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

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${(\frac{3 {ν^{2}}_{0}}{2 α})}^{\frac{1}{3}}$

$\sqrt{\frac{3 {ν^{2}}_{0}}{2 α}}$

${(\frac{3 ν_{0}}{2 α})}^{\frac{1}{3}}$

$\sqrt{\frac{3 ν_{0}}{2 α}}$

answer is D.

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Detailed Solution

According to question we get,

$= v \frac{d v}{d x} = - a x^{2} = \int_{v_{0}}^{0} v d v = - α \int_{0}^{x} x^{2} d x = \frac{- v_{0}^{2}}{2} = - α \frac{x^{3}}{3} . x = {(\frac{3 v_{0}^{2}}{2 α})}^{\frac{1}{3}}$

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