Distance to market \\\\(s=2.5 km=2.5 \ imes 10^3=2500 m\\\\)Speed with which he goes to market \\\\(=5 km/h=5 \\dfrac{10^3}{3600}=\\dfrac{25}{18} m/s\\\\)Speed with which he comes back \\\\(=7.5 km/h=7.5 \ imes \\dfrac{10^3}{3600}=\\dfrac{75}{36} m/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:\\\\(\\dfrac{2.5}{5}=1/2 h=30\\\\) minutes.Average speed over this interval \\\\(=5 km/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 \ imes \\dfrac{1}{3}=2.5 km\\\\)His average speed in 0 to 50 minutes: \\\\(V_{avg}=\\dfrac{distance \\quad traveled}{time}\\\\)\\\\(=\\dfrac{2.5+2.5}{(50/60)}=6 km/h\\\\)(iii)In 40-30=10 minutes he travels a distance of :\\\\(7.5 \ imes \\dfrac{1}{6}=1.25 km\\\\)\\\\(V_{avg}=\\dfrac{2.5+1.25}{(40/60)}=5.625 km/h\\\\)