Answer
Since average velocity comes out to be the same as instantaneous velocity.
Hence verified.
Work Step by Step
We know that average velocity is equal to $\dfrac{\Delta y}{\Delta x}$, where $\Delta y$ is change in position and $\Delta x$ is change in time.
Now to find change in time substract $t_\circ +\Delta t$ and $t_\circ -\Delta t$.
We get, change in time equals to $t_\circ +\Delta t-t_\circ +\Delta t$ or $2\Delta t$.
Now to calculate change in position, substract $s(t_\circ +\Delta t)$ and $s(t_\circ -\Delta t)$.
We get, change in position equals to $s(t_\circ +\Delta t)-s(t_\circ -\Delta t)$.
Now calculate $s(t_\circ +\Delta t)$, by substituting $t=t_\circ +\Delta t$ in $s(t)=-\dfrac{1}{2}at^2+c$.
We get, $s(t_\circ +\Delta t)=-\dfrac{1}{2}a(t_\circ +\Delta t)^2+c=-\dfrac{1}{2}at_\circ^2-\dfrac{1}{2}a\Delta t^2-at_\circ \Delta t+c$
Now calculate $s(t_\circ -\Delta t)$, by substituting $t=t_\circ -\Delta t$ in $s(t)=-\dfrac{1}{2}at^2+c$.
We get, $s(t_\circ -\Delta t)=-\dfrac{1}{2}a(t_\circ -\Delta t)^2+c=-\dfrac{1}{2}at_\circ^2-\dfrac{1}{2}a\Delta t^2+at_\circ \Delta t+c$
Now substitute,
$s(t_\circ +\Delta t)=-\dfrac{1}{2}at_\circ^2-\dfrac{1}{2}a\Delta t^2-at_\circ \Delta t+c$ and $s(t_\circ -\Delta t)=-\dfrac{1}{2}at_\circ^2-\dfrac{1}{2}a\Delta t^2+at_\circ \Delta t+c$ in $s(t_\circ +\Delta t)-s(t_\circ -\Delta t)$.
$s(t_\circ +\Delta t)-s(t_\circ -\Delta t)=-\dfrac{1}{2}at_\circ^2-\dfrac{1}{2}a\Delta t^2-at_\circ \Delta t+c
\hspace{200px}-(-\dfrac{1}{2}at_\circ^2-\dfrac{1}{2}a\Delta t^2+at_\circ \Delta t+c)$
$\implies s(t_\circ +\Delta t)-s(t_\circ -\Delta t)=-\dfrac{1}{2}at_\circ^2-\dfrac{1}{2}a\Delta t^2-at_\circ \Delta t+c
\hspace{200px}+\dfrac{1}{2}at_\circ^2+\dfrac{1}{2}a\Delta t^2-at_\circ \Delta t-c$
$\implies s(t_\circ +\Delta t)-s(t_\circ -\Delta t)=-2at_\circ \Delta t$
Divide change in position by change in time to get average velocity.
Average velocity $=\dfrac{-2at_\circ \Delta t}{2\Delta t}=-at_\circ$.
Now to find the instantaneous velocity at $t=t_\circ$ for the position function $s(t)=-\dfrac{1}{2}at^2+c$.
Differentiate $s(t)=-\dfrac{1}{2}at^2+c$ with respect to time $t$.
We get, $s'(t)=-\dfrac{1}{2}a\times2t=-at$
Now substitute $t=t_\circ$.
We get, $s'(t_\circ)=-at_\circ=$ average velocity over the time interval $[t_\circ+\Delta t,t_\circ-\Delta t]$.
Hence verified
