Answer
Distance $=\frac{13}{3} m$
Displacement $=4 \mathrm{m}$
Work Step by Step
We have $ v(0)=-1$, a(t)=3 at $0 \leq t \leq 2$
\[
\begin{aligned}
\therefore v(t) &=\int_{0}^{2} a(t) d t \\
&=3 \int_{0}^{2} d t=C+3 t \\
v(0)=-1 &=C+0 \\
-1=C & \\
\therefore v(t) &=-1+3 t
\end{aligned}
\]
Displacement
\[
s(t)=\int_{0}^{2}(-1+3 t) d t=\left.\left(\frac{-t+3 t^{2}}{2}\right)\right|_{0} ^{2}=4
\]
Distance
\[
\begin{aligned}
\int_{0}^{2}|-1+3 t| d t &=-\int_{0}^{\frac{1}{3}}(-1+3 t) d t+\int_{\frac{1}{3}}^{2}(-1+3 t) d t \\
&=-\left.\left(-t+\frac{3 t^{2}}{2}\right)\right|_{0} ^{\frac{1}{3}}+\left.\left(-t+\frac{3 t^{2}}{2}\right)\right|_{\frac{1}{3}} ^{2}
\end{aligned}
\]
\[
\begin{array}{l}
=\left(\frac{1}{6}+4\right)+\frac{1}{6} \\
=\frac{25}{6}+\frac{1}{6}
\end{array}
\]
$=\left[\frac{13}{3}\right] m$
