804. A point moves in a straight line according to the law $s=4 t+t^{3}$. Find the velocity and acceleration of the point at the moments of time $t_{0}=0, t_{1}=1, t_{2}=2$ ( $s$ is given in centimeters, $t$ - in seconds).
Solution. The velocity $v=s_{t}^{\prime}=\left(4 t+t^{3}\right)^{\prime}=4+3 t^{2}$; acceleration $a=s_{t}^{\prime \prime}=\left(4+3 t^{2}\right)^{\prime}=6 t$. Therefore, at the moment $t_{0}=0 \quad v_{0}=4 \mathrm{~cm} /$ sec, $a_{0}=0 \mathrm{~cm} /$ sec $^{2}$. At $t_{1}=1 \quad v_{1}=s^{\prime}(1)=$
$=7 \mathrm{cm} /$ sec, $a_{1}=6 \mathrm{cm} /$ sec $^{2}$. At $t_{2}=2 v_{2}=s^{\prime}(2)=16 \mathrm{cm} /$ sec, $a_{2}=12 \mathrm{~cm} /$ sec $^{2}$.