37&45&50 - TAM 212 Spring, 2004 1.37 Given: x 0 =...

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TAM 212 Spring, 2004 1.37 Given: m x 10 0 = at t = 0; 2 2 t v = m/s (t is in seconds) Required: ( 29 ( 29 t x t a , for s t s 5 2 Solution: ( 29 ( 29 ( 29 ( 29 2 2 2 / / 2 / / / s m t s m t dt d dt t dv t a = = = ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 m t m t x m dt t t x t x dt t v dt dt t x d t v dt t x d t t t t t t 6 / 10 2 / 3 0 2 0 0 0 = - = - = = = = = ( 29 ( 29 m m t t x 10 6 / 3 + = 1.45 Given: 2 / 12 s m t x P i i a = = (t is in seconds); ( 29 s m s t P / 2 1 i v = = ; ( 29 m s t OP i r 3 2 = = (a) Required: ( 29 ( 29 s t s t OP OP 0 5 = - = r r Solution: We start by finding the velocity and position as a function of time and then plug in the time we want. We can use either direct or indirect integration to do this: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 6 6 2 6 1 12 2 1 2 1 1 1 - = - = = - = = = = = = = t t x t t x t x tdt dt t x dt dt t x d t x dt t x d t t t t t t t t ( 29
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This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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37&45&50 - TAM 212 Spring, 2004 1.37 Given: x 0 =...

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