Answer
(a) $-\frac{3}{2}~m$
(b) $\frac{41}{6}~m$
Work Step by Step
(a) We can find the displacement:
$\int_{0}^{3}v(t)~dt$
$=\int_{0}^{3}(3t-5)~dt$
$=(\frac{3t^2}{2}-5t)~\vert_{0}^{3}$
$=[\frac{3~(3)^2}{2}-5(3)]- [\frac{3(0)^2}{2}-5(0)]$
$=(\frac{27}{2}-15)- (0)$
$= -\frac{3}{2}~m$
(b) Note that the function $~3t-5~$ is negative on the interval $0 \leq t \lt \frac{5}{3}$
We can find the distance traveled:
$\int_{0}^{3}\vert v(t) \vert~dt$
$=\int_{0}^{3}\vert 3t-5 \vert~dt$
$=\int_{0}^{5/3}\vert 3t-5 \vert~dt+\int_{5/3}^{3}\vert 3t-5 \vert~dt$
$=\int_{0}^{5/3}(5-3t)~dt+\int_{5/3}^{3}(3t-5)~dt$
$=(5t-\frac{3t^2}{2})~\vert_{0}^{5/3}+(\frac{3t^2}{2}-5t)~\vert_{5/3}^{3}$
$=[5~(\frac{5}{3})-\frac{3~(\frac{5}{3})^2}{2}]- [5~(0)-\frac{3(0)^2}{2}]+[\frac{3~(3)^2}{2}-5(3)]- [\frac{3(\frac{5}{3})^2}{2}-5(\frac{5}{3})]$
$=(\frac{25}{3}-\frac{25}{6})- (0)+(\frac{27}{2}-15)- (\frac{25}{6}-\frac{25}{3})$
$=(\frac{25}{6})-(\frac{3}{2})+ (\frac{25}{6})$
$=(\frac{25}{6})-(\frac{9}{6})+ (\frac{25}{6})$
$= \frac{41}{6}~m$
