Explanation

(d) $Given,\ initial\ velocity = v_{0}$
$final\ velocity = 0$
$deceleration\ a = - \alpha x^{2}\ \ldots.(i)$
Let the distance travelled by the particle be $s,$
$Now,\ we\ know\ that$
$a = \frac{dv}{dt} = \frac{dv}{dt} \times \frac{dx}{dx} = \frac{vdv}{dx}$
$or\ \ a = v\frac{dv}{dx}\ \ldots(ii)$
$From\ Eqs.(i)and\ (ii)$
$v\frac{dv}{dx} = - \alpha x^{2}$
$or\ \ vdv = - \alpha x^{2}dx$
On integrating with limit $v_{0} \rightarrow 0\ and\ 0 \rightarrow s$
$\therefore\ \ \int_{v_{0}}^{0}{vdv = \int_{0}^{s}{- \alpha x^{2}dx}}$
$or\ \ \left\lbrack \frac{v^{2}}{2} \right\rbrack_{v_{0}}^{0} = - \alpha\left\lbrack \frac{x^{3}}{3} \right\rbrack_{0}^{s}$
$- \frac{v_{0}^{2}}{2} = - \frac{\alpha(s)^{3}}{3}$
$or\ \ \frac{v_{0}^{2}}{2} = \frac{\alpha s^{3}}{3}$
$or\ \ \frac{{3v}_{0}^{2}}{2\alpha} = s^{3}$
$or\ \ s = \left\lbrack \ \frac{{3v}_{0}^{2}}{2\alpha} \right\rbrack^{1/3}$