å. From the result for Problem 23, we have 9.51025 Å. 1.0 ly = (9.51015 m)/(10–10 m/Å) =. 40. We use the values for the masses from Table 1–3. (a) Nbacterium = (10–15 kg)/(10–27 kg/proton) = 1012 protons (or neutrons). –17 –27 10 (b) NDNA = (10 kg)/(10 kg/proton) = 10 protons (or neutrons). 2 –27 (c) Nhuman = (10 kg)/(10 kg/proton) = 1029 protons (or neutrons). 41 –27 (d) Ngalaxy = (10 kg)/(10 kg/proton) = 1068 protons (or neutrons). 41. (a) 1.00 yr = (365.25 days)(24 h/day)(3600 s/h) = 3.16107 s 7 –9 16 3.1610 ns. (b) 1.00 yr = (3.1610 s)/(10 s/ns) = 3.1710–8 yr. (c) 1.00 s = (1.00 s)/(3.16107 s/yr) = 42. 1 hectare = (104 m2)[(3.28 ft)/(1 m)]2[(1 acre)/(4104 ft2)] =. ˜ p107 s.. 2.69 acres.. 43. (a) The maximum number of buses is needed during rush hour. If we assume that at any time there are 40,000 persons commuting by bus and each bus has 30 passengers, we have N = (40,000 commuters)/(30 passengers/bus) ˜ 1000 buses ˜ 1,000 drivers. (b) Left to the reader. 44. If we ignore any loss of material from the slicing, we find the number of wafers from (30 cm)(10 mm/cm)/(0.60 mm/wafer) = 500 wafers. For the maximum number of chips, we have (500 wafers)(100 chips/wafer) = 50,000 chips. 45. If we assume there is 1 automobile for 2 persons and a U. S. population of 250 million, we have 125 million automobiles. We estimate that each automobile travels 15,000 miles in a year and averages 20 mi/gal. Thus we have N = (125106 automobiles)(15,000 mi/yr/automobile)/(20 mi/gal) ˜ 11011 gal/yr. 46. We let D represent the diameter of a gumball. Because there are air gaps around the gumballs, we estimate the volume occupied by a gumball as a cube with volume D3 . The machine has a square crosssection with sides equivalent to 10 gumballs and is about 14 gumballs high, so we have N = volume of machine/volume of gumball = (14D)(10D)2/D3 ˜ 1.4103 gumballs.. Page 5.

(6) Chapter 1. 47. We will convert all units to meters. The volume used in one year is V = [(40,000 persons)/(4 persons/family)](1200 L/family · day)  (365 days/yr)(10–3 m3/L) = 4.4106 m3/yr.           If we let d represent the loss in depth, we have d = V/area = (4.4106 m3/yr)/(50 km2)(103 m/km)2 ˜ 0.09 m ˜ 9 cm/yr. 48. For the volume of a 1-ton rock, we have V = (2000 lb)/(3)(62 lb/ft3) ˜ 11 ft3. If we assume the rock is a sphere, we find the radius from 11 ft3 = )pr3, which gives r ˜ 1.4 ft, so the diameter would be. ˜ 3 ft ˜ 1 m.. 49. We find the amount of water from its volume: m = (5 km)(8 km)(1.0 cm)(105 cm/km)2(10–3 kg/cm3)/(103 kg/t) = 40104 t ˜. 4105 t.. 50. We will use a pencil with a diameter of 5 mm and assume that it is held 0.5 m from the eye. Because the triangles AOD and BOC are similar, we can equate the ratio of distances: BC/AD = OQ/OP; BC/(0.005 m) = (3.8105 km)/(0.5 m), which gives BC ˜ 4103 km.. 51. We assume that we can walk an average of 15 miles a day. If we ignore the impossibility of walking on water and travel around the equator, the time required is time = 2prEarth/speed = 2p(6103 km)(0.621 mi/km)/(15 mi/day)(365 days/yr) ˜ 4 yr. 52. If we use 0.5 m for the cubit, for the dimensions we have 150 m long, 25 m wide, and 15 m high. 53. From the diagram we see that d/L = tan ; (120 yd)/L = tan 30°, which gives L = 2.1102 yd = (2.1102 yd)(3 ft/yd)/(3.28 ft/m) =. 1.9102 m.. 54. We assume the oil slick is circular with diameter D and thickness d. For the volume we have V = d(pr2) = #dpD2; (1 L)(1000 cm3/L)/(1102 cm/m)3 = #(210–10 m)pD2, which gives D ˜ 3103 m.. Page 6.

(7) Chapter 1. 55. The trigonometric function is not a linear function of the angle, so we can find the uncertainty in the sine by calculating two values. 3%. (a) Percent uncertainty in  = (0.5°/15.0°) 100 = We find the percent uncertainty in the sine from sin 15.0° = 0.2588; sin 15.5° = 0.2672; Percent uncertainty in sin  = [(0.2672 – 0.2588)/0.2588] 100 = 3%. 0.7%. (b) Percent uncertainty in  = (0.5°/75.0°) 100 = We find the percent uncertainty in the sine from sin 75.0° = 0.9659; sin 75.5° = 0.9681; Percent uncertainty in sin  = [(0.9681 – 0.9659)/0.9659] 100 = 0.2%. Note that it is possible to approximate uncertainties with differential quantities, if the angle is not very small (i. e., sin  is not small). We have d(sin ) = cos  d; d(sin )/(sin ) = d/(tan ). The angle must be in radians, so we get d(sin )/(sin ) = [(0.5°)(p/180°)]/tan 15.0° = 0.03 = 3%; d(sin )/(sin ) = [(0.5°)(p/180°)]/tan 75.0° = 0.002 = 0.2%. 56. Because you are lying on the sand, your line of sight is tangent to the surface of the Earth. If the height of the hull is h, from the triangle on the diagram we have r2 + d2 = (r + h)2; d2 = 2rh + h2 = 2(6.38106 m)(2.5 m) + (2.5 m)2, which gives d = 5.6103 m = 5.6 km.. Page 7.

(8) Chapter 2. CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1.. We find the time from average speed = d/t; 15 km/h = (75 km)/t , which gives. t = 5.0 h.. 2.. We find the average speed from average speed = d/t = (280 km)/(3.2 h) =. 3.. We find the distance traveled from average speed = d/t; (110 km/h)/(3600 s/h) = d/(2.0 s), which gives. 88 km/h.. d= 6.110–2 km = 61 m.. 4.. We find the average velocity from æ = (x2 – x1)/(t2 – t1) = (– 4.2 cm – 3.4 cm)/(6.1 s – 3.0 s) =. 5.. We find the average velocity from æ = (x2 – x1)/(t2 – t1) = (8.5 cm – 3.4 cm)/[4.5 s – (– 2.0 s)] = 0.78 cm/s (toward + x). Because we do not know the total distance traveled, we cannot calculate the average speed.. 6.. (a) We find the elapsed time before the speed change from speed = d1/t1 ; 65 mi/h = (130 mi)/t1 , which gives t1 = 2.0 h. Thus the time at the lower speed is t2 = T – t1 = 3.33 h – 2.0 h = 1.33 h. We find the distance traveled at the lower speed from speed = d2/t2 ; 55 mi/h = d2/(1.33 h), which gives d2 = 73 mi. The total distance traveled is D = d1 + d2 = 130 mi + 73 mi = 203 mi. (b) We find the average speed from average speed = d/t = (203 mi)/(3.33 h) = 61 mi/h. Note that the average speed is not !(65 mi/h + 55 mi/h). The two speeds were not maintained for equal times.. 7.. Because there is no elapsed time when the light arrives, the sound travels one mile in 5 seconds. We find the speed of sound from speed = d/t = (1 mi)(1610 m/1 mi)/(5 s) ˜ 300 m/s.. 8.. (a) We find the average speed from average speed = d/t = 8(0.25 mi)(1.61103 m/mi)/(12.5 min)(60 s/min) = (b) Because the person finishes at the starting point, there is no displacement; thus the average velocity is æ = x/t = 0.. 9.. (a) (b). – 2.5 cm/s (toward – x).. 4.29 m/s.. We find the average speed from average speed = d/t = (160 m + 80 m)/(17.0 s + 6.8 s) = 10.1 m/s. The displacement away from the trainer is 160 m – 80 m = 80 m; thus the average velocity is æ = x/t = (80 m)/(17.0 s + 6.8 s) = + 3.4 m/s, away from trainer.. 10. Because the two locomotives are traveling with equal speeds in opposite directions, each locomotive will travel half the distance, 4.25 km. We find the elapsed time from Page 1.

