The position x of a particle varies with time as $x=at^2-bt^3$. The acceleration of the particle will be zero at time t equal to

Question

$\begin{array}{1 1} (a)\;\frac{a}{b}\\ (b)\;\frac{2a}{3b} \\(c)\;\frac{a}{3b}\\ (d)\;zero\end{array}$