137. A body moves with a velocity $v=\left(3 t^{2}-1\right) \mathrm{m} / \mathrm{s}$. Find the law of motion $s(t)$, if at the initial moment the body was 5 cm away from the origin.
Solution. $1^{0}$. Since $d s=v d t=\left(3 t^{2}-1\right) d t$, then $s=\int\left(3 t^{2}-1\right) d t=$ $=t^{3}-t+C$.

$2^{0}$. From the condition, if $t=0$, then $s=5 \text{~cm}=0.05 \text{m}$. Substituting these values into the obtained equation, we have $s=t^{3}-t+C$, from which $0.05=C$.

$3^{0}$. Then the desired equation will take the form $s=t^{3}-t+0.05$.