(9) Chapter 2. speed1 = d1/t ; (95 km/h)/(60 min/h) = (4.25 km)/t, which gives. t = 2.7 min.. x (m). 11. (a) We find the instantaneous velocity from the slope of the straight line from t = 0 to t = 10.0 s: 20 v10 = ?x/?t = (2.8 m – 0)/(10.0 s – 0) = 0.28 m/s. (b) We find the instantaneous velocity from the slope 10 of a tangent to the line at t = 30.0 s: v30 = ?x/?t = (22 m – 10 m)/(35 s – 25 s) = 1.2 m/s. 0 0 (c) The velocity is constant for the first 17 s (a straight line), so the velocity is the same as the velocity at t = 10 s: æ05 = 0.28 m/s. (d) For the average velocity we have æ2530 = ?x/?t = (16 m – 8 m)/(30.0 s – 25.0 s) = 1.6 m/s. (e) For the average velocity we have æ4050 = ?x/?t = (10 m – 20 m)/(50.0 s – 40.0 s) = – 1.0 m/s.. 10. 20. 30. 12. (a) (b) (c) (d). Constant velocity is indicated by a straight line, which occurs from t = 0 to 17 s. The maximum velocity is when the slope is greatest: t = 28 s. Zero velocity is indicated by a zero slope. The tangent is horizontal at t = 38 s. Because the curve has both positive and negative slopes, the motion is in both directions.. 13. (a). We find the average speed from 13.4 m/s. average speed = d/t = (100 m + 50 m)/[8.4 s + @(8.4 s)] = The displacement away from the trainer is 100 m – 50 m = 50 m; thus the average velocity is æ = x/t = (50 m)/[8.4 s + @(8.4 s)] = + 4.5 m/s, away from master.. (b). 40. t (s). 14. (a) We find the position from the dependence of x on t: x = 2.0 m – (4.6 m/s)t + (1.1 m/s2)t2. x1 = 2.0 m – (4.6 m/s)(1.0 s) + (1.1 m/s2)(1.0 s)2 = – 1.5 m; x2 = 2.0 m – (4.6 m/s)(2.0 s) + (1.1 m/s2)(2.0 s)2 = – 2.8 m; 2 2 x3 = 2.0 m – (4.6 m/s)(3.0 s) + (1.1 m/s )(3.0 s) = – 1.9 m. (b) For the average velocity we have æ13 = ?x/?t = [(– 1.9 m) – (– 1.5 m)]/(3.0 s – 1.0 s) = – 0.2 m/s (toward – x). (c) We find the instantaneous velocity by differentiating: v = dx/dt = – (4.6 m/s) + (2.2 m/s2)t; v2 = – (4.6 m/s) + (2.2 m/s2)(2.0 s) = – 0.2 m/s (toward – x); v3 = – (4.6 m/s) + (2.2 m/s2)(3.0 s) = + 2.0 m/s (toward + x). 15. Because the velocities are constant, we can use the relative speed of the car to find the time: t = d/vrel = [(0.100 km)/(90 km/h – 75 km/h)](60 min/h) = 0.40 min = 24 s. 16. We find the total time for the trip by adding the times for each leg: T = t1 + t2 = (d1/v1) + (d2/v2) = [(2100 km)/(800 km/h)] + [(1800 km)/(1000 km/h)] = 4.43 h. We find the average speed from average speed = (d1 + d2)/T = (2100 km + 1800 km)/(4.43 h) = 881 km/h. Note that the average speed is not !(800 km/h + 1000 km/h). The two speeds were not maintained for equal times.. Page 2. 50.

(10) Chapter 2. 17. We find the time for the outgoing 200 km from t1 = d1/v1 = (200 km)/(90 km/h) = 2.22 h. We find the time for the return 200 km from t2 = d2/v2 = (200 km)/(50 km/h) = 4.00 h. We find the average speed from average speed = (d1 + d2)/(t1 + tlunch + t2) = (200 km + 200 km)/(2.22 h + 1.00 h + 4.00 h) = 55 km/h. Because the trip finishes at the starting point, there is no displacement; thus the average velocity is æ = x/t = 0. 18. If. vAG is the velocity of the automobile with respect to the ground, vTG the velocity of the train with respect to the ground, and vAT the velocity of the automobile with respect to the train, then vAT = vAG – vTG . If we use a coordinate system in the reference frame of the train with the origin at the back of the train, we find the time to pass from x1 = vATt1 ; 1.10 km = (90 km/h – 80 km/h)t1 , which gives t1 = 0.11 h = 6.6 min. With respect to the ground, the automobile will have traveled x1G = vAGt1 = (90 km/h)(0.11 h) = 9.9 km. If the automobile is traveling toward the train, we find the time to pass from x2 = vATt2 ; 1.10 km = [90 km/h – (– 80 km/h)]t2 , which gives t2 = 0.00647 h = 23.3 s. With respect to the ground, the automobile will have traveled x2G = vAGt2 = (90 km/h)(0.00647 h) = 0.58 km.. 19. We find the time for the sound to travel the length of the lane from tsound = d/vsound = (16.5 m)/(340 m/s) = 0.0485 s. We find the speed of the ball from v = d/(T – tsound) = (16.5 m)/(2.50 s – 0.0485 s) = 6.73 m/s. 20. We find the average acceleration from Æ = v/t = [(95 km/h)(1 h/3.6 ks) – 0]/(6.2 s) =. 4.3 m/s2.. 21. We find the time from Æ = v/t ; 1.6 m/s2 = (110 km/h – 80 km/h)(1 h/3.6 ks)/t, which gives. Page 3. t =. 5.2 s..

(11) Chapter 2. 22.. v (m/s). 40 30 20 10 0. 0. 10. 20. 30. 40. 50. 60. 70. 80. 90 100 110 120. t (s) (a) The maximum velocity is indicated by the highest point, which occurs at (b) Constant velocity is indicated by a horizontal slope, which occurs from (c) Constant acceleration is indicated by a straight line, which occurs from t = 0 to 30 s, and t = 90 s to 107 s. (d) The maximum acceleration is when the slope is greatest: t = 75 s. 23. We find the acceleration (assumed to be constant) from v2 = v02 + 2a(x2 – x1); 0 = [(100 km/h)/(3.6 ks/h)]2 + 2a(55 m), which gives The number of g’s is N =  a/g = (7.0 m/s2)/(9.80 m/s2) = 0.72.. t = 50 s. t = 90 s to 107 s.. a = – 7.0 m/s2.. . 24. For the average acceleration we have Æ2 = ?v/?t = (24 m/s – 14 m/s)/(8 s – 4 s) = 2.5 m/s2; Æ4 = ?v/?t = (44 m/s – 37 m/s)/(27 s – 16 s) = 0.6 m/s2.. 50 5th gear 4th gear. v (m/s). 40 3rd gear. 30 20. 2nd gear. 10 0. 1st gear 0. 10. 20. t (s). 25. (a) For the average acceleration we have Æ1 = ?v/?t = (14 m/s – 0)/(3 s – 0) = 4.7 m/s2. (b) For the average acceleration we have Æ3 = ?v/?t = (37 m/s – 24 m/s)/(14 s – 8 s) = 2.2 m/s2. (c) For the average acceleration we have Æ5 = ?v/?t = (52 m/s – 44 m/s)/(50 s – 27 s) = 0.3 m/s2. (d) For the average acceleration we have Page 4. 30. 40.

(12) Chapter 2. Æ14 = ?v/?t = (44 m/s – 0)/(27 s – 0) = 1.6 m/s2. Note that we cannot add the four average accelerations and divide by 4.. 27. We find the velocity and acceleration by differentiating x = (6.0 m/s)t + (8.5 m/s2)t2: v = dx/dt = (6.0 m/s) + (17 m/s2)t; a = dv/dt = 17 m/s2. 28. The position is given by x = At + 6Bt3. (a) All terms must give the same units, so we have A ~ x/t = m/s; and B ~ x/t3 = m/s3. (b) We find the velocity and acceleration by differentiating: v = dx/dt = A + 18Bt2; a = dv/dt = 36Bt. Page 5. 6 5 4 3 2 1 0. a (m/s 2 ). v (m/s). 26. (a) We take the average velocity during a time interval as the instantaneous velocity at the midpoint of the time interval: vmidpoint = æ = x/t. Thus for the first interval we have v0.125 s = (0.11 m – 0)/(0.25 s – 0) = 0.44 m/s. (b) We take the average acceleration during a time interval as the instantaneous acceleration at the midpoint of the time interval: amidpoint = Æ = v/t. Thus for the first interval in the velocity column we have a0.25 s = (1.4 m/s – 0.44 m/s)/(0.375 s – 0.125 s) = 3.8 m/s2. The results are presented in the following table and graph. t(s) x(m) t(s) v(m/s) t(s) a(m/s2) 0.0 0.0 0.0 0.0 acceleration 0.125 0.44 0.25 0.11 0.25 3.8 0.375 1.4 0.50 0.46 0.50 4.0 30 0.625 2.4 0.75 1.06 0.75 4.5 0.875 3.5 1.00 1.94 1.06 4.9 20 1.25 5.36 1.50 4.62 1.50 5.0 1.75 7.85 2.00 8.55 2.00 5.2 2.25 10.5 10 2.5013.79 2.50 5.3 velocity 2.75 13.1 3.0020.36 3.00 5.5 3.25 15.9 3.50 28.31 3.50 5.6 0 3.75 18.7 0 1 2 3 4 5 6 4.00 37.65 4.00 5.5 t (s) 4.25 21.4 4.5048.37 4.50 4.8 4.75 23.9 5.0060.30 5.00 4.1 5.25 25.9 5.5073.26 5.50 3.8 5.75 27.8 6.0087.16 Note that we do not know the acceleration at t = 0..

(13) Chapter 2. (c) For the given time we have v = dx/dt = A + 18Bt2 = A + 18B(5.0 s)2 = A + (450 s2)B; a = dv/dt = 36Bt = 36B(5.0 s) = (180 s)B. (d) We find the velocity by differentiating: v = dx/dt = A – 3Bt–4.. 29. We find the acceleration from v = v0 + a(t – t0); 21 m/s = 12 m/s + a(6.0 s), which gives We find the distance traveled from x = !(v + v0)t = !(21 m/s + 12 m/s)(6.0 s) =. a = 1.5 m/s2.. 99 m.. 30. We find the acceleration (assumed constant) from v2 = v02 + 2a(x2 – x1); 0 = (25 m/s)2 + 2a(75 m), which gives a = – 4.2 m/s2. 31. We find the length of the runway from v2 = v02 + 2aL; (32 m/s)2 = 0 + 2(3.0 m/s2)L, which gives 32. We find the average acceleration from v2 = v02 + 2Æ(x2 – x1); (44 m/s)2 = 0 + 2Æ(3.5 m), which gives. Æ = 2.8102 m/s2.. 33. We find the average acceleration from v2 = v02 + 2Æ(x2 – x1); (11.5 m/s)2 = 0 + 2Æ(15.0 m), which gives We find the time required from x = !(v + v0)t ; 15.0 m = !(11.5 m/s + 0), which gives. L = 1.7102 m.. Æ = 4.41 m/s2.. t = 2.61 s.. 34. The average velocity is æ = (x – x0)/(t – t0), so we must find the position as a function of time. Because the acceleration is a function of time, we find the velocity by integrating a = dv/dt : ? dv = ? a dt = ? (A + Bt) dt; v = At + (Bt2/2) + C. If v = v0 when t = 0, C = v0; so we have v = At + (Bt2/2) + v0. We find the position by integrating v = dx/dt : ? dx = ? v dt = ? [At + (Bt2/2) + v0] dt; x = (At2/2)+ (Bt3/6) + v0t + D. If x = x0 when t = 0, D = x0; so we have x = (At2/2)+ (Bt3/6) + v0t + x0. Thus the average velocity is æ = (x – x0)/(t – t0) = {[(At2/2)+ (Bt3/6) + v0t + x0] – x0}/(t – 0) = (At/2)+ (Bt2/6) + v0. If we evaluate (v + v0)/2, we get (v + v0)/2 = {[At + (Bt2/2) + v0] + v0}/2 = (At/2)+ (Bt2/4) + v0 , which is not the average velocity. Page 6.

