433. A point moves in a straight line according to the law $s=3 t^{2}-3 t+$ +8 . Find the velocity and acceleration of the point at the moment $t=4$.
Solution. To determine the velocity, we need to find the first derivative of the given function at $t=4$. We have

$$
v=s^{\prime}=\left(3 t^{2}-3 t+8\right)^{\prime}=6 t-3 ; v_{t=4}=6 \cdot 4-3=21(\mathrm{~m} / \mathrm{s})
$$

The acceleration is equal to the second derivative of the function at $t=4$, i.e.,

$$
a=s^{\prime \prime}=\left(s^{\prime}\right)^{\prime}=(6 t-3)^{\prime}=6
$$

The magnitude of the acceleration turned out to be constant for any value of $t$; this means that the motion of the point according to the given law occurs with constant acceleration.