Answer
Distance: $-2 \sqrt{2} +\frac{10}{3}\mathrm{m}$
Displacement: $\frac{1}{3} \mathrm{m}$
Work Step by Step
Displacement is an integral of the velocity that was evaluated over the given interval.
\[
\text { Displacement }=\int_{1}^{3}\left(-\frac{1}{t^{2}}+\frac{1}{2}\right) d t=\frac{1}{3} \mathrm{m}
\]
The distance traveled is an integral of the absolute value of the velocity that was evaluated over the given interval.
\[
\begin{array}{l}
\text { Distance }=\int_{1}^{3}\left|-\frac{1}{t^{2}}+\frac{1}{2} \right| d t \\
=-\int_{1}^{\sqrt{2}}\left(-\frac{1}{t^{2}}+\frac{1}{2}\right) d t+\int_{\sqrt{2}}^{3}\left(-\frac{1}{t^{2}}+\frac{1}{2}\right) d t \\
=-2 \sqrt{2}+\frac{10}{3} \mathrm{m}
\end{array}
\]
