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Q.

A particle is projected with velocity $ν_{ο}$ along $x - axis$ . The deceleration on the particle is proportional to the square of the distance from the origin i.e., $a = {αx}^{3} .$ The distance at which the particle stops is

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a

${(\frac{3 v_{o}}{2 α})}^{\frac{1}{4}}$

b

$\sqrt{\frac{3 υ_{0}}{2 α}}$

c

$\sqrt{\frac{3 υ {}_{0}^{2}}{2 α}}$

d

none of these

answer is D.

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Detailed Solution

$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx} = - {αx}^{3}$ (given)
$\begin{aligned} \Rightarrow \int_{i_{0}}^{0} vdv = - α \int_{0}^{s} x^{3} dx \Rightarrow {[\frac{v^{2}}{2}]}_{v_{0}}^{0} = - α {[\frac{x^{4}}{4}]}_{0}^{5} \\ \Rightarrow \frac{v_{0}^{2}}{2} = \frac{{αS}^{4}}{4} \Rightarrow S = {(\frac{2 v_{0}^{2}}{α})}^{\frac{1}{4}} \end{aligned}$

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