A boy walks on a straight road from his home to a market of $2.5 \mathrm{km}$ with a speed of $5 \mathrm{km} {\mathrm{h}}^{-1}$. Finding the market closed he instantly turns and walks back with a speed of $7.5 \mathrm{km} {\mathrm{h}}^{-1}$. What are the average speed and average velocity of the boy between $\mathrm{t}=0$ to $\mathrm{t}=50 \min$ respectively?

Time taken by the boy to go from his home to the market, ${\mathrm{t}}_1=\frac{2.5 \mathrm{km}}{5 \mathrm{km} {\mathrm{h}}^{-1}}=\frac{1}{2}\mathrm{h}$
Time taken by the boy to return back from the market to his home, ${\mathrm{t}}_2=\frac{2.5 \mathrm{km}}{7.5 \mathrm{km} {\mathrm{h}}^{-1}}=\frac{1}{3}\mathrm{h}$
$\therefore \mathrm{Total} \mathrm{time} \mathrm{taken}={\mathrm{t}}_1+{\mathrm{t}}_2=\frac{1}{2}\mathrm{h}+\frac{1}{3}\mathrm{h}=\frac{5}{6}\mathrm{h}=50 \min$
In $\mathrm{t}=0$ to $50 \min$,

$$\begin{gathered} \mathrm{Total} \mathrm{distance} \mathrm{travelled}=2.5 \mathrm{km}+2.5 \mathrm{km}=5 \mathrm{km} \\ \mathrm{Displacement}=0\left(\mathrm{As} \mathrm{the} \mathrm{boy} \mathrm{returns} \mathrm{back} \mathrm{home}\right) \\ \Rightarrow \mathrm{Average} \mathrm{velocity} =0 \\ \therefore \mathrm{Average} \mathrm{speed}=\frac{\mathrm{Distance} \mathrm{travelled}}{\mathrm{Time} \mathrm{taken}}=\frac{5 \mathrm{km}}{{\displaystyle \left(\frac{5}{6}\mathrm{h}\right)}}=6 \mathrm{km} {\mathrm{h}}^{-1} \end{gathered}$$