405. The path traveled by a material point is given by the following function of time: $s=3 t^{2}-2 t+4$. Find the speed of the point at the end of the 5th second.
Solution. We find the derivative: $\frac{d s}{d t}=6 t-2$; at $t=5$ we get $\frac{d s}{d t}=6 \cdot 5-2=28(\mathrm{m} / \mathrm{s})$. Therefore, $v=\left.\frac{d s}{d t}\right|_{t=5}=28 \mathrm{m} / \mathrm{s}$.