65. The speed of a body coming out of a state of rest is equal to $5 t^{2} \mathrm{m} / \mathrm{s}$ after $t$ seconds. Determine the distance the body will travel in 3 s (see problem 20).
Solution. Using the condition, we form the differential equation: $\frac{d s}{d t}=5 t^{2}$, since velocity $v=\frac{d s}{d t}$.

Let's find the general solution of this equation:

$$
d s=5 t^{2} d t ; \int d s=5 \int t^{2} d t ; s=\frac{5}{3} t^{3}+C
$$

Now, let's find the particular solution of this equation. The initial conditions are determined by the fact that the body starts from rest, i.e., $s=0$ when $t=0$. Substituting these values into the general solution, we get $C=0$. Therefore, the particular solution is $s=\frac{5}{3} t^{3}$.

Let's determine the distance traveled by the body in $3 \mathrm{~s}$:

$$
s=\frac{5}{3} \cdot 3^{3}=\frac{5}{3} \cdot 27=45(\mathrm{~m})
$$