a train is travelling at a speed of 90 km per hour brakes are applied so as to produce a uniform acceleration of -0.5 metre per second square how far the train will go before it brought to rest
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Explanation:
Given :
▪ Initial speed of train = 90kmph
▪ Retardation of train = 5m/s^2
To Find :
▪ Distance covered by train before coming at rest.
Concept used :
๏ This question is completely based on concept of stopping distance.
Formula Derivation :
✒ v^2 - u^2 = 2as
✒ (0)^2 - u^2 = 2(-a)s
✒ -u^2 = -2as
✒ s = u^2/2a
s denotes stopping distance
u denotes initial velocity
a denotes retardation
Conversion :
✳ 1kmph = 5/18mps
✴ 90kmph = 90×5/18 = 25mps
Calculation :
→ s = u^2/2a
→ s = (25)^2/(2×5)
→ s = 625/10
→ s = 62.5m
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