Answer
(a) $\frac{2}{3}~m$
(b) $4~m$
Work Step by Step
(a) We can find the displacement:
$\int_{2}^{4}v(t)~dt$
$=\int_{2}^{4}(t^2-2t-3)~dt$
$=(\frac{t^3}{3}-t^2-3t)~\vert_{2}^{4}$
$=[\frac{4^3}{3}-4^2-3(4)]- [\frac{2^3}{3}-2^2-3(2)]$
$=(\frac{64}{3}-28)- (\frac{8}{3}-10)$
$=(\frac{64}{3}-\frac{84}{3})- (\frac{8}{3}-\frac{30}{3})$
$=(-\frac{20}{3})+ (\frac{22}{3})$
$=\frac{2}{3}~m$
(b) Note that the function $~(t^2-2t-3)~$ is negative on the interval $2 \leq t \lt 3$
We can find the distance traveled:
$\int_{2}^{4}\vert v(t) \vert~dt$
$=\int_{2}^{4}\vert t^2-2t-3 \vert~dt$
$=\int_{2}^{3}\vert t^2-2t-3 \vert~dt+\int_{3}^{4}\vert t^2-2t-3 \vert~dt$
$=\int_{2}^{3} (-t^2+2t+3)~dt+\int_{3}^{4}(t^2-2t-3)~dt$
$=(-\frac{t^3}{3}+t^2+3t)~\vert_{2}^{3}+(\frac{t^3}{3}-t^2-3t)~\vert_{3}^{4}$
$=[-\frac{3^3}{3}+3^2+3(3)]- [-\frac{2^3}{3}+2^2+3(2)]+[\frac{4^3}{3}-4^2-3(4)]- [\frac{3^3}{3}-3^2-3(3)]$
$=(-9+9+9)- (-\frac{8}{3}+10)+(\frac{64}{3}-28)- (9-9-9)$
$=(9)- (\frac{22}{3})+(\frac{64}{3}-\frac{84}{3})+ (9)$
$=(\frac{27}{3})- (\frac{22}{3})-(\frac{20}{3})+ (\frac{27}{3})$
$= \frac{12}{3}$
$= 4~m$
