Question

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 magnitude of average speed of 0 to 50 minutes​

Answer 1

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

Speed with which he goes to market =5km/h=53600103=1825m/s

Speed with which he comes back =7.5km/h=7.5×3600103=3675m/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:52.5=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 :7.5×31=2.5km

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

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

(iii)In 40-30=10 minutes he travels a distance of :7.5×61=1.25km

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