Velocity problem

Integrating $\displaystyle \int_{0}^{8}v(t)dt$ will give the displacement of point P, but not its total distance traveled. For that, we need to determine the intervals on which $v$ is negative and positive. Setting $v = 0$ , we find that the velocity changes sign at $t = 3$ , so we'll perform the separate integrations $\displaystyle \int_{0}^{3}v(t)dt$ and $\displaystyle \int_{3}^{8}v(t)dt$ .

$\displaystyle \int_{0}^{3}4t - 12dt = -18$

This means that P covered a distance of 18cm in the negative direction, but we still count this distance as positive and we'll add it to

$\displaystyle \int_{3}^{8}4t - 12dt = 50$

for a total distance of 68cm.