Q:

Rate of Change with Derivatives

We just started with rate of change while using derivatives and I am stuck on a question, hope you can help.
A particle moves on a vertical line so that its altitude at time $t$ is $y=t^3−12·t+3$, where 
$t≥0$. Find the distance that the particle travels in the time interval $0≤t≤3$.

A:

Hint: If the particle's position function was monotonic, then this question would be easy; all you would have to do is compute the net displacement $|y(3) - y(0)|$. Unfortunately, the particle changes direction. To find this turning point, compute the derivative and solve for the time value $t_0 \in (0,3)$ where $y'(t_0) = 0$. Then the distance will be given by:
$$
|y(3) - y(t_0)| + |y(t_0) - y(0)|
$$

