A man walks on a straight …

Distance to market s=2.5km=2.5×10³ =2500m

Speed with which he goes to market =5km/h=$5 \frac{10^3}{3600}=\frac{25}{18} \mathrm{~m} / \mathrm{s}$

Speed with which he comes back = $7.5 \mathrm{~km} / \mathrm{h}=7.5 \times \frac{10^3}{3600}=\frac{75}{36} \mathrm{~m} / \mathrm{s}$

(a)Average velocity is zero since his displacement is zero.

(b) (i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market: $\frac{2.5}{5}=1 / 2 \mathrm{~h}=30 \text { minutes. }$

Average speed over this interval =5km/h

(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes :$7.5 \times \frac{1}{3}=2.5 \mathrm{~km}$

His average speed in 0 to 50 minutes: V_avg ​ =distance traveled/time $=\frac{2.5+2.5}{(50 / 60)}=6 \mathrm{~km} / \mathrm{h}$

(iii)In 40-30=10 minutes he travels a distance of : $7.5 \times \frac{1}{6}=1.25 \mathrm{~km}$

$\mathrm{V}_{\mathrm{avg}}=\frac{2.5+1.25}{(40 / 60)}=5.625 \mathrm{~km} / \mathrm{h}$