Explanation

(d) Given, acceleration $a = - kv^{3}$
Initial velocity at cut-off, $v_{1} = v_{0}$
Initial time of cut-off, $t = 0$
And final time after cut-off, $t_{2} = t$
Again, $a = \frac{dv}{dt} = - kv^{3}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dv}{v^{3}} = \text{-}kdt$
Integrating both sides, with in the condition of motion.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int_{v_{0}}^{v}\frac{dv}{v^{3}} = - \int_{0}^{t}{k\ dt}$
Or $\ \ \ \ \left\lbrack - \frac{1}{2v^{2}} \right\rbrack_{v_{0}}^{v} = - \lbrack kt\rbrack_{0}^{t}$
Or $\ \ \ \ \frac{1}{2v^{2}} - \frac{1}{2v_{0}^{2}} = kt$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \frac{v_{0}}{\sqrt{1 + 2\ kt\ v_{0}^{2}}}$