191. The acceleration of a material point moving in a straight line depends on time and is expressed by the formula $a(t)=$ $=6 t-2$. Find the law of motion, if at the initial moment of time $t=0$ the velocity $v=1 \mathrm{m} / \mathrm{s}$, and the path $s=0$.
Solution. $1^{\text {" }}$. The law of motion of a point is expressed by the function $s(t)$. Then $v(t)=\frac{d s}{d t}$ - the velocity of the point; $a(t)=\frac{d v}{d t}=\frac{d^{2} s}{d t^{2}}$ - the acceleration of the motion. According to the condition, we form the differential equation

$$
\frac{d^{2} s}{d t^{2}}=6 t-2
$$

$2^{0}$. The obtained equation is the simplest second-order differential equation. We reduce its order; since $\frac{d^{2} s}{d t^{2}}=\frac{d v}{d t}$, then

$$
\frac{d v}{d t}=6 t-2 ; d v=(6 t-2) d t ; v=3 t^{2}-2 t+C_{1}
$$

From this

$$
\frac{d s}{d t}=3 t^{2}-2 t+C_{1} ; d s=\left(3 t^{2}-2 t+C_{1}\right) d t ; s=t^{3}-t^{2}+C_{1} t+C_{2}
$$

$3^{0}$. Using the initial conditions, we find $C_{1}=1, C_{2}=0$; therefore, $s=t^{3}-t^{2}+t$ - the sought law of motion.