86_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 12 Solution: a t bs = v 0 v 1 v v d 0 s 1 s d = v 1 2 2 v 0 2 2 b 2 s 1 2 = v 1 v 0 2 1 2 + = v 1 4.583 m s = a t1 1 = a n1 v 1 2 ρ = a 1 a t1 2 a n1 2 + = a 1 0.653 m s 2 = * Problem 12–116 The particle travels with a constant speed v along the curve. Determine the particle’s acceleration when it is located at point x = x 1 . Given: v 300 mm s = k 20 10 3 × mm 2 = x 1 200 mm = Solution: yx () k x = y' x x d d = y'' x x y' x d d = x
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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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