# A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h−1 . Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h−1. What is the

(a) magnitude of average velocity, and

(b) average speed of the man over the interval of time

(i) 0 to 30 min,

(ii) 0 to 50 min

(iii) 0 to 40 min?

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

Speed with which he goes to market =5km/h=51033600=2518m/s

Speed with which he comes back =7.5km/h=7.5×1033600=7536m/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:2.55=1/2h=30 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 :7.5×13=2.5km

His average speed in 0 to 50 minutes: Vavg=distancetraveledtime

=2.5+2.5(50/60)=6km/h

(iii)In 40-30=10 minutes he travels a distance of :7.5×16=1.25km

Vavg=2.5+1.25(40/60)=5.625km/h