(14) Chapter 2. 35. For the constant acceleration the average speed is !(v + v0), thus x = !(v + v0)t: = !(0 + 22.0 m/s)(5.00 s) =. 55.0 m.. 36. We find the speed of the car from v2 = v02 + 2a(x1 – x0); 0 = v02 + 2(– 7.00 m/s2)(75 m), which gives. v0 = 32 m/s.. 37. We convert the units for the speed: (55 km/h)/(3.6 ks/h) = 15.3 m/s. (a) We find the distance the car travels before stopping from v2 = v02 + 2a(x1 – x0); 0 = (15.3 m/s)2 + 2(– 0.50 m/s2)(x1 – x0), which gives x1 – x0 = 2.3102 m. (b) We find the time it takes to stop the car from v = v0 + at ; 0 = 15.3 m/s + (– 0.50 m/s2)t, which gives t = 31 s. (c) With the origin at the beginning of the coast, we find the position at a time t from x = v0t + !at2. Thus we find x1 = (15.3 m/s)(1.0 s) + !(– 0.50 m/s2)(1.0 s)2 = 15 m; x4 = (15.3 m/s)(4.0 s) + !(– 0.50 m/s2)(4.0 s)2 = 57 m; x5 = (15.3 m/s)(5.0 s) + !(– 0.50 m/s2)(5.0 s)2 = 70 m. During the first second the car travels 15 m – 0 = 15 m. During the fifth second the car travels 70 m – 57 m = 13 m. 38. We find the average acceleration from v2 = v02 + 2Æ(x2 – x1); 0 = [(95 km/h)/(3.6 ks/h)]2 + 2Æ(0.80 m), which gives The number of g’s is Æ = (4.4102 m/s2)/[(9.80 m/s2)/g] = 44g.. Æ = – 4.4102 m/s2.. 39. We convert the units for the speed: (90 km/h)/(3.6 ks/h) = 25 m/s. With the origin at the beginning of the reaction, the location when the brakes are applied is x0 = v0t = (25 m/s)(1.0 s) = 25 m. (a) We find the location of the car after the brakes are applied from v2 = v02 + 2a1(x1 – x0); 0 = (25 m/s)2 + 2(– 4.0 m/s2)(x1 – 25 m), which gives x1 = 103 m. (b) We repeat the calculation for the new acceleration: v2 = v02 + 2a2(x2 – x0); 0 = (25 m/s)2 + 2(– 8.0 m/s2)(x2 – 25 m), which gives x2 = 64 m. 40. We find the acceleration of the space vehicle from v = v0 + at; 162 m/s = 65 m/s + a(10.0 s – 0.0 s), which gives a = 9.7 m/s2. We find the positions at the two times from x = x0 + v0t + !at2; x2 = x0 + (65 m/s)(2.0 s) + ! (9.7 m/s2)(2.0 s)2 = x0 + 149 m; Page 7.

(15) Chapter 2. x6 = x0 + (65 m/s)(6.0 s) + ! (9.7 m/s2)(6.0 s)2 = x0 + 565 m. Thus the distance moved is x6 – x2 = 565 m – 149 m = 416 m = 4.2102 m. 41. We use a coordinate system with the origin at the a v0 = 0 initial position of the front of the train. We can TRAIN find the acceleration of the train from the motion up to the point where the front of the train passes D L the worker: y=0 v12 = v02 + 2a(D – 0); 2 (25 m/s) = 0 + 2a(140 m – 0), which gives a = 2.23 m/s2. Now we consider the motion of the last car, which starts at – L, to the point where it passes the worker: v22 = v02 + 2a[D – (– L)] = 0 + 2(2.23 m/s2)(140 m + 75 m), which gives v2 = 31 m/s. 42. With the origin at the beginning of the reaction, the location when the brakes are applied is d0 = v0tR . We find the location of the car after the brakes are applied, which is the total stopping distance, from v2 = 0 = v02 + 2a(dS – d0), which gives dS = v0tR – v02/(2a). Note that a is negative. 43. (a). We assume constant velocity of v0 through the intersection. The time to travel at this speed is t = (dS + dI)/v0 = tR – (v0/2a) + (dI/v0). (b) For the two speeds we have t1 = tR – (v01/2a) + (dI/v01) = 0.500 s – [(30.0 km/h)/(3.60 ks/h)2(– 4.00 m/s2)] + [(14.4 m)(3.60 ks/h)/(30.0 km/h)] = 3.27 s. t2 = tR – (v02/2a) + (dI/v02) = 0.500 s – [(60.0 km/h)/(3.60 ks/h)2(– 4.00 m/s2)] + [(14.4 m)(3.60 ks/h)/(60.0 km/h)] = 3.45 s. Thus the chosen time is 3.45 s.. 44. We convert the units: (95 km/h)/(3.6 ks/h) = 26.4 m/s. (140 km/h/s)/(3.6 ks/h) = 38.9 m/s. We use a coordinate system with the origin where the motorist passes the police officer, as shown in the diagram. The location of the speeding motorist is given by xm = x0 + vmt = 0 + (38.9 m/s)t. The location of the police officer is given by xp = x0 + v0p(1.00 s) + v0p(t – 1.00 s) + !ap(t – 1.00 s)2. v 0p. = 0 + (26.4 m/s)t + !(2.00 m/s2)(t – 1.00 s)2. The officer will reach the speeder when these locations coincide, so we have x m = x p;. ap. vm. x=0 t=0. t = 1.00 s. (38.9 m/s)t = (26.4 m/s)t + !(2.00 m/s2)(t – 1.00 s)2. The solutions to this quadratic equation are 0.07 s and 14.4 s. Because the time must be greater than 1.00 s, the result is t = 14.4 s. 45. If the police car accelerates for 6.0 s, the time from when the speeder passed the police car is 7.0 s. From the analysis of Problem 44 we have Page 8.

(16) Chapter 2. x m = x p; vmt = v0pt + !ap(t – 1.00 s)2 ; vm(7.0 s) = (26.4 m/s)(7.0 s) + !(2.00 m/s2)(6.0 s)2, which gives vm =. 32 m/s (110 km/h).. 46. We find the assumed constant speed for the first 27.0 min from v0 = ?x/?t = (10,000 m – 1100 m)/(27.0 min)(60 s/min) = 5.49 m/s. The runner must cover the last 1100 m in 3.0 min (180 s). If the runner accelerates for t s, the new speed will be v = v0 + at = 5.49 m/s + (0.20 m/s2)t; and the distance covered during the acceleration will be x1 = v0t + !at2 = (5.49 m/s)t + !(0.20 m/s2)t2. The remaining distance will be run at the new speed, so we have 1100 m – x1 = v(180 s – t); or 1100 m – (5.49 m/s)t – !(0.20 m/s2)t2 = [5.49 m/s + (0.20 m/s2)t](180 s – t). This is a quadratic equation: 0.10 t2 – 36 t + 111.8 = 0, with the solutions t = + 363 s, + 3.1 s. Because we want a total time less than 3 minutes, the physical answer is t = 3.1 s. 47. We use a coordinate system with the origin at the top of the cliff and down positive. To find the time for the object to acquire the velocity, we have v = v0 + at ; (100 km/h)/(3.6 ks/h) = 0 + (9.80 m/s2)t, which gives t = 2.83 s. 48. We use a coordinate system with the origin at the top of the cliff and down positive. To find the height of the cliff, we have y = y0 + v0t + !at2 = 0 + 0 + !(9.80 m/s2)(2.75 s)2 =. 37.1 m.. 49. We use a coordinate system with the origin at the top of the building and down positive. (a) To find the time of fall, we have y = y0 + v0t + !at2; (b). t = 8.81 s. 380 m= 0 + 0 + !(9.80 m/s2)t2, which gives We find the velocity just before landing from v = v0 + at = 0 + (9.80 m/s2)(8.81 s) = 86.3 m/s.. 50. We use a coordinate system with the origin at the ground and up positive. (a) At the top of the motion the velocity is zero, so we find the height h from v2 = v02 + 2ah; 0 = (20 m/s)2 + 2(– 9.80 m/s2)h, which gives h = 20 m. (b) When the ball returns to the ground, its displacement is zero, so we have y = y0 + v0t + !at2 0 = 0 + (20 m/s)t + !(– 9.80 m/s2)t2, which gives t = 0 (when the ball starts up), and. t = 4.1 s.. 51. We use a coordinate system with the origin at the ground and up positive. We can find the initial velocity from the maximum height (where the velocity is zero): v2 = v02 + 2ah; 0 = v02 + 2(– 9.80 m/s2)(2.55 m), which gives v0 = 7.07 m/s. When the kangaroo returns to the ground, its displacement is zero. For the entire jump we have Page 9.

