17_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 12 Solution: av s p v d d = ks p 1 3 = 0 v v v d v 2 2 = 0 s p s p 1 3 d = 3 2 p 2 3 = v 3 k s p 1 3 = t s p d d = 3 k t 0 s p s p s p 1 3 d = 3 2 s p 2 3 = s p t () 23 k t 3 3 2 = s p t 1 41.6 m = vt t s p t d d = 1 10.39 m s = * Problem 12-28 The acceleration of a particle along a straight line is defined by a p = b t + c . At t = 0, s p = s p0 and v p = v p0 . When t = t 1 determine (a) the particle's position, (b) the total distance traveled, and (c) the velocity. Given: b 2 m s 3 = c 9 m s 2 = s p0 1m = v p0 10 m s = t 1 9s = Solution: a p bt c + = v p b 2 t 2 ct + v p0 + = s p b 6 t
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This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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