Answer:
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An electric train leaves a station starting from rest and attains a speed of 72 km/hr in 10 seconds. It travels at that speed for 120 seconds. Then it undergoes uniform retardation for 20 seconds to come to halt at the next station. Calculate a) the distance between the train stations. b) the average velocity of train.
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cristopherig5f
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1Answer :
The distance between the train stations = 2700 m.
The average velovity of train is 64.9 km/ hr.
Given :
Here the given informations are
Initial velocity (u₁) = 0
Final velocity ( v₁) = 72 km/hr = 20 m/s
Time (t₁) = 10 seconds
Initial velocity (u₂) = 72 km/hr = 20 m/s
Final velocity ( v₂) = 0
Time (t₂) = 20 seconds
Time for uniform motion (t₃) = 120 seconds.
To find :
The distance between the train stations (S)
The average velovity of train (V)
Explanation :
We can divide the travel into 3 parts.
1. Accelerated motion
2. Retarded motion
3. Uniform motion
x₁= u₁t₁+ ½ at₁²
a=v₁-u₁/t₁
=(20-0)/10
=2m/s²
x₁=0+½ ×2×10²
= 100m
x₂=u₂t₂+ ½ at₂²
a=v₂-u₂/t₂
=(0-20)/20
= - 1m/s²
x₂=20×20+½ ×-1×20²
= 200m
x₃= v₃ × t₃
= 20×120
=2400 m.
Then, he distance between the train stations (S)
= x₁+x₂+x₃
=100 + 200 + 2400
=2700 m.
Total time taken (T) = t₁+t₂+t₃
= 10+120+20
= 150 seconds.
Average velocity (V) = S/T
= 2700/150
=18 m/s
=64.8 km/hr.
To learn more questions about the topic refer to the links below:
https://brainly.in/question/7172471
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