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# Chapter 2 - Solution s Chapter 2 2.1 We assume that you are...

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2 2 m 2 10 s 0.02 s 100 m s x t - = = = × = v 20 ft x t = = v 1 yr 1 m 3.281 ft 1 yr 7 7 2 10 m s 3.156 10 s - = × × 100 ft x t = = v 1 yr 1 m 3.281 ft 1 yr 6 7 1 10 m s 3.156 10 s - = × × 3 3 10 mi x t × = = v 10 mm 1609 m yr 1 mi 3 10 mm 1 m 8 5 10 yr = × f i f i x x x t t t - = = - v Solution s Chapter 2 2.1 We assume that you are approximately 2 m tall and that the nerve impulse travels at uniform speed. The elapsed time is then 2.4 (a) or in particularly windy times, (b) The time required must have been 2.6 The average velocity over any time interval is

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10.0 m 0 5.00 m s 2.00 s 0 x t - = = = - v 5.00 m 0 1.25 m s 4.00 s 0 x t - = = = - v 5.00 m 10.0 m 2 .50 m s 4.00 s 2.00 s x t - = = = - - v 5.00 m 5.00 m 3.33 m s 7.00 s 4.00 s x t - - = = = - - v 2 1 2 1 0 0 0 8.00 s 0 x x x t t t - - = = = = - - v ( 29 ( 29 1 h Displacement 85.0 km h 35.0 min 130 km 180 km 60.0 min x = ∆ = + = ( 29 1 h 35.0 min 15.0 min 2.00 h 2.84 h 60.0 min t = + + = (a) (b) (c) (d) (e) 2.7 (a) (b) The total elapsed time is so,
180 km 63.4 km h 2.84 h x t = = = v ( 29 ( 29 4 Circumference of Orbit 2 2 Earth’s radius + 200 miles 2 3963 200 mi 2 .61 10 mi r π π π = = = + = × 4 2 .61 10 mi 1.32 h 19 800 mi h Circumference t average speed × = = = ( 29 ( 29 1 1 1 1 1 x L L t t t + = = = + v ( 29 ( 29 2 2 2 2 2 x L L t t t - = = = - v ( 29 ( 29 ( 29 ( 29 total 1 2 total 1 2 1 2 1 2 total 0 0 x x x L L t t t t t t t + + - = = = = = + + + v ( 29 ( 29 ( 29 ( 29 1 2 trip 1 2 1 2 1 2 total 2 . x x L L total distance traveled L ave speed t t t t t t t + + + - = = = = + + + 1 1 x t t = = v v ( 29 ( 29 ( 29 ( 29 1 1 1 89.5 km h 77.8 km h 0.367 h 77.8 km h 28.5 km x t t t = = + = + 11. The distance traveled by the space shuttle in one orbit is Thus, the required time is 2.12 (a) (b) (c) (d) 2.13 The total time for the trip is t = t 1 + 22.0 min = t 1 + 0.367 h, where t 1 is the time spent traveling at υ 1 = 89.5 km/h . Thus, the distance traveled is , which gives or

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( 29 1 89.5 km h 77.8 km h 28.5 km t - = 1 0.367 h 2.80 t t h = + = ( 29 ( 29 x t 77.8 km h 2.80 h 218km = = = v 0.20 m = + t h x x ( 29 120 s 0.20 m t h t t = - + v v ( 29 ( 29 ( 29 0.100 m s 2 .0 m s 120 s 0.20 m t t = - + 2 1.3 10 s t = × 2 60.0 mi h 0 1 0.447 m s 0.391 s 7 9.80 m s 1 mi h g t a g     - = = =         v 0 25.0 m s = + v 2 0.750 m s a = + From which, t 1 = 2.44 for a total time of. Therefore, . 2.14 (a) At the end of the race, the tortoise has been moving for time t and the hare for a time t - 2.0 min = t - 120 s. The speed of the tortoise is υ t = 0.100 m/s, and the speed of the hare is υ h = 20 υ t = 2.0 m/s. The tortoise travels distance x t , which is 0.20 m larger than the distance x h traveled by the hare. Hence, ` which becomes or This gives the time of the race as (b) 2.22 From a υ = υ / t, the required time is seen to be 2.26 We choose eastward as the positive direction so the initial velocity of the car is given by . (a) In this case, the acceleration is and the final velocity will be
( 29 ( 29 2 0 25.0 m s 0.750 m s 8.50 s 31.4 m s at = + = + + + = + v v 31.4 m s eastward = v 2 0.750 m s a = - ( 29 ( 29 2 0 25.0 m s 0.750 m s 8.50 s 18.6 m s at = + = + + - = + v v 18.6 m s eastward = v ( 29 2 2 2 f i a x = + v v ( 29 2 2 2 f i a x - = v v ( 29 ( 29 ( 29 ( 29 2 2 2 2 2 2 30.0 m s 20.0 m s 1.25 m s 2 2 2.00 10 m f i a x - - = = = × v v a t = ∆ v 2 30.0 m s 20.0 m s 8.00 s 1.25 m s f i t a a - - = = = = v v v or (b)

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Chapter 2 - Solution s Chapter 2 2.1 We assume that you are...

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