Question
Mon September 17, 2012 By: Meet
 

a particle move along the parabolic path x=ysquare+2y+2 in such a way that the y component of velocity vector remains 5 m/s during the motion. the magnitude of the acceleration is?

Expert Reply
Tue September 18, 2012
x = y2 + 2y +2
 
differentiating the equation with respect to time
 
thus dx/dt = 2y dy/dt + 2dy/dt
or Vx = 2y Vy + 2 Vy
 
here Vy = 5 m/s a constant
 
now Vx = 10y + 10
differentiating again with time:
d2x/dt2 = a = 10 dy/dt = 10 Vy = 50 m/s2
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