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Q.

A particle is projected with a 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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a

$\sqrt{\frac{3 v_{0}}{2 α}}$

b

${(\frac{3 v_{0}}{2 α})}^{1 / 3}$

c

$\sqrt{\frac{3 v_{0}^{2}}{2 α}}$

d

${(\frac{3 v_{0}^{2}}{2 α})}^{1 / 3}$

answer is D.

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Detailed Solution

$a = \frac{d v}{d t} \Rightarrow \frac{d v}{d x} . \frac{d x}{d t} = v \frac{d v}{d t} = α x^{2} \Rightarrow \int_{v_{0}}^{0} v d v = \int_{0}^{s} - α x^{2}; {[\frac{v^{2}}{2}]}_{v_{0}}^{0} = - α {[\frac{x^{3}}{3}]}_{0}^{s} = \frac{v_{0}^{2}}{2} = \frac{α s^{3}}{3}$

$S = {(\frac{3 v_{0}^{2}}{2 α})}^{\frac{1}{3}}$

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