University Physics with Modern Physics with Mastering Physics (11th Edition)

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2.68: One convenient way to do the problem is to do part (b) first; the time spent accelerating from rest to the maximum speed is . s 80 2 s m 2.5 s m 20 = At this time, the officer is . m 0 . 80 ) s m 5 . 2 ( 2 ) s m 20 ( 2 2 2 2 1 1 = = = a v x This could also be found from , ) 2 / 1 ( 2 1 1 t a where t 1 is the time found for the acceleration. At this time the car has moved (15 s m )(8.0 s) = 120 m, so the officer is 40 m behind the car. a) The remaining distance to be covered is 300 m –
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Unformatted text preview: x 1 and the average speed is (1/2) ( v 1 + v 2 ) = 17.5 s m , so the time needed to slow down is , s . 16 s m 5 . 17 m 80 m 360 =-and the total time is 24.0 s. c) The officer slows from 20 s m to 15 s m in 16.0 s (the time found in part (a)), so the acceleration is –0.31 2 s m . d), e)...
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