Skip to content
Textbook Guides

Solution Manual For College Physics With Masteringphysics, 7th Edition

  • Physics
  • 536 pages
  • Shared February 2025
  • 37 downloads

Solution Manual For College Physics With Masteringphysics, 7th Edition gives you the answers you need, explained in a simple and clear way.

Preview 31 of 536 pages

See the first 31 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    CHAPTER 1 MEASUREMENT AND PROBLEM SOLVING Remind students that their answers to odd - numbered exercises may be slightly different from those given here because of rounding. Refer to Problem - Solving Hint: The "Correct" Answer in Chapter 1. Multiple Choice : 1. (c). 2. (b). 3. (c). 4. (b). The gram is the mass of one cm 3 of water, so the kilogram (100 0 g) and tonne (1000 kg) are all related to a volume of water. Only the pound is not related to a volume of water. 5. (b). 6. (b). 7. (a). 8. (d). A quart is slightly less than a liter, 2000 μL is 2 mL, and 2000 mL is 2 L, so 2000 mL has the greatest volume. 9. (c). Micro - is 10 - 6 , centi - is 10 - 2 , nano - is 10 - 9 , and milli - is 10 - 3 , so nano - is the smallest. 10. (d). 11. (d). 12. (c). 13. (a). The kg is a unit of mass and the lb is a unit of force. At the surface of the Earth, 1 kg is equivalent to 2.2 lb, which means that a 1 - kg object weighs 2.2 lb. 14. (c). Because the μL is the smallest volume of those listed, you will need more of them to make up any given volume, and hence a larger number of μL.

  2. Page 2
  3. Page 3
    Read page 3 as text

    2 College Physics Seventh Edition: Instructor Solutions Manual 15. (a). 16. (c). 17. (b). 18. (a). 19. (c). 20. (d). Conceptual Questions : 1. Because there are no more fundamental units. The units of all quantities can be expressed in terms of the fundamental, or base, units. 2. Weight depends on the force of gravity, which can vary with location. 3. The mean solar day replaced the original definition. No, because this has been replaced by atomic clocks. 4. One major difference is the decimal versus duodecimal basis. Another difference is that SI basic units are meters, kilograms and seconds, whereas the British system uses feet, pounds and seconds. 5. No, because 3 cm is over an inch. Ladybugs are on the order of several mm long. Yes, a 10 - kg salmon would weigh on the order of 22 pounds, which is typical for a medium sized fish like that. 6. This is by definition 3 1 L 1000 mL and 1 L 1000 cm . = = 7. The metric ton is actually a misnomer since it is not a weight unit but a mass unit, defined as the mass of 1 cubic meter of water. But 3 l m 1000 L = and 1 L of water has a mass of 1 kg. So one metric ton is equal to 1000 kg. 8. No, it only tells if the equation is dimensionally correct. You could be missing (or have extra) dimensionless numbers such as ½ or π. 9. No, unit analysis can only tell if it is dimensionally correct. Dimensionless factors such as π may be missing. 10. By putting in units and solving for those of the unknown quantity.

  4. Page 4
    Read page 4 as text

    Chapter 1 Measurement and Problem Solving 3 11. π is dimensionless and therefore also unitless becaus e it is defined as the ratio of two lengths, the circumference to the diameter of a circle. 12. No, they are not the same. An equivalent statement is not dimensionally correct. 13. Yes, whether you multiply or divide should be consistent with unit analysis for the final answer. 14. Give him 2.54 cm and he’ll take 1.61 km, since 2.54 cm = 1.00 in. and 1.61 km = 1.00 mi. 15. To provide an estimate of the accuracy of a quantity. 16. No, there is always one doubtful digit, the last digit. 17. For (a) and (b), the result should have the least number of significant figures. For (c) and (d), the result should have the least number of decimal places. 18. Because 5 is midway between the upper and lower extremes of 9 and 1. 19. See the six steps as listed in Chapter 1. 20. No, since an order of magnitude calculation is only an estimate. 21. The accuracy of the answer is expected to be within a factor of 10 of the correct answer. 22. Approximate the area of skin - covered body parts using familiar geometric shapes. For example, use a sphere for the head and cylinders for the arms, legs, and torso. 23. Since a liter is close to a quart and there are four quarts in a gallon, this volume is about 75 gallons which is not reasonable for a car (but might be for a large truck or other large vehicle such as an RV). 24. No , since 30 km/h ≈ 19 mi/h < 25 mi/h.

  5. Page 5
    Read page 5 as text

    4 College Physics Seventh Edition: Instructor Solutions Manual Exercises : 1 . The decimal system (base 10) has a dime worth 10¢ and a dollar worth 10 dimes, or 100¢. By analogy, a duodecimal system would have a dime worth 12¢ and a dollar worth 12 “dimes,” or $1.44 in decimal d ollars. Then a penny would be 1 144 of a dollar. 2 . (a) Different ounces are used for volume and weight measurements. 16 oz = 1 pt is a volume measure and 16 oz = 1 lb is a weight measure. (b) Two different pound units are used. Avoirdupois lb = 16 oz ‚ troy lb = 12 oz. 3. (a)        6 1 MB 40,000,000 bytes 10 bytes = 40 MB (b)        1 L 0.5722 mL 1000 mL =  -4 5.722 10 L (c)        100 cm 2.684 m 1 m = 268.4 cm (d)        1 kilobuck 5, 500 bucks 1000 bucks = 5.5 kilobucks 4 . That is because 1 nautical mile = 6076 ft = 1.15 mi. A nautical mile is larger than a (statute) mile. 5.        1 g water (25 cm)(35 cm)(55 cm) 3 1 cm water = 48,100 g = 48 kg 6. Let x be the length of each side of the cube. =             3 3.788 L 1000 cm 3 (1 qt) 4 qt 1 L x = 947 cm 3 ( ) = = 1/3 947 cm 9.82 cm x 7. (a) (20 cm) 3 (1 L/1000 cm 3 ) = 8.0 L (b) ( ) = = =       3 1 m 3 1000 kg/m (8.0 L) 8.0 kg 1000 L m V 

  6. Page 6
    Read page 6 as text

    Chapter 1 Measurement and Problem Solving 5 8 . (Length) = (Length) + ( L ength ) ( Time )  ( T ime ) = (Length) + (Length) . 9 . (d). 10 . m 2 = (m) 2 = m 2 . 11 . Since ax 2 is in meters, a = m m 2 = 1/m . Since bx is in meters, b = m m = dimensionless . c is in m . 12. Yes , since [m 3 ] = [m] 3 = [m 3 ]. 13. No . V = 4  r 3 /3 = 4  (8 r 3 )/24 = 4  (2 r ) 3 /24 =  d 3 /6. So it should be V =  d 3 /6 . 14 . Since p =  v 2 , the unit of pressure is (kg/m 3 )(m/s) 2 = kg / (m  s 2 ). No , this does not prove that this relationship is physically correct , because there might be a coefficient in the equation. 15 . Yes , because m 2 = 1 2 m ( m + m ) = m 2 + m 2 . 16 . (a) Since F = ma , newton = (kg)(m/s 2 ) = kg·m/s 2 . (b) Yes , because (kg)  m 2 /s 2 m = kg·m/s 2 ( F = m v 2 / r ). 17 . (a) The unit of angular momentum is (kg)(m/s)(m) = kg  m 2 /s (b) The unit of L 2 2 mr 2 is ( kg  m 2 /s ) 2 kg  m 2 =  2 2 kg m s , which is the unit of kinetic energy, K . (c) The unit of moment of inertia is (kg)(m) 2 = kg  m 2 . 18 . (a) Since E = mc 2 , the units of energy = (kg)(m/s) 2 = kg  m 2 /s 2 . (b) Yes , because (kg)(m/s 2 )(m) = kg  m 2 /s 2 ( E = mgh ). 19 . 130 ft = (130 ft)  1 m 3.28 ft = 39.6 m .

