607. A point moves along a straight line according to the law $s=3 t^{2}+2 t$, where $t$ - time in seconds, $s$ - distance in meters. Find the velocity of the point at the moments $t=3$ and $t=4$.
Solution. Guided by point 3, we sequentially find:

1) $s+\Delta s=3(t+\Delta t)^{2}+2(t+\Delta t)$
2) $\Delta s=3(t+\Delta t)^{2}+2(t+\Delta t)-\left(3 t^{2}+2 t\right)=6 t \Delta t+2 \Delta t+3 \Delta t^{2}$;
3) $v_{\mathrm{cp}}=\frac{\Delta s}{\Delta t}=\frac{\Delta t(6 t+2+3 \Delta t)}{\Delta t}=6 t+2+3 \Delta t$

(average speed over the time interval $\Delta t$ from $t$ to $(t+\Delta t)$;

4) now we find the speed at the moment of time $t$:

$$
v=\lim _{\Delta t \rightarrow 0} \frac{\Delta s}{\Delta t}=\lim _{\Delta t \rightarrow 0}(6 t+2+3 \Delta t)=6 t+2
$$

in particular, at $t=3 v=20$ (m/sec), at $t=4 v=26$ (m/sec).