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Answer :

Given, initial velocity= 90 km/h = \( 90 × \) \( 5 \over 18 \) \( = 25 \ m/s \)

final velocity = 0 m/s

{ as train is brought to rest}

a = \( –0.5 m /s ^2 \)

By the third motion equation,

\( \therefore v^2 - u^2 = 2as \)

\( \Rightarrow 0^2 \ - \ 25^2 \ = \ 2 ×(-0.5) s \) \( \Rightarrow s = 625 \ m \)

\(\therefore \) The train will travel 625 meters before it is brought to rest.

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