435. At time $t$, the body is at a distance of $s=\frac{1}{4} t^{4}+4 t^{3}+16 t^{2}$ km from the starting point. Find its acceleration after 2 hours.
Solution. We find $v=s^{\prime}=t^{3}+12 t^{2}+32 t, a=v^{\prime}=s^{\prime \prime}=3 t^{2}+$ $+24 t+32$. For $t=2$ we have $a=3 \cdot 4+24 \cdot 2+32=92$ (km $\left./ \mathrm{u}^{2}\right)$.