129. A body moves in a straight line with acceleration $a=6 t-4$. At $t=0$, the initial path $s_{0}=0$, initial velocity $v_{0}=4$. Find the velocity and the distance traveled as functions of time.
Solution. According to the condition, we have $s^{\prime \prime}=6 t-4$. Integrating both sides of this equation, we get $\frac{d s}{d t}=3 t^{2}-4 t+C_{1}$, i.e., the velocity is expressed as a function of time.

Integrating both sides of the last equation, we express the path as a function of time:

$$
s=t^{3}-2 t^{2}+C_{1} t+C_{2}
$$

The arbitrary constants $C_{1}$ and $C_{2}$ are determined from the initial conditions: $v_{0}=4$ and $s_{0}=0$ at $t=0$. Substituting these data into the expressions $\frac{d s}{d t}$ and $s$, we find $C_{1}=4, C_{2}=0$.

Then the velocity of the body and the distance it has traveled are finally written as $v=3 t^{2}-4 t, s=t^{3}-2 t^{2}+4 t$.