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Question
A man walks on a straight road from his home to a market 2.5km away with a speed of 5km/h. Finding the market closed, he instantly turns and walk back home with a speed of 7.5 km/h. The average speed of the man over the, the interval of time 0 to 40min is equal to?
A man walks on a straight road from his home to a market 2.5km away with a speed of 5km/h. Finding the market closed, he instantly turns and walk back home with a speed of 7.5 km/h. The average speed of the man over the, the interval of time 0 to 40min is equal to?
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Solution
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
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
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
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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