Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 52

When an object is dropped, the distance it falls in $t$ seconds, assumming that air resistance is negligible, is given by
$s(t) = 4.905t^2$, where $s(t)$ is in meters (m).
Determine the velocity and acceleration of the object after it has been falling for 3 sec.

$
\begin{equation}
\begin{aligned}
v(t) = s'(t) &= \frac{d}{dt} (4.905 t^2) \\
\\
&= 4.905 (2) t^{2 - 1}\\
\\
&= 9.81t\\
\\
a(t) = v'(t) &= \frac{d}{dt} (9.81t) \\
\\
&= 9.81(1) \\
\\
&= 9.81
\end{aligned}
\end{equation}
$


Therefore, the velocity and acceleration of the store after 3s is given by

$
\begin{equation}
\begin{aligned}
v(3) &= 9.81(3) = 29.43 \frac{\rm{m}}{\rm{s}}\\
\\
a(3) &= 9.81 \frac{\rm{m}}{\rm{s}^2}
\end{aligned}
\end{equation}
$