  7. Page 7
    Read page 7 as text

    6 College Physics Seventh Edition: Instructor Solutions Manual 20. (a) The answer is ( 4 ) c entimeter , as it is the smallest unit among those listed. (b) Since 1 ft = 30.5 cm, 6.00 ft = (6.00 ft)  30.5 cm 1 ft = 183 cm . 21 . 40 000 mi = (40 000 mi)  1609 m 1 mi = 64 400 000 m. So 64 400 000 m 1.75 m = 37 000 000 times . 22 . (a) Since 1 gal = 3.785 L < 4 L, or ½ gal < 2 L, ½ gal holds ( 3 ) less soda . (b) 0.5 gal = (0.5 gal)  3.785 L 1 gal = 1.89 L. 2 L − 1.89 L = 0.11 L. So 2 L by 0.11 L more . 23 . (a) 300 ft = (300 ft)  1 m 3.28 ft = 91.5 m. 160 ft = (160 ft)  1 m 3.28 ft = 48.8 m. So the dimensions are 91.5 m by 48.8 m . (b) 11 in. = (11 .0 in.)  2.54 cm 1 in. = 27.9 cm. 11.25 in. = (11.25 in.)  2.54 cm 1 in. = 28.6 cm. So the length is 27.9 cm to 28.6 cm . 24 . From Exercise 1. 23 , the metric field is larger. A current = (91.4 m)(48.8 m) = 4.46  10 3 m 2 . A metric = (100 m)(54 m) = 5.4  10 3 m 2 . So the difference is 5.4  10 3 m 2 – 4.46  10 3 m 2 = 2 2 9.4 10  m . 25.                        3 1 qt 3.788 L 1000 cm 1 g (1 pt) 3 2 pts 4 qt 1 L 1 cm = 474 g 26.              1 gal 4 qt (10 L) 3.788 L 1 gal = 10.6 qt 27.                    2 yd 3 ft 1 m (175 fathoms) 1 fathom 1 yd 3.281 ft = 320 m 28 . 763 mi/h = (763 mi/h)  1609 m 1 mi  1 h 3600 s = 341 m/s .

  8. Page 8
    Read page 8 as text

    Chapter 1 Measurement and Problem Solving 7 (b) 300 ft = (300 ft)  1 m 3.28 ft = 91.46 m. So the time is 91.46 m 341 m/s = 0.268 s . 29 . (a) 1 km/h = (1 km/h)  1000 m 1 km  1 h 3600 s = 0.8 m/s < 1 m/s. 1 ft/s = (1 ft/s)  1 m 3.28 ft = 0.30 m/s < 1 m/s. 1 mi/h = (1 mi/h)  1609 m 1 mi  1 h 3600 s = 0.45 m/s < 1 m/s. So ( 1 ) 1 m/s represents the greatest speed. (b) 15.0 m/s = (15.0 m/s)  1 mi 1609 m  3600 s 1 h = 33.6 mi/h . 30 . (a) 10 mi/h = (10 mi/h)  1.609 km 1 mi = 16 km/h for each 10 mi/h . (b) 70 mi/h = (70 mi/h)  1.609 km 1 mi = 1 . 1  10 2 km/h . 31 . (a) 1 kg = 2.2 lb (equivalent). So 170 lb = (170 lb)  1 kg 2.2 lb = 77 .3 kg . (b) The density of water is 1000 kg/m 3 .  = m V ,  V = m  = 3 77.3 kg 1000 kg / m = 3 0.0773 m or about 77.3L . Using liter avoids the small decimals. 32 . The circumference of the Moon of diameter 3500 km , is  d d =  ( 3500 km) = 1.1  10 4 km. So the answer is yes . 1.0  10 5 km 1.1  10 4 km = 9.1 times . 33 . In one day, there are 24  60 min = 1440 min. So the volume of blood pumped per day is (6 0 beats/mi n)(1440 min/day)(75 mL/beat) = 6 .5  10 6 mL/day = 6 .5  10 3 L/day .

  9. Page 9
    Read page 9 as text

    8 College Physics Seventh Edition: Instructor Solutions Manual 34 . (a) 2 fl. oz = (2 fl. oz)  473 mL 16 fl. oz = 59.1 mL . (b) 100 g = (100 g)  14.5 oz 411 g = 3.53 oz . 35 . (250 mL/min)(4.5  10 6 /mm 3 )  1 min 60 s  1 L 10 − 3 mL  10 6 mm 3 1 L = 1.9  10 10 /s . 36 . 18 in. = (18 in.)  2.54 cm 1 in. = 45.7 cm. 5 ft, 6 in. = 66 in. = (66 in.)  2.54 cm 1 in. = 167.6 cm. So the growth per year is 167.6 cm – 45.7 cm 20 = 6.1 cm . 37. The Earth rotates 360° in 24 hr (= 24  60 min = 1440 min), so in one minute it will rotate 1/1440 of 360°, which is 0.250° = 15.0 min of arc . 38 . (a) 13.6 g/cm 3 = (13.6 g/cm 3 )  1 kg 1000 g  100 3 cm 1 m       = 1.36  10 4 kg/m 3 . (b)  = m V ,  m =  V = (13.6 g/cm 3 )(0.250 L)  1000 cm 3 1 L = 3.40  10 3 g = 3.40 kg . 39 . (a) The volume is equal to V = Ah =  r 2 h =  (125 m) 2 (10 ft)(0.305 m/ft) = 1.5  10 5 m 3 . (b) The water density of is 1000 kg/m 3 .  = m V ,  m =  V = (1000 kg/m 3 )(1.5  10 5 m 3 ) = 1.5  10 8 kg . (c) One kg is equivalent to 2.2 lb. 1.5  10 8 kg = (1.5  10 8 kg)  2.2 lb 1 kg = 3.3  10 8 lb . 40 . L = 300 cubits = (300 cubits)  0.5 yd 1 cubit  3 ft 1 yd  1 m 3.28 ft = 137 m. W = 50.0 cubits = 22.9 m, H = 30.0 cubits = 13.7 m. So the dimensions are 137 m  22.9 m  13.7 m . (b) V = LWH = (137 m)(22.9 m)(13.7 m) = 4.30  10 4 m 3 . 4 1 . 50 500  m = (50 500  m)  1 cm 10 000  m = 5.05 cm = 5.05  10 − 1 dm = 5.05  10 − 2 m . 4 2 . 0.001 m or 1 mm .

  10. Page 10
    Read page 10 as text

    Chapter 1 Measurement and Problem Solving 9 4 3 . (a) 4 . (b) 3 . (c) 5 . (d) 2 . 4 4 . (a) 1.0 m . (b) 8.0 cm . (c) 16 kg . (d) 1.5  10 − 2  s . 45. (a) 96 (b) 0.0021 (c) 9400 (d) 0.00034 46 . ( b ) and ( d ) ; ( a ) has four and ( c ) has six . 47 . A = LW = (0.274 m)(0.222 m) = 6.08  10 − 2 m 2 . 48 . V = LW H = (1.3 m)(3.281 ft/m)(1.05 m)(3.281 ft/m)(0.67 m)(3.281 ft/m) = 32 ft 3 . 49 . (a) The smallest division is ( 2 ) cm , as the last digit is estimated. (b) A = LW = (1.245 m)(0.760 m) = 0.946 m 2 . 50 . (a) ( 2 ) Three , since the height has only three significant figures. (b) The area is the sum of that of the top, the bottom, and the side. The side of the can is a rectangle with a length equal to the circumference and width equal to the height of the can. A = + 2 4  d + 2 4  d + Ch = 2 4  d + 2 4  d + (  d ) h =  ( 12.559 cm ) 2 4 +  ( 12.559 cm ) 2 4 +  (12.559 cm)(5.62 cm) = 4 69 cm 2 . 51 . (a) 12.634 + 2.1 = 14.7 . (b) 13.5 − 2.134 = 11.4 . (c)  (0.25 m) 2 = 0.20 m 2 . (d) 2.37/3.5 = 0.82 . 52 . (a) The answer is ( 1 ) z ero , since 38 m has zero decimal place s . (b) 46.9 m + 5.72 m – 38 m = 15 m . 53 . (a) v = x t = 8.5 m 2.7 s = 3.1 m/s, p = m v = (0.66 kg)(3.1 m/s) = 2.0 kg·m/s .

