Answer
(a) $\bar{s}_{0-2.0s}=1.0\mathrm{m/s}$
$\bar{s}_{2.0s-3.0s}=0\mathrm{m/s}$
$\bar{s}_{3.0s-4.5s}=1.3\mathrm{m/s}$
$\bar{s}_{4.5s-6.5s}=2.8\mathrm{m/s}$
$\bar{s}_{6.5s-7.5s}=0\mathrm{m/s}$
$\bar{s}_{7.5s-9.0s}=1.0\mathrm{m/s}$
(b)
$\bar{v}_{0-2.0s}=+1.0\mathrm{m/s}$
$\bar{v}_{2.0s-3.0s}=0\mathrm{m/s}$
$\bar{v}_{3.0s-4.5s}=+1.3\mathrm{m/s}$
$\bar{v}_{4.5s-6.5s}=-2.8\mathrm{m/s}$
$\bar{v}_{6.5s-7.5s}=0\mathrm{m/s}$
$\bar{v}_{7.5s-9.0s}=+1.0\mathrm{m/s}$
(c)
$v_{1.0s}=1.0\mathrm{m/s}$
$v_{2.5s}=0\mathrm{m/s}$
$v_{4.5s}=0\mathrm{m/s}$
$v_{6.0s}=-2.8\mathrm{m/s}$
(d)
$\bar{v}_{4.5s-9.0s}=-0.89\mathrm{m/s}$
Work Step by Step
(a) $t=0$ to $t=2.0 s$:
\begin{align*}
\bar{s}={d\over \Delta t}={2.0\over2.0}=1.0 \mathrm{m/s}
\end{align*}
$t=2.0$ to $t=3.0 s$:
\begin{align*}
\bar{s}={d\over \Delta t}={0\over1.0}=0 \mathrm{m/s}
\end{align*}
$t=3.0$ to $t=4.5 s$:
\begin{align*}
\bar{s}={d\over \Delta t}={2.0\over1.5}=1.3 \mathrm{m/s}
\end{align*}
$t=4.5$ to $t=6.5 s$:
\begin{align*}
\bar{s}={d\over \Delta t}={5.5\over2.0}=2.8 \mathrm{m/s}
\end{align*}
$t=6.5$ to $t=7.5 s$:
\begin{align*}
\bar{s}={d\over \Delta t}={0\over1.0}=0 \mathrm{m/s}
\end{align*}
$t=7.5$ to $t=9.0 s$:
\begin{align*}
\bar{s}={d\over \Delta t}={1.5\over1.5}=1.0 \mathrm{m/s}
\end{align*}
(b)
$t=0$ to $t=2.0 s$:
\begin{align*}
\bar{v}={\Delta x\over\Delta t}={2.0-0\over 2.0-0}={+2.0\over2.0}=+1.0 \mathrm{m/s}
\end{align*}
$t=2.0$ to $t=3.0 s$:
\begin{align*}
\bar{v}={\Delta x\over\Delta t}={2.0-2.0\over 3.0-2.0}={0\over1.0}=0 \mathrm{m/s}
\end{align*}
$t=3.0$ to $t=4.5 s$:
\begin{align*}
\bar{v}={\Delta x\over\Delta t}={4.0-2.0\over 4.5-3.0}={+2.0\over1.5}=+1.3 \mathrm{m/s}
\end{align*}
$t=4.5$ to $t=6.5 s$:
\begin{align*}
\bar{v}={\Delta x\over\Delta t}={-1.5-4.0\over 6.5-4.5}={-5.5\over2.0}=-2.8 \mathrm{m/s}
\end{align*}
$t=6.5$ to $t=7.5 s$:
\begin{align*}
\bar{v}={\Delta x\over\Delta t}={-1.5+1.5\over 7.5-6.5}={0\over1.0}=0 \mathrm{m/s}
\end{align*}
$t=7.5$ to $t=9.0 s$:
\begin{align*}
\bar{v}={\Delta x\over\Delta t}={0+1.5\over 9.0-7.5}={+1.5\over1.5}=+1.0 \mathrm{m/s}
\end{align*}
(c) Keeping in mind the instantaneous velocity is the slope of the position-time graph at the given instant.
At $t=1.0 s$:
\begin{align*}
v={\mathrm{rise}\over \mathrm{run}}={2.0\over 2.0}=1.0\mathrm{m/s}
\end{align*}
At $t=2.5 s$:
\begin{align*}
v={\mathrm{rise}\over \mathrm{run}}={0\over 1.0}=0\mathrm{m/s}
\end{align*}
At $t=4.5 s$:
\begin{align*}
v={\mathrm{rise}\over \mathrm{run}}=0\mathrm{m/s} \,\,(\mathrm{The\, dancer\, momentarily\, stops})
\end{align*}
At $t=6.0 s$:
\begin{align*}
v={\mathrm{rise}\over \mathrm{run}}={-5.5\over 2.0}=-2.8\mathrm{m/s}
\end{align*}
(d) Between $t=4.5$ and $t=9.0s$,
$$\bar{v}={\Delta x\over\Delta t}={0-4.0\over 9.0-4.5}={-4.0\over 4.5}=-0.89\mathrm{m/s}$$
