A particle starts from the origin at t=0 and moves in the x−y plane with constant acceleration b in the x−direction. The equation of motion is x=2cy2. The y−component of its velocity (att=0) is
A
√b4c
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B
b4c
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C
√4cb
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D
√bc
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Solution
The correct option is A√b4c Given, ax=b=dvxdt
From equation of motion vx=ux+axt vx=ux+bt
Also given, ay=0=dvydt
vy=uy+ayt vy=uy
Given equation, x=2cy2
differentiating w.r.t. time, we get
dxdt=2c(2y)(dydt)=4cyvy
Again differentiating w.r.t. time, we get dvxdt=4c(dydtvy+ydvydt)