  11. Page 11
    Read page 11 as text

    10 College Physics Seventh Edition: Instructor Solutions Manual (b) p = mx t = ( 0.66 kg ) ( 8.5 m ) 2.7 s = 2.1 kg·m/s . (c) No , the results are not the same. The difference comes from rounding difference . 54 . According to the Pythagorean theorem, c = a 2 + b 2 = ( 37 m ) 2 + ( 42.3 m ) 2 = 56 m . 55 . Since 1 m = 100 cm, (1 m) 3 = (100 cm) 3 or 1 m 3 = 10 6 cm 3 .  = m V ,  m =  V = (0.10 g/cm 3 )(1 m 3 )  10 6 cm 3 1 m 3 = 1.0  10 5 g = 100 kg . 56. ( )             3 3.281 ft 1 cord 3.0 m 1 m 8.0 ft 4.0 ft 4.0 ft = 0.83 cord 57 . (a) The percentage is ( 18 g ) ( 9 cal/g ) 310 cal = 0.52 = 52% . (b) Total fat = 18 g 0.28 = 64 g ; saturated fat = 7 g 0.35 = 20 g . 58 . (a) One sheet has two pages. 860 pages ha ve 430 sheets. The average thickness per sheet is 3.75 cm 430 sheets = 8.72  10 − 3 cm . (b) 1 cm 100 sheets = about 10 − 2 cm . 59. = = 3 4 3 E m m V R   = ( )  =   24 3 3 3 6 5.98 10 kg 5.4 10 kg/m 4 6.4 10 m 3  60 . (a) From the sketch, it is clear that the stadium is ( 4 ) south of west , relative to your house. (b) Consider the right triangle on the bottom of the sketch. The two sides perpendicular to each other are 500 m each. Using Pythagorean theorem, d = ( 500 m ) 2 + ( 500 m ) 2 = 707 m . 500 m 1500 m 1000 m d

  12. Page 12
    Read page 12 as text

    Chapter 1 Measurement and Problem Solving 11 61 . According to the Pythagorean theorem, (1.0 m) 2 = (0.50 m) 2 + d 2 . S o d = ( 1.0 m ) 2 − ( 0.50 m ) 2 = 0.87 m . 62 . The 1 2 - i n. pizza is a better buy. A better buy gives you more area (more pepperoni) per dollar , and the area of a pizza depends on the square of the diameter. For the 9.0 in.:  ( 4.5 in. ) 2 $7.95 = 8.0 in. 2 /dollar . For the 12 in.:  ( 6.0 in. ) 2 $13.50 = 8.4 in. 2 /dollar . 63. The area of a 12 - in. pizza is π R 2 = π(6.0 in.) 2 = 113 in. 2 The area of two 8 - in. pizzas is 2(π R 2 ) = 2π(4.0 in.) 2 = 101 in. 2 This is not such a good deal! 64 . For the center circle: A =  r 2 =  (0.64 0 cm) 2 = 1.3 cm 2 . For the outer ring: A =  ( r 2 2 – r 2 1 ) =  [(1.78 cm) 2 – ( 1.66 cm) 2 ] = 1.3 cm 2 . So it is the same area for both , 1.3 cm 2 if calculated to two significant figures. 65 . t = x v = 31 mi 75 mi/h = 0.41 h = 25 min . 66 . One liter has 1000 c m 3 , and each cm 3 has 1000 mm 3 , so 1 liter has 1.0  10 6 mm 3 . (1 cm 3 ) = (10 mm) 3 = 1000 mm 3 T he total number of white cells in 5.0 L of blood is (7 000 /mm 3 )(5  10 6 mm 3 ) = 3.5  10 10 white cells . The total number of platelets in 5.0 L of blood is (250 000 /mm 3 )(5  10 6 mm 3 ) = 1.3  10 12 platelets . 67 . (a) T he number of hairs lost in a month is (65 hairs/day)(30 days) = 1950 hairs . (b) 15% bald means 85% with hair. So in one day, the “bald is beautiful” person loses (0.85)(65 hairs) = 55 hairs. In one year, the total is (365)(55 hairs) = 2.0  10 4 hairs . 0.50 m 0.50 m 1.0 m d

  13. Page 13
    Read page 13 as text

    12 College Physics Seventh Edition: Instructor Solutions Manual 68. (a) Since d = (13 mi) tan 25  and tan 25  < 1 (tan 45  = 1), d is ( 1 ) less than 13 mi. (b) d = (13 mi) tan 25  = 6.1 mi . 69. (a) It will be ( 3 ) less than the 190 mi/h . More time spent at lower speeds, so affect average speed to be below average of all speeds. (b) The time intervals for each lap is, respectively, Lap 1: t 1 = 2.5 mi 160 mi/h = 0.015625 h; Lap 2: t 2 = 2.5 mi 180 mi/h = 0.013889 h; Lap 3: t 3 = 2.5 mi 200 mi/h = 0.012500 h; Lap 4: t 4 = 2.5 mi 220 mi/h = 0.011364 h. So the total time is t total = t 1 + t 2 + t 3 + t 4 = 0.053378 h. Therefore the average speed for four laps is 4 ( 2.5 mi ) 0.053378 h = 187 mi/h . 70 . 118 mi = (118 mi)  1609 m 1 mi = 10 5 m, 307 mi = (307 mi)  1609 m 1 mi = 10 5 m, 279 ft = (279 ft)  1 m 3.28 ft = 10 2 m. So V = LWD  (10 5 m)(10 5 m)(10 2 m) = about 10 12 m 3 . 71 . (a) The answer is ( 2 ) between 5  and 7  . The sketch below illustrate s why it is the answer. (b) The elevation for the 2.0 km segment is (2.0 km) sin 5  = 0.174 km. The horizontal distance for the 2.0 km segment is (2.0 km) cos 5  = 1.99 km. The elevation for the 3.0 km segment is (3.0 km) sin 7  = 0.366 km. The horizontal distance for the 3.0 km segment is (3.0 km) cos 7  = 2.98 km. So the tangent of the net angle of rise is tan  = 0.174 km + 0.366 km 1.99 km + 2.98 km = 0.109. Therefore  = tan − 1 (0.109) = 6.2  . 72 . d = x tan 30  = (50 m − x ) tan 40  = (50 m) tan 40  − x tan 40  . x = ( 50 m ) tan 40  tan 30  + tan 40  = 29.6 m. So d = (29.6 m) tan 30  = 17 m . 73. Expressing the area of the horse pasture, A H , in terms of h gives d 30  40  x 50 m − x d 25  13 mi

  14. Page 14
    Read page 14 as text

    Chapter 1 Measurement and Problem Solving 13 2 1 (200 m) 2 (200 m) 2 3 3   = − = −     H h h A h h h The total area of his lot is   =  =     2 1 200 m 3 2 (200 m)sin 60 (200 m) 2 2 4 Tot A . The area for the horse pasture, A H , is 1/3 of the total area of the triangular lot, giving 2 2 1 3 (200 m) (200 m) 3 4 3   − =       h h Solving for h gives h = 314.6 m and h = 31.78 m. Since h cannot be larger than the side of the triangle, we discard the first solution, giving h = 31.8 m . 74. (a) The volume of the drilled hole is V =  r 2 L =  (0.0100 m) 2 (8.00 in.)(0.0254 m/in.) = 6.38  10 − 5 m 3 . The density of water is 1000 kg/m 3 , so the density of lead is (11.4)(1000 kg/m 3 ) = 1.14  10 4 kg/m 3 .  = m V ,  m =  V = (1.14  10 4 kg/m 3 )( 6.38  10 − 5 m 3 ) = 0.727 kg . (b) The total volume of the brick is (2.00 in.)(0.0254 m/in.)  (4.00 in.)(0.0254 m/in.)  (8.00 in.)(0.0254 m/in.) = 1.049  10 − 3 m 3 . The percentage of the original lead remaining in the brick is 1.049  10 − 3 m 3 − 6.38  10 − 5 m 3 1.049  10 − 3 m 3 = 93.9% . (c) The mass of the plastic is m P = [(2)(1000 kg/m 3 )](6.38  10 − 5 m 3 ) = 0.1276 kg. The mass of the lead brick with the hole drilled is m L = (1.14  10 4 kg/m 3 )(1.049  10 − 3 m 3 − 6.38  10 − 5 m 3 ) = 11.23 kg. Therefore the overall density is 11.23 kg + 0.1276 kg 1.049  10 − 3 m 3 = 1.08  10 4 kg/m 3 . 75. (a) Calling r the radius of the inner surface and using the fact that V r = 0.900 V Tot , we have = 3 3 4 4 (0.900) (20.0 cm) 3 3 r   , which gives r = 19.3 cm . (b) š = = 2 2 4 (20.0 cm) 1.07 4 (19.3 cm) outer inner A A  , or the outer area is 7% larger than the inner area.

