A man walks on a straight road from his home to a market 2.5km away with a speed of 5kmh−1 . Finding the market closed, he instantly turns and walks back home with a speed of 7.5kmh−1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30min, (ii) 0 to 50min (iii) 0 to 40min?
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Updated on : 2022-09-05
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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.
(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
(iii)In 40-30=10 minutes he travels a distance of :7.5×61=1.25km
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