60_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 12 v x 2 v y 2 + v x 2 1 a 2 b 2 e 2 bx + () = v 0 2 = v x v 0 1 a 2 b 2 e 2 + = v y abe v 0 1 a 2 b 2 e 2 + = In specific case x 1 1 b ln y 1 a = v x1 v 0 1 a 2 b 2 e 2 1 + = v x1 0.398 ft s = v y1 1 v 0 1 a 2 b 2 e 2 1 + = v y1 3.980 ft s = Problem 12-75 The path of a particle is defined by y 2 = 4 kx , and the component of velocity along the y axis is v y = ct , where both k and c are constants. Determine the x and y components of acceleration. Solution: y 2 4 kx = 2 yv y 4 kv x = 2 v y 2 2 ya y + 4 ka x = v y ct = a y c = 2 2 2 yc + 4 x = a x c 2 k t 2 +
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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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