  15. Page 15
    Read page 15 as text

    14 College Physics Seventh Edition: Instructor Solutions Manual 76. (a) tan 30° = y /50 km → y = (50 km) tan 30° = 28.9 km (b) cos 30° = (50 km)/ d → d = (50 km)/cos 30° = 57.7 km (c) Calling  the angle subtended by y from point C, which is 70 km from point A, we have: tan  = y /(70 km), which gives  = 22.4° north of due west . 77. (a) At constant speed, th e distance x traveled is x = vt , which gives t = x/v . Since v is the same in both cases, taking the ratio of t for both trips gives = = = = circumf circumf circumf diam diam diam diam diam x t x x v x t x x t   t circumf = π t diam = π (30.0 s) = 94.2 s (b) The diameter of the pool is x = vt = (0.500 m/s)(30.0 s) = 15.0 m, its radius is R = 7.50 m and its depth is d = 1.50 m. The volume is V = ( π R 2 ) d = π (7.50 m) 2 (1.50 m) = 265 m 3 . Converting this result to gallons gives     =        3 4 -3 3 1 L 1 gal (265 m ) 7.00 10 gal 3.788 L 10 m 78. (a) m =  V = (9.0 g/cm 3 )(1.0 cm)(2 .0 cm)(4.0 cm) = 72 g = 0.072 kg (b) Calling x the length of each side of the cube and using the fact that V = m /  gives = = 3 3 144 g 9.0 g/cm m x  → x = 2.5 cm 50 km E d B y 30° A East

  16. Page 16
    Read page 16 as text

    CHAPTER 2 KINEMATICS: DESCRIPTION OF MOTION Multiple Choice Questions : 1. (a). 2. (d). Choice (a) can never be true; choices (b) and (c) are sometimes true; only choice (d) is always true. 3. (c). 4. (c). 5. (c). Since the magnitude of the displacement vector is a distance and the magnitude of the velocity vector is the speed, choice (c) is correct. 6. (d). 7. (d). 8. (c). A negative acceleration only means that the acceleration is pointing in the negative direction. If an object is moving in the positive x - direction, the velocity of the object decreases. But if it is moving in the negative x - direction, its velocity will increase so it will speed up. 9. (d). Any change in either magnitude or direction results in a change in velocity. The brake and gearshift change the magnitude, and the steering wheel changes the direction. 10. (b). Since acceleration is the rate of change of the velocity, a constant acceleration implies a constant rate of change of the velocity, making (b) the correct choice. 11. (c). The speed of a decelerating object is decreasing, which can only happen if the acceleration is opposite to the velocity. 12. (c). In both cases, the ratio of the velocity change to the time interval for that change is the same, which means they have the same magnitude acceleration. 13. (c). 14. (d). The graph of x as a function of t is a parabola, depending on the square of the time. 15. (a). Since 0 = + , v v at + + + = = = 0 0 (0 ) 1 . 2 2 2 v v at v at 16. (d).

  17. Page 17
    Read page 17 as text

    16 College Physics Seventh Edition: Instructor Solutions Manual 17. (d). Free fall is motion under the influence of gravity alone, and the acceleration is g. The initial velocity does not affect the acceleration. 18. (c). It accelerates at 9.80 m/s 2 , so it increases its speed by 9.80 m/s during each second. 19. (a). The acceleration is not zero; it is 9.80 m/s 2 downward. 20. (c). It always accelerates at 9.80 m/s 2 downward. Conceptual Questions : 1. Yes, for a round - trip. No; distance is always greater than or equal to the magnitude of displacement. 2. No final position can be given. It may be anywhere from 0 to 750 m from the start. 3. The distance traveled is greater than or equal to 300 m. The object could travel a variety of ways as long as it ends up at 300 m north. If the object travels straight north, then the minimum distance is 300 m. 4. No, this is generally not the case. The average velocity can be zero (e.g. a round trip), while the average speed is never zero. 5. Yes, this is possible. The jogger can jog in the opposite direction during part of the jog (negative instantaneous velocity) as long as the overall jog is in the forward direction (positive average velocity). 6. Yes, although the speed of the car is constant, its velocity is not, because of the change in direction. A change in velocity means that there is acceleration, and the velocity (a vector quantity) can change by either changing direction, magnitude or both. 7. Not necessarily. The change in velocity is the key. If a fast - moving object does not change its velocity, its acceleration is zero. However, if a slow - moving object changes its velocity, it will have some non - zero acceleration. 8. Not necessarily. A negative acceleration can cause an increase in speed if the velocity is also negative (that is, if the velocity is in the same direction as the acceleration). 9. In part (a), the object accelerates uniformly first, maintains constant velocity (zero acceleration) for a while, and then accelerates uniformly at the same rate as in the first segment. In (b), the object accelerates uniformly. 10. The final velocity is o v since an equal amount of time is spent on acceleration and deceleration and both of these have the same magnitude.

  18. Page 18
    Read page 18 as text

    Chapter 2 K inematics: Description of Motion 17 11. If (assuming uniform accelerations) we apply the formula v2 = vo2 + 2 a ( x – x0 ), we see that both cars have the same initial speed ( v0 = 0) and the same final speed v , so the quantity a ( x – x 0 ) must be the same for both of them. Since car B travels twice as far as car A, its acceleration must therefore be half as large as that of car A, which tells us that = A B 2 a a . Another way to view this problem is the following. Both cars have the same average speed but travel unequal distances. Car A will take less time to reach the line because it has less distance to travel. Since the change in velocities are the same, car A will have a higher rate of change of velocity, thus A’s acceleration is greater than B’s. Since car A travels half the distance as B at the same average speed as A, it will take half as long to finish as B. Thus A will have twice the acceleration as B. 12. It is zero because the velocity is constant. 13. Not necessarily because even if the acceleration is negative, the object can still have positive velocity (meaning it is slowing) and the result could be a positive value for x . 14. Consider the displacement ( ) o x x − as one quantity; there are four quantities involved in each of the kinematic equations (Eqs. 2.8, 2.10, 2.11, and 2.12). All but one of the four must be known before one can solve for any unknown. 15. Yes, if the displacement is negative meaning the object accelerates to the left. 16. When it reaches the highest point, its velocity is zero (velocity changes from up to down, so it is zero at that instant ), but its acceleration is 2 9.80 m/s downward because the velocity is changing direction, signifying acceleration (due to gravity). 17. No, since one value of the instantaneous velocity does not tell you if the velocity is changing. It could be zero just for an instant and not zero either before or after that instant, thus it could be changing and the object could be accelerating. You need two values of instantaneous velocity to determine if an object is accelerating. 18. The ball moves at constant velocity because there is no gravitational acceleration in deep space. If an object’s acceleration is zero, then v is a constant, in magnitude and direction, but is not necessarily zero. 19. Since the first stone has been accelerating downward for a longer time, it will always have a higher speed and thus as time goes by it will have fallen further and thus the gap between them (  y ) will increase. 20. First, the gravitational acceleration on the Moon is only 1/6 of that on the Earth, or g =g 6. M E Hence dropped objects take longer to reach the surface than on the Earth, and tossed objects will go higher and stay in flight longer than on Earth. Secondly, there is no air resistance on the Moon, which means that all objects, regardless of mass and/or shape, will accelerate at the same rate, whereas this is only an approximation that works well for small massive objects on the Earth.

  19. Page 19
    Read page 19 as text

    18 College Physics Seventh Edition: Instructor Solutions Manual Exercises : 1 . Displacement is the change in position. Therefore the magnitude of the displacement for half a lap is 300 m . For a full lap (the car returns to its starting position), the displacement is zero . 2. v av = d/t , where d = 80 km + 50 km = 130 km t 1 = d 1 / v 1 = (80 km)/(100 km/h) = 0.800 h t 2 = d 2 / v 2 = (50 km)/(75 km/h) = 0.667 h t = t 1 + t 2 = 0.800 h + 0.667 h = 1.467 h v av = d/t = (130 km)/(1.467 h) = 89 km/h 3. t = d/v av , where     =         100 yd 3 ft 1 m 9.0 s 1 yd 3.281 ft av v = 10.16 m/s t = (100 m)/(10.16 m/s) = 9.8 s 4 . (a) s = d  t = ( 0.30 km ) ( 1000 m/km ) ( 10 min ) ( 60 s/min ) = 0.50 m/s . (b) s 1 = 1.20  s = 1.20(0.50 m/s) = 0.60 m/s. So  t = d s 1 = 300 m 0.60 m/s = 500 s = 8.3 min . 5 . 1 cc = 1 mL. This is analogous to average speed.  t = d s = 500 mL 4.0 mL/min = 125 min . 6 . s = d  t = 2 ( 25 m ) [2 ( 0 .5 0 min ) + 4.0 min] ( 60 s/min ) = 0.17 m/s . 7. The time going is t going = d/v 1 = (300 km)/(75 km/h) = 4.00 h The time returning is t return = d/v 2 = (300 km)/(85 km/h) = 3.53 h The average speed is v av = 2d/t total = (600 km)/(4.00 h + 3.53 h + 0.50 h) = 75 km/h The average velocity is zero because the net displacement is zero. 8 . (a) The answer is ( 2 ) greater than R but less than 2 R . For any right triangle, the hypotenuse is always greater than any one of the other two sides ( R ) and less than the sum of the sum of the other two sides ( R + R = 2 R ). (b) d = 2 2 (50m) (50m) + = 71 m . 50 m 50 m d 150 m 150 m

