Given that displacement of the body,
$y \,\propto\,t^{3}$
or$\,\,\,\,\,y = kt^{3}$
$\therefore \,\,\,\,\,v = \frac{dy}{dt } = 3kt^{2}$
or $\,\,\,\,a = \frac{dv}{dt} = 6\,kt \propto$ t
$\therefore \,\,\,\,\,$ The magnitude of the acceleration of the bodyincreases with time.