First slide

Rectilinear Motion

Question

A particle is 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= $α$ x². The distance at which the particle stops is

Difficult

A

$\sqrt{\frac{3 v_{0}}{2 α}}$
B

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

$\sqrt{\frac{3 v_{0}^{2}}{2 α}}$
D

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

Solution

$\begin{array}{l} a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} = - α x^{2} (Given) \\ \Rightarrow \int_{v_{0}}^{0} v d v = - α \int_{0}^{s} x^{2} d x \Rightarrow {[\frac{v^{2}}{2}]}_{v_{0}}^{0} = - α {[\frac{x^{3}}{3}]}_{0}^{S} \\ \Rightarrow \frac{v_{0}^{2}}{2} = \frac{α S^{3}}{3} \Rightarrow S = {(\frac{3 v_{0}^{2}}{2 α})}^{\frac{1}{3}} \end{array}$

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