(17) Chapter 2. y = y0 + v0t + !at2; 0 = 0 + (7.07 m/s)t + !(– 9.80 m/s2)t2, which gives t = 0 (when the kangaroo jumps), and. t = 1.44 s.. 52. We use a coordinate system with the origin at the ground and up positive. When the ball returns to the ground, its displacement is zero, so we have y = y0 + v0t + !at2; v0 = 15 m/s. 0 = 0 + v0(3.1 s) + !(– 9.80 m/s2)(3.1 s)2, which gives At the top of the motion the velocity is zero, so we find the height h from v2 = v02 + 2ah; 0 = (15 m/s)2 + 2(– 9.80 m/s2)h, which gives h = 12 m. 53. We use a coordinate system with the origin at the ground and up positive. We assume you can throw the object 4 stories high, which is about 12 m. We can find the initial speed from the maximum height (where the velocity is zero): v2 = v02 + 2ah; 0 = v02 + 2(– 9.80 m/s2)(12 m), which gives v0 = 15 m/s.. 54. We use a coordinate system with the origin at the ground and up positive. (a) We can find the initial velocity from the maximum height (where the velocity is zero): v2 = v02 + 2ah; 0 = v02 + 2(– 9.80 m/s2)(1.20 m), which gives v0 = 4.85 m/s. (b) When the player returns to the ground, the displacement is zero. For the entire jump we have y = y0 + v0t + !at2; 0 = 0 + (4.85 m/s)t + !(– 9.80 m/s2)t2, which gives t = 0 (when the player jumps), and. t = 0.990 s.. 55. We use a coordinate system with the origin at the ground and up positive. When the package returns to the ground, its displacement is zero, so we have y = y0 + v0t + !at2; 0 = 115 m + (5.60 m/s)t + !(– 9.80 m/s2)t2. The solutions of this quadratic equation are t = – 4.31 s, and t = 5.44 s. Because the package is released at t = 0, the positive answer is the physical answer: 5.44 s.. +. v0 y0. a. H y=0. 56. We use a coordinate system with the origin at the release point and down positive. Because the object starts from rest, v0 = 0. The position of the object is given by y = y0 + v0t + !at2 = 0 + 0 + !gt2. The positions at one-second intervals are y0 = 0; y1 = !g(1 s)2 = (1 s2)!g; y2 = !g(2 s)2 = (4 s2)!g; y3 = !g(3 s)2 = (9 s2)!g; … . The distances traveled during each second are Page 10.

(18) Chapter 2. d1 = y1 – y0 = (1 s2)!g; d2 = y2 – y1 = (4 s2 – 1 s2)!g = (3 s2)(!g) = 3 d1; d3 = y3 – y2 = (9 s2 – 4 s2)!g = (5 s2)(!g) = 5 d1; … . 57. We use a coordinate system with the origin at the ground and up positive. Without air resistance, the acceleration is constant, so we have v2 = v02 + 2a(y – y0); v2 = v02 + 2(– 9.8 m/s2)(0 – 0) = v02, which gives v = ± v0. The two signs represent the two directions of the velocity at the ground. The magnitudes, and thus the speeds, are the same. 58. We use a coordinate system with the origin at the ground and up positive. (a) We find the velocity from v2 = v02 + 2a(y – y0); v2 = (23.0 m/s)2 + 2(– 9.8 m/s2)(12.0 m – 0), which gives v = ± 17.1 m/s. The stone reaches this height on the way up (the positive sign) and on the way down (the negative sign). (b) We find the time to reach the height from v = v0 + at; ± 17.1 m/s = 23.0 m/s + (– 9.80 m/s2)t, which gives t = 0.602 s, 4.09 s. (c) There are two answers because the stone reaches this height on the way up (t = 0.602 s) and on the way down (t = 4.09 s).. 59. We use a coordinate system with the origin at the release point and down positive. On paper the apple measures 8 mm, which we will call 8 mmp. If its true diameter is 10 cm, the conversion is 0.10 m/8 mmp. The images of the apple immediately after release overlap. We will use the first clear image which is 10 mmp below the release point. The final image is 63 mmp below the release point, and there are 7 intervals between these two images. The position of the apple is given by y = y0 + v0t + !at2 = 0 + 0 + !gt2. For the two selected images we have y1 = !gt12; (10 mmp)(0.10 m/8 mmp) = !(9.8 m/s2)t12, which gives t1 = 0.159 s; y2 = !gt22; (63 mmp)(0.10 m/8 mmp) = !(9.8 m/s2)t22, which gives t2 = 0.401 s. Thus the time interval between successive images is ?t = (t2 – t1)/7 = (0.401 s – 0.159 s)/7 = 0.035 s. 60. We use a coordinate system with the origin at the ground and up positive. (a) We find the velocity when the rocket runs out of fuel from v12 = v02 + 2a(y1 – y0); v12 = 0+ 2(3.2 m/s2)(1200 m – 0), which gives v1 = 87.6 m/s = 88 m/s. (b) We find the time to reach 1200 m from v1 = v0 + at1; 87.6 m/s = 0 + (3.2 m/s2)t1, which gives t1 = 27.4 s = 27 s. (c) After the rocket runs out of fuel, the acceleration is – g. We find the maximum altitude (where the velocity is zero) from v22 = v12 + 2(– g)(h – y1); 0 = (87.6 m/s)2 + 2(– 9.80 m/s2)(h – 1200 m), which gives h = 1590 m. (d) We find the time from v2 = v1 + (– g)(t2 – t1) Page 11.

(19) Chapter 2. (e). (f). 0 = 87.6 m/s + (– 9.80 m/s2)(t2 – 27.4 s), which gives t2 = 36 s. We consider the motion after the rocket runs out of fuel: v32 = v12 + 2(– g)(y3 – y1); v32 = (87.6 m/s)2 + 2(– 9.80 m/s2)(0 – 1200 m), which gives v3 = – 177 m/s = We find the time from v3 = v1 + (– g)(t3 – t1) – 177 m/s = 87.6 m/s + (– 9.80 m/s2)(t3 – 27.4 s), which gives t3 = 54 s.. – 1.8102 m/s.. 61. We use a coordinate system with the origin at the top of the window and down positive. We can find the velocity at the top of the window from the motion past the window: y = y0 + v0t + !at2; 2.2 m = 0 + v0(0.30 s) + !(9.80 m/s2)(0.30 s)2, which gives v0 = 5.86 m/s. For the motion from the release point to the top of the window, we have v02 = vrelease2 + 2g(y0 – yrelease); (5.86 m/s)2 = 0 + 2(9.80 m/s2)(0 – yrelease), which gives yrelease = – 1.8 m. The stone was released 1.8 m above the top of the window.. v=0 a. y=0. v0 H. 62. We use a coordinate system with the origin at the nozzle and up positive. For the motion of the water from the nozzle to the ground, we have y = y0 + v0t + !at2; – 1.5 m = 0 + v0(2.0 s) + !(– 9.80 m/s2)(2.0 s)2, which gives. v0 = 9.1 m/s.. 63. If the height of the cliff is H, the time for the sound to travel from the ocean to the top is tsound = H/vsound. The time of fall for the rock is T – tsound. We use a coordinate system with the origin at the top of the cliff and down positive. For the falling motion we have y = y0 + v0t + !at2; H = 0 + 0 + !a(T – tsound)2 = !(9.80 m/s2)[3.4 s – H/(340 m/s)]2 . This is a quadratic equation for H: 4.2410–5 H2 – 1.098H + 56.64 = 0, with H in m; which has the solutions H = 52 m, 2.58104 m. The larger result corresponds to tsound greater than 3.4 s, so the height of the cliff is 52 m. 64. We use a coordinate system with the origin at the ground, up positive, and t = 0 when the first object is thrown. (a) For the motion of the rock we have y1 = y0 + v01t + !at2 = 0 + (12.0 m/s)t + !(– 9.80 m/s2)t2. For the motion of the ball we have y2 = y0 + v02(t – 1.00 s) + !a(t – 1.00 s)2 = 0 + (20.0 m/s)(t – 1.00 s) + !(– 9.80 m/s2)(t – 1.00 s)2. When the two meet we have y1 = y2;. (b) (c). (12.0 m/s)t + !(– 9.80 m/s2)t2 = (20.0 m/s)(t – 1.00 s) + !(– 9.80 m/s2)(t – 1.00 s)2, which gives t = 1.40 s. We find the height from y1 = y0 + v01t + !at2 = 0 + (12.0 m/s)(1.40 s) + !(– 9.80 m/s2)(1.40 s)2 = 7.18 m. If we reverse the order we have Page 12. +.

(20) Chapter 2. y1 = y2; (20.0 m/s)t + !(– 9.80 m/s2)t2 = (12.0 m/s)(t – 1.00 s) + !(– 9.80 m/s2)(t – 1.00 s)2, which gives t = 9.38 s. We find the height from y1 = y0 + v01t + !at2 = 0 + (20.0 m/s)(9.38 s) + !(– 9.80 m/s2)(9.38 s)2 = – 244 m. This means they never collide. The rock, thrown later, returns to the ground before the ball does. To confirm this we find the time when the rock strikes the ground: y2 = y0 + v02(t – 1.00 s) + !a(t – 1.00 s)2 0 = 0 + (12.0 m/s)(t – 1.00 s) + !(– 9.80 m/s2)(t – 1.00 s)2, which gives t = 3.45 s. At this time the position of the ball is y1 = y0 + v01t + !at2 = 0 + (20.0 m/s)(3.45 s) + !(– 9.80 m/s2)(3.45 s)2 = 10.7 m. 65. We use a coordinate system with the origin at the ground, up positive, with t1 the time when the rocket reaches the bottom of the window and t2 = t1 + 0.15 s the time when the rocket reaches the top of the window. A very quick burn means we can assume that the rocket has an initial velocity at the ground. The position of the rocket is given by y = v0t + !at2. For the positions at the bottom and top of the window we have 10.0 m = v0t1 + !(– 9.80 m/s2)t12; 12.0 m = v0t2 + !(– 9.80 m/s2)t22 = v0(t1 + 0.15 s) + !(– 9.80 m/s2)(t1 + 0.15 s)2. Thus we have two equations for the two unknowns: v0 and t1 . The results of combining the equations are t1 = 0.590 s (where we have taken the positive time) and v0 = 19.8 m/s. We find the maximum height from v2 = v02 + 2ah; 0 = (19.8 m/s)2 + 2(– 9.80 m/s2)h, which gives h = 20.0 m.. 66. We find the total displacement by integration:. x=. t2 t1. v dt =. t2 t1. 2. 25 m/ s + 18 m / s t dt = 25 m / s t 2 – t1 + 9.0 m/ s. = 25 m / s 3.5 s – 1.5 s + 9.0 m / s 2. 3.5 s. 2. – 1.5 s. 2. =. 2. 2. 2. t2 – t1. 140 m.. 67. (a) If we make the suggested change of variable, we have u = g – kv, and du = – k dv, and a = g – kv = u. Thus from the definition of acceleration, we have a = dv/dt; u = – du/k dt, or du/u = – k dt. When we integrate, we get g – kv g. d u = t – k d t , which gives ln g – kv = – kt, or g – kv = g e – kt. u g 0. Thus the velocity as a function of time is v = (g/k)(1 – e – kt). (b) When the falling body reaches its terminal velocity, the acceleration will be zero, so we have a = g – kvterm = 0, or vterm = g/k.. Note that this is the terminal velocity from the velocity expression because as t  8, e – kt  0. Page 13.

