61_Dynamics 11ed - Engineering Mechanics Dynamics Chapter 12 when x = x1 a = 400 ft Given ft s v0 = 2 a0 = 0 ft x1 = 20 ft 2 s Solution y= x

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Engineering Mechanics - Dynamics Chapter 12 when x = x 1 . Given: a 400 ft = v 0 2 ft s = a 0 0 ft s 2 = x 1 20 ft = Solution: Velocity : Taking the first derivative of the path yx x 2 a = we have, v y v x 1 2 x a = v 0 1 2 x a = v x1 v 0 = v y1 v 0 1 2 x 1 a = v 1 v x1 2 v y1 2 + = v x1 2 ft s = v y1 1.8 ft s = v 1 2.691 ft s = Acceleration : Taking the second derivative: a y a x 1 2 x a 2 v x 2 a = a 0 1 2 x a 2 v 0 2 a = a x1 a 0 = a y1 a 0 1 2 x 1 a 2 v 0 2 a = a 1 a x1 2 a y1 2 + = a x1 0 ft s 2 = a y1 0.0200 ft s 2 = a 1 0.0200 ft s 2 = Problem 12–77
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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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