Q. A particle is 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., $a = - ax^{2}.$The distance at which the particle stops is
A) $\sqrt{\begin{array}{r}
\frac{3v_{0}}{2\alpha} \\
\
\end{array}}$
B) $\left( \begin{array}{r}
\frac{3v_{0}}{2\alpha} \\
\
\end{array} \right)^{\frac{1}{3}}$
C) $\sqrt{\begin{array}{r}
\frac{3v_{0}^{2}}{2\alpha} \\
\
\end{array}}$
D) $\left( \begin{array}{r}
\frac{3v_{0}^{2}}{2\alpha} \\
\
\end{array} \right)^{\frac{1}{3}}$
Explanation
(d) $a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} = - \alpha x^{2}\ \lbrack Given\rbrack$
$\Rightarrow \ \int_{v_{0}}^{0}{vdv = \alpha\int_{0}^{S}x^{2}}dx \Rightarrow {\begin{array}{r}
\left\lbrack \begin{array}{r}
\frac{v^{2}}{2} \\
\
\end{array} \right\rbrack_{v_{0}}^{0} \\
\
\end{array}^{\ }}^{\ } = - \alpha\left\lbrack \begin{array}{r}
\frac{x^{3}}{3} \\
\
\end{array} \right\rbrack_{0}^{S}$
$\Rightarrow \frac{v_{0}^{2}}{2} = \frac{\alpha S^{3}}{3} \Rightarrow S = \ \left( \begin{array}{r}
\frac{3v^{2}}{2\alpha} \\
\
\end{array} \right)^{\frac{1}{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.