Answer
$$
\begin{array}{c}
\text { The distance }=\frac{17}{3} \quad \mathrm{m} \\
\text { The displacement }=-\frac{10}{3} \mathrm{m}
\end{array}
$$
Work Step by Step
\begin{aligned}
&\text { You have } a(t)=t-2, v_{0}=0,1 \leq t \leq 5\\
&\text { We will find the velocity : }\\
\int(t-2) d t=\frac{t^{2}}{2}-2 t+C=&v(t)\\
&v(0)=0=0-0+C \Rightarrow 0=C\\
&\therefore \frac{t^{2}}{2}-2 t=v(t)
\end{aligned}
The displacement =:
$$
s(t)=\int_{1}^{5}\left(\frac{t^{2}}{2}-2 t\right) d t=\left.\left(\frac{t^{3}}{6}-t^{2}\right)\right|_{1} ^{5}=\left(\frac{125}{6}-25\right)-\left(\frac{1}{6}-1\right)=\sqrt{\frac{-10}{3} m}
$$
The distance:
$$
\begin{array}{c}
\int_{1}^{5}\left|\left(\frac{t^{2}}{2}-2 t\right)\right| d t=-\int_{1}^{4}\left(\frac{t^{2}}{2}-2 t\right) d t+\int_{4}^{5}\left(\frac{t^{2}}{2}-2 t\right) d t \\
=-\left.\left(\frac{t^{3}}{6}-t^{2}\right)\right|_{1} ^{4}+\left.\left(\frac{t^{3}}{6}-t^{2}\right)\right|_{4} ^{5} \\
=\frac{7}{6}+\frac{27}{6} \\
\left.=\frac{17}{3} m\right] \\
\end{array}
$$
