10_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 12 v avespeed s p t 1 () s p t 2 s p t 2 ( ) s p t 0 ( ) + t 1 = v avespeed 3 m s = a 1 at 1 = a 1 2 m s 2 = Problem 12–15 A particle is moving along a straight line such that when it is at the origin it has a velocity v 0 . If it begins to decelerate at the rate a = bv 1/2 determine the particle’s position and velocity when t = t 1 . Given: v 0 4 m s = b 1.5 m s 3 = t 1 2s = av bv = Solution: = t v d d = v 0 v v 1 v d2 v v 0 = bt = vt v 0 1 2 + 2 = 1 0.25 m s = s p t 0 t t d = s p t 1 3.5 m = * Problem 12-16 A particle travels to the right along a straight line with a velocity
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