(21) Chapter 2. 68. (a) We find the speed by integration: v v0. dv =. t 0. a dt =. t 0. At 1/ 2 d t, wh i chgi ves. v – v 0 = 23 At 3/ 2 – 0, or v = v 0 + 23 At 3/ 2 = (b) We find the displacement by integration: x 0. dx =. t. v dt =. 0 2 2 3 5. 5/ 2. t 0. 10 m/ s +. 2 3. 2.0 m / s 5/ 2 t 3/ 2 .. v 0 + 23 At 3/ 2 dt, wh i chgi ves 5/ 2. 4 4 x = v 0t + A t , or x = v 0 t + 15 At = 10 m / s t + 15 2.0 m / s 5/ 2 t (c) For the given time we have a = (2.0 m/s5/2)(5.0 s)1/2 = 4.5 m/s2; 5/2 v = (10 m/s) + %(2.0 m/s )(5.0 s)3/2 = 25 m/s; x = (10 m/s)(5.0 s) + (4/15)(2.0 m/s5/2)(5.0 s)5/2 = 80 m.. 5/ 2. .. 69. The height reached is determined by the initial velocity. We assume the same initial velocity of the object on the Moon and Earth. With a vertical velocity of 0 at the highest point, we have v2 = v02 + 2ah; 0 = v02 + 2(– g)h, so we get v02 = 2gEarthhEarth = 2gMoonhMoon , or hMoon = (gEarth/gMoon)hEarth = 6hEarth.. 70. For the falling motion, we use a coordinate system with the origin y 01 = 0 at the fourth-story window and down positive. For the stopping motion in the net, we use a coordinate system with the origin at the original position of the net and down positive. (a) We find the velocity of the person at the unstretched net + (which is the initial velocity for the stretching of the net) H from the free fall: v022 = v012 + 2a1(y1 – y01) = 0 + 2(9.80 m/s2)(15.0 m – 0), which gives v02 = 17.1 m/s. We find the acceleration during the stretching of the net from v22 = v022 + 2a2(y2 – y02); y 02 = 0 0 = (17.1 m/s)2 + 2a2(1.0 m – 0), which gives a2 = – 1.5102 m/s2. (b) To produce the same velocity change with a smaller acceleration requires a greater displacement. Thus the net should be loosened.. v 01 = 0. a1. a2. v 02 = v 1. 71. We assume that the seat belt keeps the occupant fixed with respect to the car. The distance the occupant moves with respect to the front end is the distance the front end collapses, so we have v2 = v02 + 2a(x – x0); Page 14.

(22) Chapter 2. 0 = [(100 km/h)/(3.6 ks/h)]2 + 2(– 30)(9.80 m/s2)(x – 0), which gives x = 72. If the lap distance is D, the time for the first 9 laps is t1 = 9D/(199 km/h), the time for the last lap is t2 = D/æ, and the time for the entire trial is T = 10D/(200 km/h). Thus we have T = t1 + t2 ; 10D/(200 km/h) = 9D/(199 km/h) + D/æ, which gives. 1.3 m.. æ = 209.5 km/h.. 73. We use a coordinate system with the origin at the release point and down positive. (a) The speed at the end of the fall is found from v2 = v02 + 2a(x – x0) = 0 + 2g(H– 0), which gives v = (2gH)1/2. (b) To achieve a speed of 50 km/h, we have v = (2gH)1/2; (50 km/h)/(3.6 ks/h) = [2(9.80 m/s2)H50]1/2, which gives H50 = 9.8 m. (c) To achieve a speed of 100 km/h, we have v = (2gH)1/2; (100 km/h)/(3.6 ks/h) = [2(9.80 m/s2)H100]1/2, which gives H100 = 39 m.. Page 15. 30 A B. 25 20. x (m). 74. For the motion from A to B, (a) The object is moving in the negative direction. The slope (the instantaneous velocity) is negative; the x-value is decreasing. (b) Because the slope is becoming more negative (greater magnitude of the velocity), the object is speeding up. (c) Because the velocity is becoming more negative, the acceleration is negative. For the motion from D to E, (d) The object is moving in the positive direction. The slope (the instantaneous velocity) is positive; the x-value is increasing. (e) Because the slope is becoming more positive (greater magnitude of the velocity), the object is speeding up. (f) Because the velocity is becoming more positive, the acceleration is positive. (g) The position is constant, so the object is not moving, the velocity and the acceleration are zero.. E. 15 10 5 0. D. C. 0. 1. 2. 3. t (s). 4. 5. 6.

(23) Chapter 2. 75. (a). (b). (c). At the top of the motion the velocity is zero, so we find the maximum height of the second child from v2 = v022 + 2ah2; 0 = (5.0 m/s)2 + 2(– 9.80 m/s2)h2 , which gives h2 = 1.28 m = 1.3 m. If the first child reaches a height h1 = 1.5h2 , we find the initial speed from v2 = v012 + 2ah1 = v012 + 2a(1.5h2) = v012 – 1.5v022 = 0; v012 = (1.5)(5.0 m/s)2, which gives v01 = 6.1 m/s. We find the time for the first child from y = y0 + v0t + !at2 0 = 0 + (6.1 m/s)t + !(– 9.80 m/s2)t2, which gives t = 0 (when the child starts up), and. t = 1.2 s.. 76. We use a coordinate system with the origin at the a initial position of the front of the train. We can v0 = 0 find the acceleration of the train from the motion TRAIN up to the point where the front of the train passes D the worker: L v12 = v02 + 2a(D – 0); y=0 (20 m/s)2 = 0 + 2a(180 m – 0), which gives a = 1.11 m/s2. Now we consider the motion of the last car, which starts at – L, to the point where it passes the worker: v22 = v02 + 2a[D – (– L)] = 0 + 2(1.11 m/s2)(180 m + 90 m), which gives v2 = 24 m/s.. 77. We use a coordinate system with the origin at the roof of the building and down positive, and call the height of the building H. (a) For the first stone, we have y1 = y01 + v01t1 + !at12; H = 0 + 0 + !(g)t12, or H = !gt12. For the second stone, we have y2 = y02 + v02t2 + !at22;. t1. +. = (30.0 m/s)(t1 – 2.00 s) + !(g)(t1 – 2.00 s)2 = (30.0 m/s)t1 – 60.0 m + !gt12 – (2.00 s)gt1 + (2.00 s2)g. When we eliminate H from the two equations, we get 0 = (30.0 m/s)t1 – 60.0 m – (2.00 s)gt1 + (2.00 s2)g, which gives t1 = 3.88 s. We use the motion of the first stone to find the height of the building: Page 16. v 02. y = 0 v 01 = 0. H = 0 + (30.0 m/s)t2 + !(g)t22. (b). t2 = t1 – 2.00 s. a. H.

(24) Chapter 2. 78. We convert the units: (110 km/h)/(3.6 ks/h) = 30.6 m/s. We use a coordinate system with the origin where the motorist passes the police officer, as shown in the diagram. (b) The location of the speeding motorist is given by xm = x0 + vmt, which we use to find the time required: 700 m = (30.6 m/s)t, which gives t = 22.9 s. (c) The location of the police car is given by xp = x0 + v0pt + !apt2 = 0 + 0 + !apt2, which we use to find the acceleration: ap = 2.67 m/s2. 700 m = !ap(22.9 s)2, which gives (d) We find the speed of the officer from vp = v0p + apt; = 0 + (2.67 m/s2)(22.9 s) = 61.1 m/s = 220 km/h. 48.4 m/s. (a). x. (c). H = !gt12 = !(9.80 m/s2)(3.88 s)2 = 73.9 m. We find the speeds from v1 = v01 + at1 = 0 + (9.80 m/s2)(3.88 s) = 38.0 m/s; v2 = v02 + at2 = 30.0 m/s + (9.80 m/s2)(3.88 s – 2.00 s) =. Speeder. Police officer. t. (about 135 mi/h!).. 79. We convert the maximum speed units: vmax = (90 km/h)/(3.6 ks/h) = 25 m/s. (a) There are (36 km)/0.80 km) = 45 trip segments, which means 46 stations (with 44 intermediate stations). In each segment there are three motions. a3 v max a1 Motion 1 is the acceleration to vmax. Station We find the time for this motion from vmax = v01 + a1t1; L1 L2 L3 25 m/s = 0 + (1.1 m/s2)t1 , which gives t1 = 22.7 s. x = 0 x1 = 0 x2 = 0 3 We find the distance for this motion from 2 x1 = x01 + v01t + !a1t1 ; L1 = 0 + 0 + !(1.1 m/s2)(22.7 s)2 = 284 m. Motion 2 is the constant speed of vmax , for which we have Page 17. t1. Station.

(25) Chapter 2. x2 = x02 + vmaxt2 ; L2 = 0 + vmaxt2 . Motion 3 is the acceleration from vmax to 0. We find the time for this motion from 0 = vmax + a3t3; 0 = 25 m/s + (– 2.0 m/s2)t3 , which gives t3 = 12.5 s. We find the distance for this motion from x3 = x03 + vmaxt + !a3t32 ;. (b). L3 = 0 + (25 m/s)(12.5 s) + !(– 2.0 m/s2)(12.5 s)2 = 156 m. The distance for Motion 2 is L2 = 800 m – L1 – L3 = 800 m – 284 m – 156 m = 360 m, so the time for Motion 2 is t2 = L2/vmax = (360 m)/(25 m/s) = 14.4 s. Thus the total time for the 45 segments and 44 stops is T = 45(t1 + t2 + t3) + 44(20 s) = 45(22.7 s + 14.4 s + 12.5 s) + 44(20 s) = 3112 s = 52 min. There are (36 km)/3.0 km) = 12 trip segments, which means 13 stations (with 11 intermediate stations.) The results for Motion 1 and Motion 3 are the same: t1 = 22.7 s, L1 = 284 m, t3 = 12.5 s, L3 = 156 m. The distance for Motion 2 is L2 = 3000 m – L1 – L3 = 3000 m – 284 m – 156 m = 2560 m, so the time for Motion 2 is t2 = L2/vmax = (2560 m)/(25 m/s) = 102 s. Thus the total time for the 12 segments and 11 stops is T = 12(t1 + t2 + t3) + 11(20 s) = 12(22.7 s + 102 s + 12.5 s) + 11(20 s) = 1870 s = 31 min. This means there is a higher average speed for stations farther apart.. 80. d2. d3. d4. a2. v1. d1. x=0. Page 18. d5. d6.

