A particle projected with velocity $V_{0}$ along x-axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e $a = - α x^{2}$ . The distance at which the particle stops is

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${(\frac{3 V_{0}}{2 α})}^{1 / 3}$

${(\frac{3 V_{0}^{2}}{2 α})}^{1 / 3}$

$\sqrt{\frac{3 V_{0}}{2 α}}$

$\sqrt{\frac{3 V_{0}^{2}}{2 α}}$

answer is D.

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Detailed Solution

$a = \frac{d v}{d t} = \frac{d v}{d x} . \frac{d x}{d t} = V . \frac{d v}{d x} = - α x^{2}$

$\int_{v_{0}}^{0} V d v = - α \int_{0}^{s} x^{2} d x$

${[\frac{V_{0}^{2}}{2}]}_{v_{0}}^{0} = - α {[\frac{x^{3}}{3}]}_{0}^{s}$

$\frac{V_{0}^{2}}{2} = \frac{α S^{3}}{3}$

$S = {[\frac{3 V_{0}^{2}}{2 α}]}^{1 / 3}$

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