Determine the instantaneous speed of a car three quarters of an hour after starting on a trip where its distance-time function is $d(t)=t^3-15t^2+75t$.

Since the instantaneous speed is the instantaneous rate of change of $d(t)=t^3-15t^2+75t$ all we have to do is find $d\,'(t)$ and plug in $0.75$ hours. We know from here that
\[
d\,'(t)=3t^2-30t+75.
\]
Now plugging in $0.75$ hours we get
\begin{eqnarray}
d\,'(0.75)&=&3(0.75)^2-30(0.75)+75\\
&=&1.6875-22.5+75\\
&=&54.1875.
\end{eqnarray}
So the car is going approximately $54$ miles per hour.