(26) Chapter 2. We convert the units for the speed limit: (50 km/h)/(3.6 ks/h) = 13.9 m/s. (a) If we assume that we are traveling at the speed limit, the time to pass through the farthest intersection is t1 = (d1 + d2 + d3 + d4 + d5 + d6)/v1 = (10 m + 15 m + 50 m + 15 m + 70 m + 15 m)/(13.9 m/s) = 12.6 s. Because this is less than the time while the lights are green, yes, you can make it through. (b) We find the time required for the second car to reach the speed limit: vmax = v02 + a2t2a; 13.9 m/s = 0 + (2.0 m/s2)t2a , which gives t2a = 6.95 s. In this time the second car will have traveled x2a = v02t1 + !a2t2a2 = 0 + !(2.0 m/s2)(6.95 s)2 = 48 m. The time to travel the remaining distance at constant speed is t2b = (d2 + d3 + d4 + d5 + d6 – x2a)/vmax = (15 m + 50 m + 15 m + 70 m + 15 m – 48 m)/(13.9 m/s) = 8.42 s. Thus the total time is ttotal = t2a + t2b = 6.95 s + 8.42 s = 15.4 s. No, the second car will not clear all the lights. 81. We use a coordinate system with the origin at the ground and up positive. (a) We find the initial speed from the motion to the window: v12 = v02 + 2a(y1 – y0); (14 m/s)2 = v02 + 2(– 9.80 m/s2)(25 m – 0), which gives v0 = 26 m/s. (b) We find the maximum altitude from v22 = v02 + 2a(y2 – y0); 0 = (26.2 m/s)2 + 2(– 9.80 m/s2)(y2 – 0), which gives y2 = 35 m. (c) We find the time from the motion to the window: v1 = v0 + at1 14 m/s = 26.2 m/s + (– 9.80 m/s2)t1 , which gives t1 = 1.2 s. Thus it was thrown 1.2 s before passing the window. (d) We find the time to reach the street from y = y0 + v0t + !at2;. +. y2. v2 = 0. y1. v1 H. y=0. 0 = 0 + (26.2 m/s)t + !(– 9.80 m/s2)t2. This is a quadratic equation for t, which has the solutions t = 0 (the initial throw), 5.3 s. Thus the time after the baseball passed the window is 5.3 s – 1.2 s = 4.1 s.. 82. (a). We find the time required for the fugitive to reach his maximum speed: vmax = v0f + aftf1; 8.0 m/s = 0 + (4.0 m/s2)tf1 , which gives tf1 = 2.0 s. In this time the fugitive will have traveled xf1 = v0ftf1 + !aftf12 = 0 + !(4.0 m/s2)(2.0 s)2 = 8.0 m. From this time the fugitive will run at constant speed. When he reaches the box car, we have xtrain = xf ; vtraint = xf1 + vmax(t – tf1); Page 19. v0. a.

(27) Chapter 2. (b). (6.0 m/s)t = 8.0 m + (8.0 m/s)(t – 2.0 s), which gives t = We can find the distance from the motion of the train: xtrain = vtraint = (6.0 m/s)(4.0 s) = 24 m.. 4.0 s.. 83. We use a coordinate system with the origin at the top of the cliff and up positive. (a) For the motion of the stone from the top of the cliff to the ground, we have y = y0 + v0t + !at2;. (b). (c). – 65.0 m = 0 + (10.0 m/s)t + !(– 9.80 m/s2)t2. This is a quadratic equation for t, which has the solutions t = – 2.76 s, 4.80 s. Because the stone starts at t = 0, the time is 4.80 s. We find the speed from v = v0 + at = 10.0 m/s + (– 9.80 m/s2)(4.80 s) = – 37.0 m/s. The negative sign indicates the downward direction, so the speed is 37.0 m/s. The total distance includes the distance up to the maximum height, down to the top of the cliff, and down to the bottom. We find the maximum height from v2 = v02 + 2ah; 0 = (10.0 m/s)2 + 2(– 9.80 m/s2)h, which gives h = 5.10 m. The total distance traveled is d = 5.10 m + 5.10 m + 65.0 m = 75.2 m.. 84. The instantaneous velocity is the slope of the x vs. t graph:. 1.5. v (m/s). 20. x (m). 2.0. 10. 1.0 0.5. 0. 0. 10. 20. 30. 40. 50. 0. t (s) –0.5 –1.0 –1.5 –2.0. 85. We use a coordinate system with the origin where the. Page 20. 10. 20. 30. t (s). 40. 50.

(28) Chapter 2. initial action takes place, as shown in the diagram. x=0 The initial speed is (50 km/h)/(3.6 ks/h) = 13.9 m/s. t=0 If she decides to stop, we find the minimum stopping distance v0 from v12 = v02 + 2a1(x1 – x0); L1 L2 2 2 0 = (13.9 m/s) + 2(– 6.0 m/s )x1 , which gives x1 = 16 m. v=0 x1 a1 Because this is less than L1 , the distance to the intersection, she can safely stop in time. x2 a2 If she decides to increase her speed, we find the acceleration from the time to go from 50 km/h to 70 km/h (19.4 m/s): v = v0 + a2t ; 19.4 m/s = 13.9 m/s + a2(6.0 s), which gives a2 = 0.917 m/s2. We find her location when the light turns red from x2 = x0 + v0t2 + !a2t22 = 0 + (13.9 m/s)(2.0 s) + !(0.917 m/s2)(2.0 s)2 = 30 m. Because this is L1 , she is at the beginning of the intersection, but moving at high speed. She should decide to stop! 86. We use a coordinate system with the origin at the water and up positive. We find the time for the pelican to reach the water from y1 = y0 + v0t + !at12; 0 = 16.0 m + 0 + ! (– 9.80 m/s2)t2, which gives t1 = 1.81 s. This means that the fish must spot the pelican 1.81 s – 0.20 s = 1.61 s after the pelican starts its dive. We find the height of the pelican at this time from y2 = y0 + v0t + !at22; = 16.0 + 0 + !(– 9.80 m/s2)(1.61 s)2 =. 3.3 m.. 87. In each case we use a coordinate system with the origin at the beginning of the putt and the positive direction in the direction of the putt. The limits on the putting distance are 6.0 m < x < 8.0 m. For the downhill putt we have: v2 = v0down2 + 2adown(x – x0); 0 = v0down2 + 2(– 2.0 m/s2)x. When we use the limits for x, we get 4.9 m/s < v0down < 5.7 m/s, or ?v0down = 0.8 m/s. For the uphill putt we have: v2 = v0up2 + 2aup(x – x0); 0 = v0up2 + 2(– 3.0 m/s2)x. When we use the limits for x, we get 6.0 m/s < v0up < 6.9 m/s, or ?v0up = 0.9 m/s. The smaller spread in allowable initial velocities makes the downhill putt more difficult.. Page 21.

(29) Chapter 2. 88. We use a coordinate system with the origin at the initial position of the car. The passing car’s position is given by L x1 = x01 + v0t + !a1t2 = 0 + v0t + !a1t2. v0 v0 The truck’s position is given by 1 xtruck = x0truck + v0t = D + v0t. The oncoming car’s position is given by D x2 = x02 – v0t = L – v0t. x=0 For the car to be safely past the truck, we must have x1 – xtruck = 10 m;. v0 2. v0t + !a1t2 – (D + v0t) = !a1t2 – D = 10 m, which allows us to find the time required for passing: !(1.0 m/s2)t2 – 30 m = 10 m, which gives t = 8.94 s. At this time the car’s location will be x1 = v0t + !a1t2 = (25 m/s)(8.94 s) + !(1.0 m/s2)(8.94 s)2 = 264 m from the origin. At this time the oncoming car’s location will be x2 = L – v0t = 400 m – (25 m/s)(8.94 s) = 176 m from the origin. Because this is closer to the origin, the two cars will have collided, so the passing attempt should not be made. 89. We use a coordinate system with the origin at the roof of the building and down positive. We find the time of fall for the second stone from v2 = v02 + at2 ; 12.0 m/s = 0 + (9.80 m/s2)t2 , which gives t2 = 1.22 s. During this time, the second stone fell y2 = y02 + v02t2 + !at22 = 0 + 0 + !(9.80 m/s2)(1.22 s)2 = 7.29 m. The time of fall for the first stone is t1 = t2 + 1.50 s = 1.22 s + 1.50 s = 2.72 s. During this time, the first stone fell y1 = y01 + v01t1 + !at12 = 0 + 0 + !(9.80 m/s2)(2.72 s)2 = 36.3 m. Thus the distance between the two stones is y1 – y2 = 36.3 m – 7.29 m = 29.0 m. 90. For the vertical motion of James Bond we use a coordinate system with the origin at the ground and up positive. We can find the time for his fall to the level of the truck bed from y = y0 + v0t + !at2;. y0 v truck. 1.5 m = 10 m + 0 + !(– 9.80 m/s2)t2, which gives t = 1.32 s. D During this time the distance the truck will travel is x = x0 + vtruckt = 0 + (30 m/s)(1.32 s) = 39.6 m. Because the poles are 20 m apart, he should jump when the truck is there is a pole at the bridge.. Page 22. y=0. 2 poles. away, assuming that.