  20. Page 20
    Read page 20 as text

    Chapter 2 K inematics: Description of Motion 19 9. (a) = = =             150 km 1.43 h = 1.4 h mi 1 km 65 h 0.6214 mi d t v (b) = = =             150 km 1.165 h mi 1 km 80 h 0.6214 mi d t v The time you would save is 1.43 h – 1.165 h = 0.27 h or about 16 min . 10 . (a) The average velocity is ( 1 ) zero , because the displacement is zero for a complete lap. (b) s = d  t = 2  r  t = 2  ( 500 m ) 50 s = 63 m/s . 11 . (a) The magnitude of the displacement is ( 3 ) between 40 m and 60 m , because any side of a triangle cannot be greater than the sum of the other two sides. In this case, looking at the triangle shown, the two sides perpendicular to each other are 20 m and 40 m, respectively. The magnitude of the displacement is the hypotenuse of the right triangle, so it cannot be smaller than the longer of the sides perpendicular to each other. (b) d = ( 40 m ) 2 + ( 50m − 30 m ) 2 = 45 m .  = tan − −       1 50 m 30 m 40 m = 27  west of north . 12 . (a) s = d  t = 2 ( 7.1 m ) 2.4 s = 5.9 m/s . (b) A verage velocity is zero , because the ball is caught at the initial height so displacement is zero . 13 . (a) s = d  t = 27 m + 21 m ( 30 min ) ( 60 s/min ) = 2.7 cm/s . (b) The displacement is  x = ( 27 m ) 2 + ( 21 m ) 2 = 34.2 m. v =  x  t = 34.2 m ( 30 min ) ( 60 s/min ) = 1.9 cm/s . 14 . (a) v =  x  t , s o AB v = 1.0 m – 1.0 m 1.0 s – 0 = 0 ; BC v = 7.0 m – 1.0 m 3.0 s – 1.0 s = 3.0 m/s ; CD v = 9.0 m – 7.0 m 4.5 s – 3. 0 s = 1.3 m/s ; DE v = 7.0 m – 9.0 m 6.0 s – 4.5 s = − 1.3 m/s ; EF v = 2.0 m – 7.0 m 9.0 s – 6.0 s = − 1.7 m/s ; FG v = 2.0 m – 2.0 m 11.0 s – 9. 0 s = 0 ; BG v = 2.0 m − 1.0 m 11.0 s − 1.0 s = 0.10 m/s . (b) The motion of BC ‚ CD ‚ and DE are not uniform , since they are not straight lines. (c) The object changes its direction of motion at point D. So it has to stop momentarily , and v = 0 . 50 m d  40 m 30 m

  21. Page 21
    Read page 21 as text

    20 College Physics Seventh Edition: Instructor Solutions Manual 15 . Use s = d  t and v =  x  t . (a) 0-2.0 s s = 2.0 m – 0 2.0 s – 0 = 1.0 m/s ; 2.0 s-3.0 s s = 2.0 m – 2.0 m 3.0 s – 2.0 = 0 ; 3.0 s-4.5 s s = 4.0 m – 2.0 m 4.5 s – 3 .0 s = 1.3 m/s ; 4.5 s-6.5 s s = 4.0 m – ( – 1. 5 m ) 6.5 s – 4.5 s = 2.8 m/s ; 6.5 s-7.5 s s = – 1.5 m – ( – 1.5 m ) 7.5 s – 6.5 s = 0 ; 7.5 s-9.0 s s = 0 – ( – 1. 5 m ) 9.0 s – 7.5 s = 1.0 m/s ; (b) 0-2.0 s v = 2.0 m – 0 2.0 s – 0 = 1.0 m/s ; 2.0 s-3.0 s v = 2.0 m – 2.0 m 3.0 s – 2.0 = 0 ; 3.0 s-4.5 s v = 4.0 m – 2. 0 4.5 s – 3.0 s = 1.3 m/s ; 4.5 s-6.5 s v = – 1. 5 m – 4.0 m 6.5 s – 4.5 s = − 2.8 m/s ; 6.5 s-7.5 s v = – 1.5 m – ( – 1. 5 m ) 7.5 s – 6.5 s = 0 ; 7.5 s-9.0 s v = 0 – ( – 1. 5 m ) 9.0 s – 7.5 s = 1.0 m/s . (c) v 1.0 s = 0-2.0 s s 0 - 2.0 s = 1.0 m/s ; v 2.5 s = 2.0 s-3.0 s s = 0 ; v 4.5 s = 0 since the object reverses its direction of motion; v 6.0 s = 4.5 s-6.5 s s = − 2.8 m/s . (d) v 4.5 s - 9.0 s = 0 – 4.0 m 9.0 s – 4.5 s = − 0.89 m/s . 16 . (a) The magnitude of the displacement is d = ( 90.0 ft ) 2 + ( 10.0 ft ) 2 = 90.6 ft .  = 1 10.0 tan 90.0 −       = 6.3  above horizontal . (b)  v = 90.6 ft at 63  2.5 s = 36.2 ft/s at 6.3  . (c) Average speed depends on the total path length, which is not given. The ball might take a curved path. 17. (a) At t = 0, x = a = 10 m (b)  x = x 4 – x 2 = a – b (4.0 s) 2 – [ a – b (2.0 s) 2 ] = (0.50 m/s 2 )[(2.0 s) 2 – (4.0 s) 2 ]  x = − 6.0 m (c) At the origin, x = 0: 0 = a – bt 2 = = 2 10 m 0.50 m/s a t b = 4.5 s 18. (a)  x = 3(2.0 s) 2 m – 0 = 12 m, and  t = 2.0 s, so v av =  x /  t = (12 m)/(2.0 s) = 6.0 m/s (b)  x = 3(4.0 s) 2 m – 3(2.0 s) 2 m = 36 m, and  t = 2.0 s, so v av =  x /  t = (36 m)/(2.0 s) = 18 m/s 3333 30.0 yd = 90.0 ft 3333 d 3333  3333 10.0 ft

  22. Page 22
    Read page 22 as text

    Chapter 2 K inematics: Description of Motion 21 19 . d = 3.5 cm − 1.5 cm = 2.0 cm. s = d  t ,   t = 2.0 cm 2. 0 c m /mo = 1 month . 20 . The minimum speed is s = d  t = 675 km 7.00 h = 96.4 km/h = 59.9 mi/h . No , she does not have to exceed the 65 mi/h speed limit. 21 . (a) See the sketch on the right. d = 2 2 (400km) (300km) + = 500 km .  = 1 300 tan 400 −       = 37  east of north . (b)  t = 45 min + 30 min = 75 min = 1 .25 h. v =  x  t = 5 00 km  37  east of north 1.25 h = 400 km/h  37  east of north . (c ) s = d  t = 400 km + 300 km 1.25 h = 5 6 0 km/h . (d) Since speed involves total distance  which is greater than the magnitude of the displacement , the average speed is not equal to the magnitude of the average velocity. 22 . To the runner on the right, the runner on the left is running at a velocity of +4.50 m/s – ( − 3.50 m/s) = +8.00 m/s. So it takes  t =  x v = 100 m 8.00 m/s = 12.5 s . They meet at (4.50 m/s)(12.5 s) = 56.3 m ( relative to runner on left ) . 23 . 1 5.0 km/h = ( 1 5.0 km/h)  1000 m 1 km  1 h 3600 s = 4 . 167 m/s, 65.0 km/h = 18.06 m/s. So a =  v  t = 18.06 m/s – 4 . 167 m/s 6.00 s = 2 . 32 m/s 2 . 24 . 60 mi/h = (60 mi/h)  1609 m 1 mi  1 h 3600 s = 26.8 m/s. a =  v  t = 26.8 m/s – 0 3.9 s = 6.9 m/s 2 . 25 .  t = 26.8 m/s – 0 7.2 m/s 2 = 3.7 s . 26 . (a) The direction of the acceleration vector is ( 2 ) opposite to velocity as the object slows down. d d 1 d 2 

