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Q.9 A man walks on a straight road 1from his home to a market 6 kmaway with a speed of 12 km/hourthe time taken by the man to gofrom his home to market, equals- |
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Answer» Answer: Distance to market s=2.5km=2.5×10 3 =2500m Speed with which he goes to market =5km/h=5 3600 10 3
= 18 25
m/s Speed with which he comes BACK =7.5km/h=7.5× 3600 10 3
= 36 75
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: 5 2.5
=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× 3 1
=2.5km His average speed in 0 to 50 minutes: V avg
= time distancetraveled
= (50/60) 2.5+2.5
=6km/h (iii)In 40-30=10 minutes he travels a distance of :7.5× 6 1
=1.25km V avg
= (40/60) 2.5+1.25
=5.625km/h |
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