7_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 12 t 2 1.5 s = Given v p t 2 () 0 = t 2 Find t 2 = t 2 2s = v ave s p t 1 s p t 0 t 1 = v ave 4 m s = v avespeed s p t 2 s p t 0 s p t 1 s p t 2 + t 1 = v avespeed 6 m s = a 1 a p t 1 = a 1 18 m s 2 = Problem 12–10 A particle is moving along a straight line such that its acceleration is defined as a = kv . If v = v 0 when d = 0 and t = 0, determine the particle’s velocity as a function of position and the distance the particle moves before it stops. Given: k 2 s = v 0 20 m s = Solution: a p v k v = v s v d d k v = v 0 v v 1 d k s p = Velocity as a function of position vv 0 ks p = Distance it travels before it stops 0 v 0 p = s p v 0 k
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