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A particle projected with velocity V0 along x-axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e a=αx2 . The distance at which the particle stops is

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a
3V02α
b
(3V02α)1/3
c
3V022α
d
(3V022α)1/3

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detailed solution

Correct option is D

a=dvdt=dvdx.dxdt=V.dvdx=−αx2 ∫v00Vdv=−α∫0sx2dx [V022]v00=−α[x33]0s V022=αS33 S=[3V022α]1/3


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