  23. Page 23
    Read page 23 as text

    22 College Physics Seventh Edition: Instructor Solutions Manual (b) 40.0 km/h = (40 km/h)  1000 m 1 km  1 h 3600 s = 11.1 m/s. So a =  v  t = 0 – 11.1 m/s 5.0 s = − 2.2 m/s 2 or − 2.2 m/s each s econd . The negative sign indicates that the acceleration vector is in opposite direction of velocity . 27 . 75 km/h = (75 km/h)  1000 m 1 km  1 h 3600 s = 20.8 m/s, 30 km/h = 8. 33 m/s. So a =  v  t = 8.33 m/s – 20.8 m/s 6.0 s = − 2. 1 m/s 2 . The negative sign indicates that the acceleration vector is in opposite direction of velocity. 28. (a) Once the object is released, the only force acting on it is gravity, so its acceleration is 2 9.80 m/s downward. (b) At the instant the object is dropped, the height of the balloon is v 2 = v o 2 + 2 a ( x – x 0 ) (15 m/s) 2 = 0 + 2(3.0 m/s 2 ) h → h = 37.5 m When the ball hits the ground, its vertical position is zero and its initial position was 37.5 m. Therefore v 2 = v o 2 + 2 a ( x – x 0 ) v 2 = (15 m/s) 2 + 2( – 9.8 m/s 2 )(0 – 37.5 m) → v = 31 m/s downward 29. (a) When they meet, x and t are the same for both cars. (60 km/h) t = ½ (3.0 m/s 2 ) t 2 [(60,000 m)/(3600 s)] t = 1.5 m/s 2 t 2 → t = 11.1 s The distance down the road is x = ½ at 2 = ½ (3.0 m/s 2 )(11.1 s) 2 = 190 m (b) From part (a), we see that t = 11 s . (c) v = v 0 + at = 0 + (3.0 m/s 2 )(11.1 s) = 33 m/s 30 . v = v + v o 2 = v o + 0 2 ,  v o = 2 v = − 70.0 km/h = − 19.4 m/s . a = v – v o t = 0 – ( – 1 9.4 m/s ) 7.00 s = +2.78 m/s 2 . In this case, the positive 2.78 m/s 2 indicates deceleration because the velocity is negative. 31 . (a) Given: v o = 3 5. 0 km/h = 9.72 m/s, a = 1.50 m/s 2 , x = 200 m (take x o = 0). Find: v . v 2 = v 2 o + 2 a ( x − x o ) = (9.72 m/s) 2 + 2(1.5 m/s 2 )(200 m) = 694 m 2 /s 2 ,  v = 2 6 . 3 m/s . (b) v = v o + at ,  t = v – v o a = 2 6.3 m/s – 9.72 m/s 1.50 m/s 2 = 1 1 . 1 s . 32 . Use the direction to the right as the positive direction.

  24. Page 24
    Read page 24 as text

    Chapter 2 K inematics: Description of Motion 23  a =  v  t = 11 m/s − ( − 35 m/s ) 0.095 s = 4.8  10 2 m/s 2 . This is a very large acceleration due to the change in direction of the velocity and the short contact time. 33 .  a 0 - 4.0 =  v  t = 8.0 m/s – 0 4.0 s – 0 = 2.0 m/s 2 ;  a 4.0 - 10.0 = 8.0 m/s – 8.0 m/s 10.0 s – 4.0 s = 0 ;  a 10.0 - 18.0 = 0 – 8.0 m/s 18.0 s – 10.0 s = − 1.0 m/s 2 . The object accelerates at 2.0 m/s 2 first, moves with constant velocity, then decelerates at 1.0 m/s 2 . 34 . (a)  a 0 - 1.0 s =  v  t = 0 − 0 1.0 s − 0 = 0 ;  a 1.0 s - 3.0 s = 8.0 m/s − 0 3.0 s − 1.0 s = 4.0 m/s 2 ;  a 3.0 s - 8.0 s = − 12 m/s − 8.0 m/s 8.0 s − 3.0 s = − 4.0 m/s 2 ;  a 8.0 s - 9.0 s = − 4 m/s − ( − 12.0 m/s ) 9.0 s − 8.0 s = 8.0 m/s 2 ;  a 9.0 s - 13.0 s = − 4.0 m/s − 4.0 m/s 13.0 s − 9.0 s = 0 . (b) Constant velocity of − 4.0 m/s . 35 . (a) See the sketch on the right. (b) The acceleration is negative as the object slows down (assume velocity is positive). v = v o + at = 25 m/s + ( − 5.0 m/s 2 )(3.0 s) = 10 m/s . (c) x = x 1 + x 2 + x 3 = (25 m/s)(5.0 s) + (25 m/s)(3.0 s) + 1 2 ( − 5.0 m/s 2 )(3.0 s) 2 + (10 m/s)(6.0 s) = 237.5 m = 2 .4  10 2 m . (d)  s = d  t = 237.5 m 14.0 s = 17 m/s . 36 . 72 km/h = (72 km/h)  1000 m 1 km  1 h 3600 s = 20 m/s. During deceleration,  t 1 =  v  a = 0 − 20 m/s − 1.0 m/s 2 = 20 s;  x 1 =  v 1  t 1 = 20 m/s + 0 2 (20 s) = 200 m. It would have taken the train 200 m 20 m/s = 10 s to travel 200 m. So it lost only 20 s – 10 s = 10 s during deceleration. During acceleration,  t 2 = 20 m/s – 0 0.50 m/s 2 = 40 s;  x 2 = 0 + 20 m/s 2 (40 s) = 400 m. V velocity (m/s) V 5 V 10 V 15 V 25 V 20 V 15 V 10 V 5 V time (s)

  25. Page 25
    Read page 25 as text

    24 College Physics Seventh Edition: Instructor Solutions Manual It would have taken the train 400 m 20 m/s = 20 s to travel 400 m. So it lost only 40 s – 20 s = 20 s during acceleration. Therefore, the train lost 2 min + 10 s + 20 s = 150 s in stopping at the station. 37 . The average velocity is  v =  x  t = 100 m 4.5 s = 22.2 m/s. v = o v v 2 + = v 2 . So the final velocity must be v = 2(22.2 m/s) = 44.4 m/s.  a =  v  t ,   t =  v  a = 44.4 m/s – 0 9.0 m/s 2 = 4.9 s > 4.5 s. So no , the driver did not do it. The acceleration must be 44.4 m/s – 0 4.5 s = 9.9 m/s 2 . 38 . Given: v o = 0, a = 2.0 m/s 2 , t = 5.00 s. Find: v and x (take x o = 0). (a) v = v o + a t = 0 + (2.0 m/s 2 )(5.0 s) = 10 m/s . (b) x = x o + v o t + 1 2 a t 2 = 0 + 0(5.00 s) + 1 2 (2.0 m/s 2 )(5.0 s) 2 = 25 m . 39 . Given: v o = 25 mi/h = 11 . 2 m/s, v = 0, x = 35 m (take x o = 0). Find: a and t . (a) v 2 = v o 2 + 2 a ( x – x o ),  a = v 2 − v o 2 2 x = ( 0 ) 2 − ( 11 . 2 m/s ) 2 2 ( 35 m ) = − 1.79 m/s 2 = − 1.8 m/s 2 . The negative sign indicates that the acceleration vector is in the opposite direction of the velocity. (b) v = v o + at ,  t = v − v o a = 0 − 1 1 . 2 m/s − 1 . 79 m/s 2 = 6.3 s . 40 . Given: v o = 60 km/h = 16.7 m/s, v = 40 km/h = 11.1 m/s, x = 50 m (take x o = 0). Find: a . v 2 = v o 2 + 2 a ( x − x o ),  a = v 2 − v o 2 2 x = ( 11.1 m/s ) 2 − ( 16.7 m/s ) 2 2 ( 50 m ) = − 1 .6 m/s 2 . 41 . (a) Given: v o = 100 km/h = 27.78 m/s, a = − 6.50 m/s 2 , x = 20.0 m (take x o = 0). Find: v . v 2 = v o 2 + 2 a ( x − x o ) = (27.78 m/s) 2 + 2( − 6.50 m/s 2 )(20.0 m) = 511.6 m 2 /s 2 , So v = 22.62 m/s = 81.4 km/h . (b) v = v o + at ,  t = v – v o a = 22.62 m/s – 27.78 m /s – 6.50 m/s 2 = 0.794 s . 42. (a) v = v 0 + at = 20 m/s + ( – 0.75 m/s 2 )(10 s) = 13 m/s (b) x = v 0 t + ½ at 2 = (20 m/s)(10 s) + ½ ( – 0.75 m/s 2 )(10 s) 2 = 160 m 43 . Given: v o = 250 km/h = 69.44 m/s, a = − 8.25 m/s 2 , x = 175 m (Take x o = 0). Find: t . x = x o + v o t + 1 2 a t 2 , so 13.7 s Point of Reverse Thrust

