a man walks on a straight road from his home to a market 2. 5 km away with a speed of 5 km/hr. finding market closed, he instantly turns and walks back home with a speed of 7. 5 km/ hr. the average speed of the man over the interval of time 0 to 40 min is equal to...
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Hi ...dear...
.
here is your answer....
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....
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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..
hope this helped you....!!!
#shubhendu
.
here is your answer....
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....
.
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..
hope this helped you....!!!
#shubhendu
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