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Question
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.
Sum
Solution
Initial velocity of train (u) = 90 km h−1
= `(90 xx 1000)/3600`
= 25 m/s
Final velocity due to rest (v) = 0
Acceleration (a) = −0.5 ms−2
Third equation of motion
v2 = u2 + 2as
0 = 252 + 2 (−0.5) × s
s = 625 m
Therefore, distance covered before coming to rest = 625 m
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