1324. The velocity of a point's movement is $v=0.5 t^{3}$ m/s. Find the distance $s$ traveled by the point over a time period $T=8$ s after the start of the motion. What is the average velocity over this interval?
Solution. From differential calculus, it is known that $v=\frac{d s}{d t}$, therefore, $\frac{d s}{d t}=0.5 t^{3}$, from which $d s=0.5 t^{3} d t$. Considering that $0 \leqslant t \leqslant 8$, we have:

$$
\begin{gathered}
s=\int_{0}^{8} 0.5 t^{3} d t=0.5 \left.\frac{t^{4}}{4}\right|_{0} ^{8}=512 \mathrm{M} ; \\
v_{\mathrm{cp}}=\frac{s}{T}=\frac{512}{8}=64 \mathrm{M} / \mathrm{sec}
\end{gathered}
$$