434. A material point moves according to the law $s=2 t^{3}-6 t^{2}+$ $+4 t$. Find its acceleration at the end of the 3rd second.
Solution. We find $v=s^{\prime}=6 t^{2}-12 t+4, a=s^{\prime \prime}=12 t-12$, from which at $t=3$ we get $a=12 \cdot 3-12=24\left(\mathrm{m} / \mathrm{c}^{2}\right)$.