370. The velocity of a body is given by the equation $v=$ $=\left(12 t-3 t^{2}\right) \mathrm{m} / \mathrm{s}$. Determine the distance traveled by the body from the start of the motion until it stops.
Solution. The speed of the body is zero at the beginning of its motion and at the moment of stopping. To find the moment of stopping, we set the speed to zero and solve the equation with respect to $t$; we get $12 t-3 t^{2}=0 ; 3 t(4-t)=0 ; t_{1}=0, t_{2}=4$. Therefore,

$$
s=\int_{0}^{4}\left(12 t-3 t^{2}\right) d t=\left.\left(6 t^{2}-t^{3}\right)\right|_{0} ^{4}=6 \cdot 4^{2}-4^{3}=32(\mathrm{M})
$$