  26. Page 26
    Read page 26 as text

    Chapter 2 K inematics: Description of Motion 25 175 m = 0 + (69.44 m/s) t + 1 2 ( − 8.25 m/s 2 ) t 2 . Reduce to quadratic equation , 4.125 t 2 – 69.44 t + 175 = 0. Solving, t = 3.09 s and 13.7 s . The 13.7 s answer is physically possible but not likely in reality. After 3.09 s, it is 175 m from where the reverse thrust was applied, but the rocket keeps traveling forward while slowing down. Finally it stops. However, if the reverse thrust is continuously applied (which is possible, but not likely), it will reverse its direction and go back to 175 m from the point where the initial reverse thrust was applied; a process that would take 13.7 s. 44 . (a) Given: Car A: a A = 3.00 m/s 2 , v o = 2.50 m/s, t = 10 s. Car B: a B = 3.00 m/s 2 , v o = 5.00 m/s, t = 10 s. Find:  x (taking x o = 0). From x = x o + v o t + 1 2 a t 2 , x A = 0 + (2.50 m/s)(10 s) + 1 2 (3.00 m/s) 2 (10 s) 2 = 175 m, x B = 0 + (5.00 m/s)(10 s) + 1 2 (3.00 m/s) 2 (10 s) 2 = 200 m. So  x = x B − x A = 200 m − 175 m = 25 m . (b) From v = v o + at , v A = 2.50 m/s + (3.00 m/s)(10 s) = 32.5 m/s, v B = 5.00 m/s + (3.00 m/s)(10 s) = 35.0 m/s. So car B is faster. 45 . If the acceleration is less than 4.90 m/s 2 , then there is friction. Given: v o = 0, x = 15.00 m (take x o = 0), t = 3.0 s. Find: a . x = x o + v o t + 1 2 a t 2 ,  15.00 m = 0 + 1 2 a (3.0 s) 2 . a = 3.33 m/s 2 . So the answer is no , the incline is not frictionless. 46 . (a) ( 3 ) The object will travel in the + x - direction and then reverse its direction . This is because the object has initial velocity in the + x - direction, and it takes time for the object to decelerate, stop, and then reverse direction. We take x o = 0. Given: v o = 40 m/s, a = − 3.5 m/s 2 , x = 0 (“returns to the origin”). Find: t and v . (b) x = x o + v o t + 1 2 a t 2 ,  0 = 0 + (40 m/s) t + 1 2 ( − 3.5 m/s 2 ) t 2 . Reduce to quadratic equation: 1.75 t 2 – 40 t = 0. Solving , t = 0 or 22.9 s. The t = 0 answer corresponds to the initial time. So the answer is t = 23 s . (c) v = v o + at = 40 m/s + ( − 3.5 m/s 2 )(22.9 s) = − 40 m/s = 40 m/s in the − x - direction . 47 . Given: v o = 330 m/s, v = 0, x = 25 cm = 0. 25 m (Take x o = 0). Find: a . v 2 = v 2 o + 2 a ( x − x o ),  a = v 2 – v 2 o 2 x = ( 0 ) 2 – ( 330 m/s ) 2 2 ( 0. 25 m ) = − 2 . 2  10 5 m/s 2 .

  27. Page 27
    Read page 27 as text

    26 College Physics Seventh Edition: Instructor Solutions Manual The negative sign here indicates that the acceleration vector is in the opposite direction of the velocity. 48 . 40 km/h = (40 km/h)  1000 m 1 km  1 h 3600 s = 11.11 m/s. During the reaction time , the car travels a distance of d = (11.11 m/s)(0.25 s) = 2.78 m. So the car really has only 13 m – 2.78 m = 10.2 m to come to rest. Let’s calculate the stopping distance of the car. We take x o = 0. Given: v o = 11.1 m/s, v = 0, a = − 8.0 m/s 2 . Find: x . (Take x o = 0.) v 2 = v 2 o + 2 a ( x − x o ),  x = v 2 – v 2 o 2 a = 0 – ( 11.1 m/s ) 2 2 ( – 8.0 m/s 2 ) = 7.70 m. So it takes the car only 2.78 m + 7.70 m = 10.5 m (< 13 m) to stop. Yes , the car will stop before hitting the child. 49 . Repeat the calculation of Exercise 2. 48 . d = (11.1 m/s)(0.50 s) = 5.55 m. 5.55 m + 7.70 m = 13.3 m > 13 m. No , the car will not stop before hitting the child. 50 . Given: v o = 350 m/s, v = 210 m/s, x = 4.00 cm = 0.0400 m (take x o = 0). Find: t . x = x o + v t = v o + v 2 t ,  t = 2 x v o + v = 2 ( 0.0400 m ) 350 m/s + 210 m/s = 1.43  10 − 4 s . 51 . (a) For constant acceleration, the v vs. t plot is a straight line. Point p has coordinates of (0, v o ) and point q has coordinates of ( t , v o + at ). The distance from point q to point o is therefore at . The area under the curve is the area of the triangle 1 2 ( at ) t plus the area of the rectangle v o t . So A = v o t + 1 2 a t 2 = x − x o . (Here x − x o is displacement.) (b) The total area consists of two triangles from 0 to 4.0 s and from 10.0 s to 18.0 s and a rectangle from 4.0 s to 10.0 s. x − x o = A = 1 2 (4.0 s – 0)(8.0 m/s) + (10.0 s – 4.0 s)(8.0 m/s) + 1 2 (18.0 s – 10.0 s)(8.0 m/s) = 96 m . 52 . (a) ( 3 ) t 1 > t 2 . Since the object is accelerating, it will spend less time in traveling the second 3.00 m. (b) For the first 3.00 m: Given: v o = 0, a = 2.00 m/s 2 , x = 3.00 m (take x o = 0). Find: t . x = x o + v o t + 1 2 a t 2 = 0 + 0 + 1 2 a t 2 ,  t 1 = 2 x a = 2 ( 3.00 m ) 2.00 m/s 2 = 1.73 s . At the end of the first 3.00 m, the velocity of the object is v = v o + at = 0 + (2.00 m/s 2 )(1.73 s) = 3.46 m/s. This is then the initial velocity for the second 3.00 m. v q at v o p o 0 t t

  28. Page 28
    Read page 28 as text

    Chapter 2 K inematics: Description of Motion 27 F or the second 3.00 m: Given: v o = 3.46 m/s, a = 2.00 m/s 2 , x = 3.00 m. Find: t . x = x o + v o t + 1 2 a t 2 ,  3.00 m = 0 + (3.46 m/s) t 2 + 1 2 (2.00 m/s 2 ) t 2 2 . Reducing to quadratic equation, t 2 + 3.46 t − 3.00 = 0. Solving, t 2 = 0.718 s or − 4.18 s. 53 . (a) At the end of phase 1, the change in velocity is v 1 − 0 = v 1 . At the end of phase 2, the change in velocity is v 2 − v 1 . Since the object is accelerating, it spends less time in phase 2 than in phase 1. Since the change in velocity is equal to acceleration times the time, the change in velocity is greater in phase 1 than in phase 2. Or v 1 > v 2 − v 1 . That is 2 v 1 > v 2 . Therefore, v 1 > 1 2 v 2 . The answer is ( 3 ) v 1 > 1 2 v 2 . (b) For phase 1 : v o = 0, a = 0.850 m/s 2 , x = 50.0 m (take x o = 0). Find: v . v 2 = v 2 o + 2 a ( x − x o ) = 0 2 + 2(0.850 m/s 2 )(50.0 m) = 85.0 m 2 /s 2 ,  v 1 = 9.22 m/s . For phase 2: v o = 9.22 m/s, a = 0.850 m/s 2 , x = 50.0 m (take x o = 0). Find: v . v 2 = v 2 o + 2 a ( x − x o ) = (9.22 m/s) 2 + 2(0.850 m/s 2 )(50.0 m) = 170 m 2 /s 2 ,  v 2 = 13.0 m/s . So v 1 = 9.22 m/s > 1 2 v 2 = 1 2 (13.0 m/s) = 6.50 m/s. 54 . Take x o = 0. During acceleration: v o = 0, a = 1.5 m/s 2 , t = 6.0 s. x 1 = x o + v o t + 1 2 a t 2 = 0 + 0 + 1 2 (1.5 m/s 2 )(6.0 s) 2 = 27 m,

