The displacement of a particle is proportional to the cube of the time. Then magnitude of its acceleration 

Given that  $S\alpha t^3$
velocity  $V=\frac{ds}{dt}$
$\begin{array}{c}\begin{array}{rl} & =\frac{d}{dt}\left(t^3\right)\\ & \Rightarrow V\alpha 3t^2\end{array}\\ \therefore \text{ acceleration }a=\frac{dv}{dt}\\ =\frac{d}{dt}\left(3t^2\right)\\ \Rightarrow a\alpha 6t\end{array}$

                           $\Rightarrow a\alpha t$