A particle moves along a straight line and its position at time $t$ is given by $s(t)= 2t^{3}-21t^{2}+36t$ where $s$ is measured in feet and $t$ in seconds.

Find the velocity (in ft/sec) of the particle at time $t=0$:

Answer:

The particle stops moving (i.e. is in a rest) twice, once when $t=A$ and again when $t=B$ where $A < B$. Determine $A$ and $B$.

$A=$
$B=$

What is the position of the particle at time $14$?
Answer:

Finally, what is the total distance the particle travels between time $0$ and time $14$?
Answer: