A man walks on a straight road from his home to a market 2.5km away with speed 5km/hr. finding the market close he instanteniously turns and walks back home with speed of 7.5km/hr. what is the 1) magnitude of average velocity 2)average speed of man over the interval of time 1)0to 30 minutes 2)0 to 50minutes and 3)0to 40 minute
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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
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
ankit3643:
thank you very much for helping
Answered by
5
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
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
thanks for your help
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