Answer
$ \text {(a) Displacement = 2m and Distance = $\frac{10}{3}$}$
$ \text {(b) Displacement = $\frac{10}{4}$ and Distance = $\frac{10}{4}$}$
Work Step by Step
$ \text {Given the velocity and the time period and displacement/distance is unknown.} $
$ \text {(a)}$
\begin{align}
v(t) = 3t -2; \ 0\leq t \leq2
\end{align}
$ \text {Displacement:}$
\begin{align}
s(t) = \int_{0}^{2} (3t-2) \ dt = \left[{\frac{3t^2}{2}} - 2t\right]_{0}^{2} = 2 m
\end{align}
$ \text {Distance:}$
\begin{align}
d(t) = \int_{0}^{2} |3t-&2| \ dt = \int_{0}^{\frac{2}{3}} (-3t+2) \ dt \ + \int_{\frac{2}{3}}^{2} (3t-2) \ dt = \\ & = \left[{-\frac{3t^2}{2}} + 2t\right]_{0}^{\frac{2}{3}} + \left[{\frac{3t^2}{2}} - 2t\right]_{\frac{2}{3}}^{2} = \frac{2}{3} + \frac{8}{3} = \frac{10}{3}
\end{align}
$\text{(b)}$
\begin{align}
v(t) = |1-2t|; \ 0\leq t \leq2
\end{align}
$ \text {Displacement:}$
\begin{align}
s(t) = \int_{0}^{2} |1-&2t| \ dt = \int_{0}^{\frac{1}{2}}(1-2t) \ dt + \int_{\frac{1}{2}}^{2} (-1+2t) \ dt = \\ & = \left[t - t^2 \right]_{0}^{\frac{1}{2}} + \left[-t + t^2 \right]_{\frac{1}{2}}^{2} = \frac{1}{4} + \frac{9}{4} = \frac{10}{4}
\end{align}
\begin{align}
d(t) = \int_{0}^{2} |1-&2t| \ dt = \int_{0}^{\frac{1}{2}}(1-2t) \ dt + \int_{\frac{1}{2}}^{2} (-1+2t) \ dt = \\ & = \left[t - t^2 \right]_{0}^{\frac{1}{2}} + \left[-t + t^2 \right]_{\frac{1}{2}}^{2} = \frac{1}{4} + \frac{9}{4} = \frac{10}{4}
\end{align}
