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Question

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 (at t=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)

ax=4c(v2y+yay)

Since, it is given at t=0, x=0 , y=0
Therefore,

ax=4cv2y=b

vy=b4c

Hence, option (B) is the right choice.

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