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. $\sqrt{\frac{3v_0}{2a}}$
• B. $⟮\frac{3v_0}{2a}{⟯}^{\frac{1}{3}}$
• C. $\sqrt{\frac{3v_0^2}{2a}}$
• D. $⟮\frac{3v_0^2}{2a}{⟯}^{\frac{1}{3}}$

Answer 1

The correct option is B $⟮\frac{3v_0}{2a}{⟯}^{\frac{1}{3}}$

$\begin{array}{c}a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=-d{x}^2\\ or,{\int }_{V_0}^{0}vdv=-\alpha {\int }_0^s{x}^{2}dx\\ or,{\left[\frac{V^2}{2}\right]}_{V_0}^0=-\alpha {\left[\frac{n^2}{3}\right]}_0^s\\ or,\frac{V_0^2}{2}=\frac{\alpha s^3}{3}\\ \therefore S=\left[\frac{3V_0^2}{2d}\right]\end{array}$

Hence, Option $B$ is correct.