  29. Page 29
    Read page 29 as text

    28 College Physics Seventh Edition: Instructor Solutions Manual v = v o + at = 0 + (1.5 m/s 2 )(6.0 s) = 9.0 m/s. During constant velocity: x 2 = (9.0 m/s)(8.0 s) = 72 m. So v =  x  t = 27 m + 72 m 14 s = 7.1 m/s . 55 . (a) v (8.0 s) = − 12 m/s ; v (11.0 s) = − 4.0 m/s . (b) Use the result of Exercise 2. 5 1 a. The total area consists of a rectangle from 0 to 1.0 s, a triangle from 1.0 s to 5.0 s, a trapezoid from 5.0 s to 11.0 s, and a triangle from 6.0 s to 9.0 s with baseline at – 4.0 m/s. x − x o = A = 0 + 1 2 (5.0 s – 1.0 s)(8.0 m/s) + ( 11.0 s – 6. 0 s ) + ( 11.0 s – 5.0 s ) 2  ( – 4.0 m/s) + 1 2 (9.0 s – 6.0 s)[( – 12.0 m/s) – ( – 4.0 m/s)] = − 18 m . (c) The total distance (not displacement) is the addition of the absolute values of the areas. d =  A i = 0 + 1 2 (5.0 s – 1.0 s)(8.0 m/s) + ( 11.0 s – 6.0 s ) + ( 11.0 s – 5.0 s ) 2  (4.0 m/s) + 1 2 (9.0 s – 6.0 s)[(12.0 m/s – 4.0 m/s] = 50 m . 56 . (a) v 2 = v o 2 + 2 a ( x − x o ) ,  x − x o = v 2 − v o 2 2 a = 0 2 − v o 2 2 a = − 2 0 a . 2 v Taking x o = 0, so ( x − x o ) = x is proportional to v o 2 . If v o doubles, then x becomes 4 times as large. The answer is then ( 3 ) 4 x . (b) x 2 x 1 = 2 2o v 2 1o v = 60 2 40 2 = 2.25. So x 2 = 2.25 x 1 = 2.25 (3.00 m) = 6.75 m . 57 . (a) Given: a = 3.00 m/s 2 , t = 1.40 s, x = 20.0 m (take x o = 0). Find: v o . x = x o + v o t + 1 2 a t 2 ,  20.0 m = 0 + v o (1.40 s) + 1 2 (3.00 m/s 2 )(1.40 s) 2 . Solving, v o = 12.2 m/s . v = v o + at = 12.2 m/s + (3.00 m/s 2 )(1.40 s) = 16.4 m/s . (b) Given: v o = 0, a = 3.00 m/s 2 , v = 12.2 m/s. Find: x (take x o = 0). v 2 = v o 2 + 2 a ( x − x o ) ,  x − x o = v 2 − v o 2 2 a = ( 12.2 m/s ) 2 − 0 2 2 ( 3.00 m/s 2 ) = 24.8 m . (c) v = v o + at ,  t = v − v o a = 12.2 m/s − 0 3.00 m/s 2 = 4.07 s . 58 . 75 .0 mi/h = 33.5 m/s. (a) Given: v o = 33.5 m/s , a = − 1 .00 m/s 2 , x = 100 m (take x o = 0). Find: v . v 2 = v o 2 + 2 a ( x − x o ) = (33.5 m/s) 2 + 2( − 1.00 m/s 2 )(100 m) = 922 m 2 /s 2 . So v = 30.4 m/s .

  30. Page 30
    Read page 30 as text

    Chapter 2 K inematics: Description of Motion 29 (b) The initial velocity on dry concrete is then 30.4 m/s. Consider on dry concrete. Given: v o = 30.4 m/s, a = − 7.00 m/s 2 , v = 0 m. Find: x . v 2 = v o 2 + 2 ax ,  x = v 2 − v o 2 2 a = 0 2 − ( 30.4 m/s ) 2 2 ( − 7 .00 m/s 2 ) = 66.0 m. So the total distance is 100 m + 66.0 m = 166 m . (c ) Use v = v o + at. On ice: t 1 = v − v o a = 30.4 m/s − 33.5 m/s − 1.00 m/s 2 = 3.10 s, On dry concrete: t 2 = 0 − 30.4 m/s − 7.00 m/s 2 = 4.34 s. So the total time is 3.10 s + 4.34 s = 7.44 s . 59 . (a) Given: v o = 0, t = 2.8 s. Find: v (take y o = 0). v = v o – gt = 0 – (9.80 m/s 2 )(2.8 s) = − 27 m/s . (b) y = y o + v o t – 1 2 g t 2 = 0 + 0 − 1 2 (9.80 m/s 2 )(2.8 s) 2 = – 38 m . 60 . (a) We take y o = 0. y = y o + v o t − 1 2 g t 2 = − 1 2 g t 2 . So y is proportional to the time squared. Therefore twice the time means ( 3 ) four times the height. Given: v o = 0, t = 1.80 s. Find: y A and y B . (b) y A = − 1 2 (9.80 m/s 2 )(1.80 s) 2 = − 15.9 m. So the height of cliff A above the water is 15.88 m = 15.9 m . y B = y A 4 = 15. 88 m 4 = 3 . 97 m . 61 . (a) A straight line (linear) , slope = − g . (b) A parabola. 62 . Given: v o = 0, y = − 0.157 m (take y o = 0). Find: t . y − y o = v o t − 1 2 g t 2 = − 1 2 g t 2 ,  t = 2 y − g = 2 ( − 0.157 m ) − 9.80 m/s 2 = 0.18 s < 0.20 s.

  31. Page 31
    Read page 31 as text

    30 College Physics Seventh Edition: Instructor Solutions Manual It takes less than the average human reaction time for the dollar bill to fall. So the answer is no  not a good deal . 63. If the ball is in the air for twice as long on the second toss as it is on the first toss, the time for it to fall from its maximum height on the second toss will be twice as long as the time for it to fall from its maximum height on the first toss. Realizing that t 2 = 2 t 1 , the maximum heights reached are = 2 1 1 1 2 h gt = = = 2 2 2 1 1 2 2 1 1 2 2 (2 ) 2 h gt g t gt Taking the ratio of the heights gives = = 2 2 1 2 1 1 1 2 2 4 h gt h gt Therefore 2 1 = 4 h h , so it must be tossed 4 times as high . 64 . Given: v o = 15 m/s, v = 0 (maximum height). Find: y . (Take y o = 0.) v 2 = v o 2 − 2 g ( y − y o ),  y = v o 2 − v 2 2 g = ( 15 m/s ) 2 − ( 0 ) 2 2 ( 9.80 m/s 2 ) = 11 m . 65 . From Exercise 2. 6 4 , y = v o 2 − v 2 2 g = ( 15 m/s ) 2 − ( 0 ) 2 2 ( 1.67 m/s 2 ) = 67 m . 66 . Taking y o = 0, y = y o + v o t – 1 2 g t 2 = 0 + 0 − 1 2 gt 2 = – 1 2 g t 2 , so t = – 2 y g . For y = − 452 m, t = 9.604 s; for y = − 443 m, t = 9.508 s. So  t = 9.604 s – 9.508 s = 0.096 s . 67. First convert 100 km/h to m/s, giving 100 km/h = 27.78 m/s. Now use the formula v 2 = v 0 2 + 2 a ( x – x 0 ) and solve for x . (27.78 m/s) 2 = 0 + 2(9.80 m/s 2 ) x → x = 39.4 m 68 . Given: v o = 6.0 m/s, y = − 12 m (take y o = 0). Find: t and v . (a) y = y o + v o t − 1 2 g t 2 ,  − 12 m = 0 + (6.0 m/s) t – 1 2 (9.80 m/s 2 ) t 2 . Or 4.9 t 2 – 6.0 t – 12 = 0. Solving, t = 2.3 s or – 1.1 s. The negative time is discarded. (b) v = v o − g t = 6.0 m/s – (9.80 m/s 2 )(2.29 s) = − 16 m/s . 69 . (a) When the ball rebounds, it is a free fall with an initial upward velocity. At the maximum height, the velocity is zero. Taking y o = 0, v 2 = v 2 o – 2 g ( y − y o ),  y = v 2 o – v 2 2 g . So y max = v 2 o 2 g . Therefore, the height depends on the initial velocity squared. 95% = 0.95 and 0.95 2 = 0.90 < 0.95.

505 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Physics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.