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. $a = \alpha x^{2}$, the distance at which the particle stops is
A) $\sqrt{\frac{3V_{0}}{2\alpha}}$
B) $\left( \frac{3V_{0}}{2\alpha} \right)^{\frac{1}{3}}$
C) $\sqrt{\frac{2V_{0}^{2}}{3\alpha}}$
D) $\left( \frac{3V_{0}^{2}}{2\alpha} \right)^{\frac{1}{3}}$
Explanation
(d) Acceleration $a = \alpha x^{2} \Rightarrow \frac{dV}{dt} = \alpha x^{2}$
$\Rightarrow dV = \alpha x^{2}dt \Rightarrow dV = \alpha x^{2}dx\frac{dt}{dx}$
$\Rightarrow \int_{0}^{v_{0}}{VdV} = \int_{0}^{x}{\alpha x^{2}dx}\ \Rightarrow \frac{V_{0}^{2}}{2} = \frac{\alpha.x^{3}}{3}$
$\Rightarrow x = \left\lbrack \frac{3V_{0}^{2}}{2\alpha} \right\rbrack^{- 1/3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.