A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of –0.5 m s-2. Find how far the train will go before it is brought to rest.
ScienceClass 9CBSEAnswered 27 Mar 2026
Answer
Solution
Given, initial velocity (u) = 90 km/hour = 25 m.s-1
Terminal velocity (v) = 0 m.s-1
Acceleration (a) = -0.5 m.s-2
As per the third motion equation, v2-u2=2as
Therefore, distance traveled by the train (s) =(v2-u2)/2a
s = (02-252)/2(-0.5) meters = 625 meters
The train must travel 625 meters at an acceleration of -0.5 ms-2 before it reaches the rest position.
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