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a boy goes from his house to a market 2.6 km away point in the market close he return back to his house finding the distance covered by the boy what is the displacement |
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Answer» Answer: Distance to market s=2.5km=2.5×10 =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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