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A man walks on a straight road from his home to market 2.5km away with a speed of 5km\h. finding the market closed, he instntly turns and backs home with a speed of 7.5km .the average speed of the man over the interval of time 0 to 40 min. is equal to |
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Answer» Explanation: Time taken by the MAN to reach the market from home,t1 = 2.5/5 = 1/2 h = 30 min Time taken by the man to reach home from the market, t2 = 2.5/7.5 = 1/3 h = 20 min Total time taken in the whole journey = 30 + 20 = 50 min (i) 0 to 30 min Average velocity = Displacement/Time = 2.5/(1/2) = 5 km/h Average speed = Distance/Time = 2.5/(1/2) = 5 km/h (II) 0 to 50 min Time = 50 min = 50/60 = 5/6 h Net displacement = 0 Total distance = 2.5 + 2.5 = 5 km Average velocity = Displacement / Time = 0 Average speed = Distance / Time = 5/(5/6) = 6 km/h (iii) 0 to 40 min Speed of the man = 7.5 km/h Distance travelled in first 30 min = 2.5 km Distance travelled by the man (from market to home) in the next 10 min = 7.5 × 10/60 = 1.25 km Net displacement = 2.5 – 1.25 = 1.25 km Total distance travelled = 2.5 + 1.25 = 3.75 km Average velocity = Displacement / Time = 1.25 / (40/60) = 1.875 km/h Average speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h |
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