Distance went by the Train in...

A train starts it's journey from still and travels a distance of 2.5 k m 2.5 km within 3 m i n u t e s 3 minutes and comes back to still . Within the journey it obtains a maximum velocity of 25 m s 1 25ms^{-1} . If it's value of acceleration and deceleration is equal, find the distance that the train traveled in it's maximum speed in meters ?


The answer is 500.

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1 solution

Saya Suka
May 27, 2019

Answer
= 25 * [2 * (2.5 *1000 / 25) - 3 * 60].
= 25 * [200 - 180].
= 500 m


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