Question

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. $a=-a x^{2}$. The distance at which the particle stops is

• A. $\sqrt{\frac{3 v_{0}}{2 a}}$
• B. $\left(\frac{3 v_{0}}{2 a}\right)^{\frac{1}{3}}$
• C. $\sqrt{\frac{3 v _{0}^{2}}{2 a }}$
• D. $\left(\frac{3 v_{0}^{2}}{2 a}\right)^{\frac{1}{3}}$

Answer 1

Answer: (D)

$a=\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}$
$=v \frac{d v}{d x}=-d x^{2}$
or, $\int\limits_{V_{0}}^{0} v d v=-\alpha \int\limits_{0}^{s} x^{2} d x$
or, $\left[\frac{V^{2}}{2}\right]_{V_{0}}^{0}=-\alpha\left[\frac{n^{2}}{3}\right]_{0}^{s}$
or,$\frac{V_{0}^{2}}{2}=\frac{d s^{3}}{3}$
$\therefore S=\left[\frac{3 V_{0}^{2}}{2 d}\right]$