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Question

A particle starts from the origin of co-ordinates at time t=0 and moves in the xy plane with a constant acceleration α in the y – direction. Its equation of motion y=βx2. Its velocity component in the x-direction is

A
αβ
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B
2αβ
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C
α2β
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D
α2β
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Solution

The correct option is D α2β
y=βx2
Differentiating w.r.t t,
dydt=2βxdxdt=2βxvx
Differenciating w.r.t t again,
d2ydt2=2βvxdxdt+2xβdvxdt
d2ydt2=2βv2x+2xβdvxdt
dvxdt=0 ( There is no acceleration in the horizontal direction
d2ydt2=α (given)α=2βv2xvx=α2β
Alternate solution:
y=βx2
Comparing with the equation of the trajectory,
y=xtanθgx22u2cos2θ
tanθ=0θ=0 ...(1)g2u2cos2θ=β ...(2)
acceleration in y direction, g=α ..(3)
Putting (1) and (3) in (2), we get
α2u2(cos20)=βu2=α2βu=α2β

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