problem02_20 solution

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

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2.20: a) The bumper’s velocity and acceleration are given as functions of time by 5 6 2 ) s m 600 . 0 ( ) s m 60 . 9 ( t t dt dx v x - = = . ) s m 000 . 3 ( ) s m 60 . 9 ( 4 6 2 t dt dv a x - = = There are two times at which v = 0 (three if negative times are considered), given by t = 0 and t 4 = 16 s 4 . At t = 0, x = 2.17 m and
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Unformatted text preview: a x = 9.60 s m 2 . When t 4 = 16 s 4 , x = (2.17 m) + (4.80 s m 2 ) ) s 16 ( 4 – (0.100) 6 s m )(16 s 4 ) 3/2 = 14.97 m, a x = (9.60 s m 2 ) – (3.000 6 s m )(16 s 4 ) = –38.4 s m 2 . b)...
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• Velocity, S4, 0.600 m, 2.17 m, 3.000 m, 4.80 m

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