Principle. Average velocity uses net displacement; average speed uses path length. The market is reached at 30 min.
\[\begin{aligned}0\text{--}30:\ |\bar v| &= \frac{2.5}{0.5}\\&= 5.0,\ \bar s\\&= 5.0\ \mathrm{km\,h^{-1}}\end{aligned}\]
\[\begin{aligned}0\text{--}50:\ |\bar v| &= 0,\ \bar s\\&= 5/(\frac{5}{6})\\&= 6.0\ \mathrm{km\,h^{-1}}\end{aligned}\]
\[\begin{aligned}0\text{--}40:\ \text{return distance} &= 7.5(\frac{1}{6})\\&= 1.25\ \mathrm{km};\ \Delta x\\&= 1.25\ \mathrm{km},\ s\\&= 3.75\ \mathrm{km}\end{aligned}\]
\[\begin{aligned}|\bar v| &= 1.25/(\frac{2}{3})\\&= 1.875,\quad\bar s\\&= 3.75/(\frac{2}{3})\\&= 5.625\ \mathrm{km\,h^{-1}}\end{aligned}\]
Final results3\ (1.875,5.625)\ \mathrm{km\,h^{-1}}\text{ as (velocity magnitude, speed).}\]