From Mechanics

From part a)

\displaystyle a=\frac{dv}{dt}=36-6t

We see \displaystyle a decreases as the time \displaystyle t increases. Thus the maximum acceleration is at \displaystyle t=0 giving

\displaystyle \text{a}\ \text{=}\ \text{36 m}{{\text{s}}^{-\text{2}}}