378. Two bodies simultaneously leave one point: one with a speed of $v_{1}=5 t \mathrm{~m} / \mathrm{c}$, the other with a speed of $v_{2}=3 t^{2} \mathrm{~m} / \mathrm{c}$. At what distance from each other will they be after $20 \mathrm{c}$, if they move in a straight line in the same direction?
Solution. $s_{1}=\int_{0}^{20} 5 t d t=5 \cdot \frac{t^{2}}{2} \left\lvert\, \frac{20}{\bar{\sigma}} 5 \cdot 200=1000(\mathrm{m})\right. ;$

$$
s_{2}=\int_{0}^{20} 3 t^{2} d t=\left.t^{3}\right|_{0} ^{20}=8000(\mathrm{m}) ; s=s_{2}-s_{1}=8000-1000=7000(\mathrm{m})
$$