18_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 12 t 3 c c 2 2 bv p0 + b = t 3 7.702 s = s p2 b 6 t 2 3 c 2 t 2 2 + v p0 t 2 + s p0 + = s p2 7.127 m = s p3 b 6 t 3 3 c 2 t 3 2 + v p0 t 3 + s p0 + = s p3 36.627 m = ds p2 s p0 s p2 s p3 + s p1 s p3 + = d 56.009 m = c ) The velocity v p1 b 2 t 1 2 ct 1 + v p0 + = v p1 10 m s = Problem 12–29 A particle is moving along a straight line such that its acceleration is defined as a = k s 2 . If v = v 0 when s = s p0 and t = 0, determine the particle’s velocity as a function of position. Given: k 4 1 ms 2 = v 0 100 m s = s p0 10 m = Solution: av s p v d d = ks p 2 = v 0 v v v d s p0 s p s p p 2 d = 1 2 v 2 v 0 2 () 1 3 p 3 s p0 3 = vv 0 2 2 3 p 3 s p0 3 + = Problem 12–30 A car can have an acceleration and a deceleration
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