A train is travelling at a speed of 60km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s2 . How far the train will go before it is brought to rest.
Answers
Answered by
303
Given:-
- Initial velocity of train = 60km/h = 60 × 5/18 =50/3 m/s
- Acceleration of train = -0.5 m/s²
- Final velocity = 0 m/s ( as train come to rest)
To Find:-
- Distance covered by train before it is brought to be rest.
Solution :-
By using 3rd equation of motion
➦ v² = u² +2as
➭ 0² = (50/3)² + 2 × (-0.5) × s
➭ 0 = 2500/9 + (-1 )× s
➭ s = 2500/9
➭ s = 277.7 m
∴ The distance covered by train before it brought to rest is 277.7 m .
Answered by
50
Answer:
velocity of train = 60km/h
= 60 × 5/18
=50/3 m/s
Acceleration of train = -0.5 m/s²
Final velocity = 0 m/s ( as train come to rest)
Distance covered by train before it is brought to be rest.
By using 3rd equation of motion
v² = u² +2as
0² = (50/3)² + 2 × (-0.5) × s
0 = 2500/9 + (-1 )× s
s = 2500/9
s = 277.7 m
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