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Question

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h1 . Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h1. 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?

Solution
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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

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A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h1 . Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h1. What is the
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