(30) Chapter 3. CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1.. 2.. We choose the west and south coordinate system shown. D 2x For the components of the resultant we have W RW = D1 + D2 cos 45° D 2y D 2 = (200 km) + (80 km) cos 45° = 257 km; RS = 0 + D2 = 0 + (80 km) sin 45° = 57 km. We find the resultant displacement from R = (RW2 + RS2)1/2 = [(257 km)2 + (57 km)2]1/2 = 263 km; tan  = RS/RW = (57 km)/(257 km) = 0.222, which gives  = 13° S of W. We choose the north and east coordinate system shown. For the components of the resultant we have RE = D2 = 10 blocks; RN = D1 – D3 = 18 blocks – 16 blocks = 2 blocks. We find the resultant displacement from R = (RE2 + RN2)1/2 = [(10 blocks)2 + (2 blocks)2]1/2 = 10 blocks; tan  = RN/RE = (2 blocks)/(10 blocks) = 0.20, which gives  = 11° N of E.. 3.. From Fig. 3–6c, if we write the equivalent vector addition, we have V1 + Vwrong = V2 , or Vwrong = V2 – V1.. 4.. We find the vector from V = (Vx2 + Vy2)1/2 = [(8.80)2 + (– 6.40)2]1/2 = 10.9; tan  = Vy/Vx = (– 6.40)/(8.80) = – 0.727, which gives  = 36.0° below the x-axis.. The resultant is. 13.6 m, 18° N of E. North. V2. V1 V3. VR. . East. Page 1. . Ry. R Rx S. N D2 D1 D3. . R. E. y Vx. x.  V. 5.. D1. Vy.

(31) Chapter 3. 6.. (a) V1x = – 6.0, V1y = 0; y V2x = V2 cos 45° = 4.5 cos 45° = 3.18 = 3.2, R V2y = V2 sin 45° = 4.5 sin 45° = 3.18 = 3.2. V2 (b) For the components of the sum we have   Rx = V1x + V2x = – 6.0 + 3.18 = – 2.82; x V1 Ry = V1y + V2y = 0 + 3.18 = 3.18. We find the resultant from R = (Rx2 + Ry2)1/2 = [(– 2.82)2 + (3.18)2]1/2 = 4.3; tan  = Ry/Rx = (3.18)/(2.82) = 1.13, which gives  = 48° above – x-axis. Note that we have used the magnitude of Rx for the angle indicated on the diagram.. 7.. (a) y (b) For the components of the vector we have Vx = – V cos  = – 14.3 cos 34.8° = – 11.7; Vy = V sin  = 14.3 sin 34.8° = 8.16. V (c) We find the vector from Vy 2 2 1/2 2 2 1/2 V = (Vx + Vy ) = [(– 11.7) + (8.16) ]  = 14.3; x tan  = Vy/Vx = (8.16)/(11.7) = 1.42, which gives Vx  = 34.9° above – x-axis. This is within significant figures. Note that we have used the magnitude of Vx for the angle indicated on the diagram.. 8.. Because the vectors are parallel, the direction can be indicated by the sign. (a) C = A + B = 6.8 + (– 5.5) = 1.3 in the + x-direction.. (a) C = A + B C. B. x. A (b) C = A – B. (b) C = A – B = 6.8 – (– 5.5) = 12.3 in the + x-direction.. C A. (c) C = B – A = – 5.5 – (6.8) = – 12.3 in the + x-direction or. (c). C=B–A. 12.3 in the – x-direction.. C –A. 9.. (a) Using the given angle, we find the components from VN = V cos 38.5° = (635 km/h) cos 41.5° = 476 km/h; VW = V sin 38.5° = (635 km/h) sin 41.5° = 421 km/h. (b) We use the velocity components to find the displacement components: dN = VN t = (476 km/h)(3.00 h) = 1.43103 km; dW = VW t = (421 km/h)(3.00 h) = 1.26103 km.. Page 2. x –B. B. x.

(32) Chapter 3. 10. The vectors are V1 = – 6.0i + 8.0j, V2 = 4.5i – 5.0j. (a) For the magnitude of V1 we have  V1 = (V1x2 + V1y2)1/2 = [(– 6.0)2 + (8.0)2]1/2 = 10.0. We find the direction from tan 1 = V1y/V1x = (8.0)/(– 6.0) = – 1.33. From the signs of the components, we have 1 = 53° above – x-axis. (b) For the magnitude of V2 we have  V2 = (V2x2 + V2y2)1/2 = [(4.5)2 + (– 5.0)2]1/2 = 6.7. We find the direction from tan 2 = V2y/V2x = (– 5.0)/(4.5) = – 1.11. From the signs of the components, we have 2 = 48° below + x-axis. (c) For the sum V1 + V2 we have V1 + V2 = – 1.5i + 3.0j. For the magnitude of V1 + V2 we have  V1 + V2 = [(– 1.5)2 + (3.0 )2]1/2 = 3.4. We find the direction from tan 1+2 = (3.0 )/(– 1.5) = – 2.0. From the signs of the components, we have 1+2 = 63° above – x-axis. (d) For the difference V2 – V1 we have V2 – V1 = 10.5i – 13.0j. For the magnitude of V1 + V2 we have  V2 – V1 = [(10.5)2 + (– 13.0 )2]1/2 = 16.7. We find the direction from tan 2–1 = (– 13.0 )/(10.5) = – 1.24. From the signs of the components, we have 2–1 = 51° below + x-axis. . . . . 11. The vectors are V1 = 4i – 8j, V2 = i + j, V3 = – 2i + 4j. (a) For the sum V1 + V2 + V3 we have V1 + V2 + V3 = 3i – 3j. For the magnitude of V1 + V2 + V3 we have  V1 + V2 + V3 = [(3)2 + (– 3 )2]1/2 = 4.2. We find the direction from tan a = (– 3)/(3) = – 1.0. a = 45° below + x-axis. From the signs of the components, we have (b) For V1 – V2 + V3 we have V1 – V2 + V3 = i – 5j. For the magnitude of V1 – V2 + V3 we have  V1 – V2 + V3 = [(1)2 + (– 5 )2]1/2 = 5.1. We find the direction from tan b = (– 5)/(1) = – 5.0. b = 79° below + x-axis. From the signs of the components, we have . . Page 3.

(33) Chapter 3. 12. (a) For the components we have Rx = Ax + Bx + Cx = 44.0 cos 28.0° – 26.5 cos 56.0° + 0 = 24.0; Ry = Ay + By + Cy = 44.0 sin 28.0° + 26.5 sin 56.0° – 31.0 = 11.6. (b) We find the resultant from R = (Rx2 + Ry2)1/2 = [(24.0)2 + (11.6)2]1/2 = 26.7; tan  = Ry/Rx = (11.6)/(24.0) = 0.483, which gives  = 25.8° above + x-axis.. y R.  . C. B. x. A.  13. (a) For the components we have y Rx = Bx – Ax –A B = – 26.5 cos 56.0° – 44.0 cos 28.0° = – 53.7; Ry = By – Ay R = 26.5 sin 56.0° – 44.0 sin 28.0° = 1.3.  We find the resultant from R = (Rx2 + Ry2)1/2 = [(– 53.7)2 + (1.3)2]1/2 = 53.7; tan  = Ry/Rx = (1.3)/(53.7) = 0.0245, which gives  = 1.40° above – x-axis. Note that we have used the magnitude of Rx for the angle indicated on the diagram. (b) For the components we have y Rx = Ax – Bx A = 44.0 cos 28.0° – (– 26.5 cos 56.0°) = 53.7; –B Ry = Ay – By = 44.0 sin 28.0° – 26.5 sin 56.0° = – 1.3. We find the resultant from  R R = (Rx2 + Ry2)1/2 = [(53.7)2 + (– 1.3)2]1/2 = 53.7; tan  = Ry/Rx = (1.3)/(53.7) = 0.0245, which gives  = 1.40° below + x-axis, which is opposite to the result from (a).. Page 4. x. x.

(34) Chapter 3. 14. (a) For the components we have Rx = Ax – Bx + Cx = 44.0 cos 28.0° – (– 26.5 cos 56.0°) + 0 = 53.7; Ry = Ay – By + Cy = 44.0 sin 28.0° – 26.5 sin 56.0° – 31.0 = – 32.3. We find the resultant from R = (Rx2 + Ry2)1/2 = [(53.7)2 + (– 32.3)2]1/2 = 62.7; tan  = Ry/Rx = (32.3)/(53.7) = 0.602, which gives  = 31.0° below + x-axis. Note that we have used the magnitude of Ry for the angle indicated on the diagram. (b) For the components we have Rx = Ax + Bx – Cx = 44.0 cos 28.0° + (– 26.5 cos 56.0°) – 0 = 24.0; Ry = Ay + By – Cy = 44.0 sin 28.0° + 26.5 sin 56.0° – (– 31.0) = 73.6. We find the resultant from R = (Rx2 + Ry2)1/2 = [(24.0)2 + (73.6)2]1/2 = 77.4; tan  = Ry/Rx = (73.6)/(24.0) = 3.07, which gives  = 71.9° above + x-axis. (c) For the components we have Rx = Cx – Ax – Bx = 0 – 44.0 cos 28.0° – (– 26.5 cos 56.0°) = – 24.0; Ry = Cy – Ay – By = – 31.0 – 44.0 sin 28.0° – 26.5 sin 56.0° = – 73.6. We find the resultant from R = (Rx2 + Ry2)1/2 = [(– 24.0)2 + (– 73.6)2]1/2 = 77.4; tan  = Ry/Rx = (73.6)/(24.0) = 3.07, which gives  = 71.9° below – x-axis. Page 5. y A. –B. . . x.  C R. y. –C. R. B.  . A. . x. y x.  –A –B . . R. C.

(35) Chapter 3. 15. (a) For the components we have Rx = Bx – 2Ax = – 26.5 cos 56.0° – 2(44.0 cos 28.0° ) = – 92.5; Ry = By – 2Ay = 26.5 sin 56.0° – 2(44.0 sin 28.0°) = – 19.3. We find the resultant from R = (Rx2 + Ry2)1/2 = [(– 92.5)2 + (– 19.3)2]1/2 = 94.5; tan  = Ry/Rx = (19.3)/(92.5) = 0.209, which gives  = 11.8° below – x-axis. (b) For the components we have Rx = 2Ax – 3Bx + 2Cx = 2(44.0 cos 28.0°) – 3(– 26.5 cos 56.0°) + 2(0) = 122.2; Ry = 2Ay – 3By + 2Cy = 2(44.0 sin 28.0° ) – 3(26.5 sin 56.0°) + 2(– 31.0) = – 86.6. We find the resultant from R = (Rx2 + Ry2)1/2 = [(122.2)2 + (– 86.6)2]1/2 = 150; tan  = Ry/Rx = (86.6)/(122.2) = 0.709, which gives  = 35.3° below + x-axis.. Page 6.

