problem02_51 solution

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

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2.51: a) From Eqs. (2.17) and (2.18), with v 0 =0 and x 0 =0, 3 4 3 3 2 2 0 2 ) s m 040 . 0 ( ) s m 75 . 0 ( 3 2 ) ( t t t B t A dt Bt At v t x - = - = - = . ) s m 010 . 0 ( ) s m 25 . 0 ( 12 6 3 2 4 4 3 3 4 3 0 3 2 t t t B t A dt t B t A x t - = - = - = b) For the velocity to be a maximum, the acceleration must be zero; this occurs at t =0 and s 5 . 12 = =
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Unformatted text preview: B A t . At t =0 the velocity is a minimum, and at t =12.5 s the velocity is s. m 1 . 39 s) 5 . 12 )( s m 040 . ( s) 5 . 12 )( s m 75 . ( 3 4 2 3 =-= x v...
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