380. Two bodies simultaneously started linear motion from a certain point in the same direction with speeds $v_{1}=$ $=\left(6 t^{2}+4 t\right) \mathrm{m} /$ s and $v_{2}=4 t \mathrm{m} / \mathrm{s}$. After how many seconds will the distance between them be 250 m?
Solution. Let $t_{1}$ be the moment of the meeting. Then

$$
s_{1}=\int_{0}^{t_{1}}\left(6 t^{2}+4 t\right) d t=\left.\left(2 t^{3}+2 t^{2}\right)\right|_{0} ^{t_{1}}=2 t_{1}^{3}+2 t_{1}^{2} ; \quad s_{2}=\int_{0}^{t_{1}} 4 t d t=2 t_{1}^{2}
$$

Since $s_{1}-s_{2}=250$, we get the equation $2 t_{1}^{3}+2 t_{1}^{2}-2 t_{1}^{2}=250$ or $2 t_{1}^{3}=250$, from which $t_{1}^{3}=125$, i.e., $t=5$ (s).