Here, distance of the market from the home of the man, $S = 2.5 \text{ km}$ Speed of the man, while going from his home to the market, $v_{1}=5\,\text{km}\,\text{h}^{-1}$ Therefore, time taken to reach the market, $t_{1}=\frac{S}{v_{1}}=\frac{2.5\,\text{km}}{5\,\text{km}\,\text{h}^{-1}}$ $=0.5 \text{ h}=30 \text{ min}$ Speed of the man, while returning from the market, $v_{2}=7.5\,\text{km}\,\text{h}^{-1}$ Therefore, time taken to return the home, $t_{2}=\frac{S}{v_{2}}=\frac{2.5\,\text{km}}{7.5\,\text{km}\,\text{h}^{-1}}$ $= \frac{1}{3} \text{ h}=20 \text{ min}$ (i) Over the interval of time 0 to 30 min :During this interval, the man covers distance from his home upto themarket.Therefore, displacement = 2.5 km; and distance covered = 2.5 kmNow, time taken = 30 min = 0.5 hTherefore, magnitude of the average velocity$= \frac{\text{displacement}}{\text{time}}$$= \frac{2.5\,\text{km}}{0.5\,\text{h}} = 5\,\text{km}\,\text{h}^{-1}$ andaverage speed $= \frac{\text{distance}}{\text{time}}$$= \frac{2.5\,\text{km}}{0.5\,\text{h}} = 5\,\text{km}\,\text{h}^{-1}$(ii) Over the interval of time 0 to 50 min :During this time interval, the man returns his home.Therefore, displacement = 0and distance covered = 2.5 + 2.5 = 5 kmNow, time taken $=50 \text{ min} = \frac{50}{60} = \frac{5}{6} \text{ h}$Therefore, magnitude of the average velocity $= \frac{0\,\text{km}}{\left(\frac{5}{6}\right)\,\text{h}} = 0$and average speed $= \frac{5.0\,\text{km}}{\left(\frac{5}{6}\right)\,\text{h}} = 6\,\text{km}\,\text{h}^{-1}$(iii) Over the interval of time 0 to 40 min :During the first 30 min of this interval, the man covers a distance of 2.5 km and in next 10 min, he covers 2.5/2 i.e. 1.25 km of the return journey.Therefore, displacement = 2.5 – 1.25 = 1.25 km;and distance covered = 2.5 + 1.25 = 3.75 kmNow, time taken = 40 min$= \frac{2}{3} \text{ h}$Therefore, magnitude of the average velocity$= \frac{1.25\,\text{km}}{\frac{2}{3}\,\text{h}} = \frac{1.25 \times 3}{2} = 1.875\,\text{km}\,\text{h}^{-1}$ and average speed $= \frac{3.75\,\text{km}}{\left(\frac{2}{3}\right)\,\text{h}} = 5.625\,\text{km}\,\text{h}^{-1}$