Answer :
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:
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