A man walks on a straight road from his home to market 2.5 km away with a speed of 5 kilometre per hour. finding the market closed he instantly turns and walk back home with speed of 7.5km/h . find the magnitude of average velocity and average speed of man over interval of time (a) 0 to 30 min(b)0 to 50 min(c) 0 to 40 min

A man walks on a straight road from his home to market 2.5 km away wit

1.	A man walks on a straight road from his home to market 2.5 km away with a speed of 5 kilometre per hour. finding the market closed he instantly turns and walk back home with speed of 7.5km/h . find the magnitude of average velocity and average speed of man over interval of time (a) 0 to 30 min(b)0 to 50 min(c) 0 to 40 min
Answer» to tell the TIRED man on his return home that his average speed was ZERO !] Answer:-Time taken by the man to reach the MARKET from home,t1 = 2.5/5 = 1/2 H = 30 minTime taken by the man to reach home from the market, t2 = 2.5/7.5 = 1/3 h = 20 minTotal time taken in the whole journey = 30 + 20 = 50 min(i) 0 to 30 minAverage velocity = Displacement/Time = 2.5/(1/2) = 5 km/hAverage speed = Distance/Time = 2.5/(1/2) = 5 km/h(ii) 0 to 50 minTime = 50 min = 50/60 = 5/6 hNet displacement = 0Total distance = 2.5 + 2.5 = 5 kmAverage velocity = Displacement / Time = 0Average speed = Distance / Time = 5/(5/6) = 6 km/h(iii) 0 to 40 minSpeed of the man = 7.5 km/hDistance travelled in first 30 min = 2.5 kmDistance travelled by the man (from market to home) in the next 10 min= 7.5 × 10/60 = 1.25 kmNet displacement = 2.5 – 1.25 = 1.25 kmTotal distance travelled = 2.5 + 1.25 = 3.75 kmAverage velocity = Displacement / Time = 1.25 / (40/60) = 1.875 km/hAverage speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h

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