(36) y –A. B. . Chapter 3. R. z V. y. A. D. y North. H. –B.  A –B.  . x. x. East. 17. (a) We find the x-component from A2 = Ax2 + Ay2; y R Ax = ± 82.9. (90.0)2 = Ax2 + (– 35.0)2; which gives (b) If we call the new vector B, we have Rx = Ax + Bx ; Ay A2 – 80.0 = + 82.9 + Bx , which gives Bx = – 162.9; Ry = Ay + By ; 0 = – 35.0 + By , which gives By = + 35.0. We find the resultant from B = (Bx2 + By2)1/2 = [(– 162.9)2 + (+ 35.0)2]1/2 = 166.6;  = 12.1° above – x-axis. tan  = By/Bx = (35.0)/(162.9) = 0.215, which gives. . –B C. x C A1. 18. We find the velocity and acceleration by differentiating: r = (7.60 m/s)ti + (8.85 m)j – (1.00 m/s2)t2k; v = dr/dt = (7.60 m/s)i – (2.00 m/s2)tk; a = dv/dt = – (2.00 m/s2)k. 19. The positions of the particle at the two times are r1 = (7.60 m/s)(1.00 s)i + (8.85 m)j – (1.00 m/s2)(1.00 s)2k = (7.60 m)i + (8.85 m)j – (1.00 m)k; r3 = (7.60 m/s)(3.00 s)i + (8.85 m)j – (1.00 m/s2)(3.00 s)2k = (22.8 m)i + (8.85 m)j – (9.00 m)k. The average velocity is vav = ?r/?t = [(15.2 m)i – (8.00 m)k]/(3.00 s – 1.00 s) = (7.60 m/s)i – (4.00 m/s)k. The instantaneous velocity at the midpoint of the interval is v2 = (7.60 m/s)i – (2.00 m/s2)(2.00 s)k = (7.60 m/s)i – (4.00 m/s)k. Note that this is the same as the average velocity because the acceleration is constant. The magnitude is v2 = [(7.60 m/s)2 + (4.00 m/s)2]1/2 = 8.59 m/s. 20. The y-position is constant. The x- and z-components are x = (7.60 m/s)t, and z = – (1.00 m/s2)t2. If we eliminate t, we get z = – (1.00 m)x2/(7.60 m)2, which is a parabola. Thus the path is parabola centered on the – z-axis at y = 8.85 m. 21. (a) Because we do not know the displacement over the given time interval, the average velocity is unknown. (b) The average acceleration is aav = ?v/?t = [(27.5 m/s)i – (– 18.0 m/s)j]/(8.00 s) = (3.44 m/s2)i + (2.25 m/s2)k. The magnitude is Page 7. x. . –A. 16. For the components we have Dx = Hx = – 4580 sin 32.4° = – 2454 m; Dy = Hy = + 4580 cos 32.4° = + 3867 m; Dz = V = + 2450 m. By extending the Pythagorean theorem, we find the magnitude from D = (Dx2 + Dy2 + Dz2)1/2 = [(– 2454 m)2 + (3867 m)2 + (2450 m)2]1/2 = 5194 m.. .

(37) Chapter 3.  aav = [(3.44 m/s2)2 + (2.25 m/s2)2]1/2 = 4.11 m/s2. We find the direction from tan  = (2.25 m/s2)/(3.44 m/s2) = 0.654, which gives  = 33.2° north of east. (c) Because we do not know the distance traveled, the average speed is unknown. . 22. (a) For the vertical component we have aV = (3.80 m/s2) sin 30.0° = 1.90 m/s2 down. (b) Because the elevation change is the vertical displacement, we find the time from the vertical motion, taking down as the positive direction: y = v0yt + !aVt2; 250 m = 0 + !(1.90 m/s2)t2, which gives. t = 16.2 s.. 23. The acceleration is a = (4.0 m/s2)i + (3.0 m/s2)j. (a) We find the velocity by integrating: v. dv =. 0. v=. t 0. t 0. a dt ;. (4.0 m / s 2)i + (3.0 m/ s 2)j dt =. (4.0 m / s 2)t i + (3.0 m / s 2)tj.. (b) The speed of the particle is  v = {[(4.0 m/s2)t]2 + [(3.0 m/s2)t]2}1/2 = (c) We find the position by integrating: . r 0. dr =. r=. t 0. t 0. (5.0 m/s2)t.. v dt; 2. 2. 2. 2. 2. 2. (4.0 m / s )ti + (3.0 m/ s )t dt = 12 (4.0 m / s )t i + 12 (3.0 m / s )t j 2. 2. = (2.0 m / s2 )t i + (1.5 m / s 2)t j. (d) For the given time we have v = (4.0 m/s2)ti + (3.0 m/s2)tj = (4.0 m/s2)(2.0 s)i + (3.0 m/s2)(2.0 s)j = (8.0 m/s)i + (6.0 m/s)j.  v = (5.0 m/s2)t = (5.0 m/s2)(2.0 s) = 10.0 m/s. r = (2.0 m/s2)t2i + (1.5 m/s2)t2j = (2.0 m/s2)(2.0 s)2i + (1.5 m/s2)(2.0 s)2j = (8.0 m)i + (6.0 m)j. . Page 8.

(38) Chapter 3. 24. The acceleration is a = (– 3.0 m/s2)i + (4.5 m/s2)j. We find the velocity by integrating: v v0. t. dv =. 0. a dt ;. v – (5.0 m/ s )i =. t 0. (– 3.0 m / s 2)i + (4.5 m/ s 2)j dt = (– 3.0 m/ s 2)ti + (4.5 m / s 2)t j, or. v = [(– 3.0 m/s2)t + 5.0 m/s]i + (4.5 m/s2)tj. We find the position by integrating: r 0. dr =. t 0. v d t;. t. (– 3.0 m/ s 2)t + 5.0 m/ s i + (4.5 m / s 2)t d t =. r=. 1 2 (–. 2. 0 2. 2. = (– 1.5 m/ s 2)t + (5.0 m/ s)t i + (2.25 m / s2 )t j. To find the time at which the particle reaches its maximum x-coordinate, we set dx/dt = 0: dx/dt = (– 3.0 m/s2)t + 5.0 m/s = 0, which gives t = 1.67 s. The velocity is v = [(– 3.0 m/s2)t + 5.0 m/s]i + (4.5 m/s2)tj = 0i + (4.5 m/s2)(1.67 s)j = (7.5 m/s)j. The position is r = [(– 1.5 m/s2)t2 + (5.0 m/s)t]i + (2.25 m/s2)t2j = [(– 1.5 m/s2)(1.67 s)2 + (5.0 m/s)(1.67 s)]i + (2.25 m/s2)(1.67 s)2j = (4.2 m)i + (6.3 m)j.. 25. The position is r = (6.0 m) cos (3.0 s–1)ti + (6.0 m) sin (3.0 s–1)tj. (a) We find the velocity by differentiating: v = dr/dt = – (6.0 m)(3.0 s–1) sin (3.0 s–1)ti + (6.0 m)(3.0 s–1) cos (3.0 s–1)tj = – (18.0 m/s) sin (3.0 s–1)ti + (18.0 m/s) cos (3.0 s–1)tj. (b) We find the acceleration by differentiating: a = dv/dt = – (18.0 m/s)(3.0 s–1) cos (3.0 s–1)ti – (18.0 m/s)(3.0 s–1) sin (3.0 s–1)tj = – (54.0 m/s2) cos (3.0 s–1)ti – (54.0 m/s2) sin (3.0 s–1)tj. (c) The magnitude of r is  r= {(6.0 m) cos (3.0 s–1)t]2 + [(6.0 m) sin (3.0 s–1)t]2]}1/2 = 6.0 m. Thus the particle is always 6.0 m from the origin, so it is traveling in a circle. (d) We see that a = – (9.0 s–2)r, so we have a = (9.0 s–2)r, with the angle between the vectors being 180°, that is, in opposite directions. (e) We see that  v = {[(18.0 m/s) sin (3.0 s–1)t]2 + [(18.0 m/s) cos (3.0 s–1)t]2}1/2 = 18.0 m/s, so v = (3.0 s–1)r, and a = (9.0 s–2)r = v2/r. . . Page 9. 2. 3.0 m/ s 2)t + (5.0 m/ s )t i + 12 (4.5 m / s 2)t j.

(39) Chapter 3. 26. We choose a coordinate system with the origin at the takeoff point, with x horizontal and y vertical, with the positive direction down. We find the time for the tiger to reach the ground from its vertical motion: y = y0 + v0yt + !ayt2; 6.5 m = 0 + 0 + !(9.80 m/s2)t2, which gives t = 1.15 s. The horizontal motion will have constant velocity. We find the distance from the base of the rock from x = x0 + v0xt; x = 0 + (4.0 m/s)(1.15 s) = 4.6 m.. O. v0. x. h. y. 27. We choose a coordinate system with the origin at the takeoff point, with x horizontal and y vertical, with the positive direction down. We find the height of the cliff from the vertical displacement: y = y0 + v0yt + !ayt2; y = 0 + 0 + !(9.80 m/s2)(3.0 s)2 = 44 m. The horizontal motion will have constant velocity. We find the distance from the base of the cliff from x = x0 + v0xt; x = 0 + (2.1 m/s)(3.0 s) = 6.3 m. 28. Because the initial and final locations are at the same level, we can use the expression for the horizontal range. The horizontal range on Earth is given by R = v02 sin(20)/g, whereas on the Moon it is RMoon = v02 sin(20)/gMoon. Because we have the same v0 and 0 , when we divide the two equations, we get RMoon/R = g/gMoon , or RMoon = (g/gMoon)R = [g/(g/6)]R = 6R, so a person could jump six times as far.. 29. Because the water returns to the same level, we can use the y v0 expression for the horizontal range: 2 R = v0 sin(2)/g; 3.0 m = (5.5 m/s)2 sin(2)/(9.80 m/s2), which gives  x sin(2) = 0.971, or 2 = 76° and 104°, O R so the angles are 38° and 52°. At the larger angle the water has a smaller horizontal velocity but spends more time in the air, because of the larger initial vertical velocity. Thus the horizontal displacement is the same for the two angles.. Page 10.

Figure

Updating...

References

Related subjects :