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A train is moving on straight track with velocity v0 = 13.5 ms-1. To stop the train at a particular station, the driver applies brakes at t = 0, which is caused of a retardation proportional to the velocity of the train. The speed of train reduces 50% in the first 2 s in t0=4 . The velocity of train (in ms-1) at t=4s (Given , e=2.7)is

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answer is 5.

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Detailed Solution

∵a=−kv or dvdt=−kv Or ∫v0v dvv=−k∫0t dt or [ln⁡v]v0v=−kt Or In v−ln⁡v0=−kt or ln⁡vv0=−kt Or vv0=e−kt For t=t0,v=v0−v0×50100=v02∴ v02=v0e−kt Or v=v0e−ln⁡2t0t or ln⁡v=ln⁡v0−ln⁡2t0t Or ln⁡vv0=−ln⁡24ln⁡2t or ln⁡vv0=−t4 or vv0=e−t4∴ v=v0e−t4=13.5e−44=13.5e=13.52.7=5ms−1

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A train is moving on straight track with velocity v0 = 13.5 ms-1. To stop the train at a particular station, the driver applies brakes at t = 0, which is caused of a retardation proportional to the velocity of the train. The speed of train reduces 50% in the first 2 s in t0=4 . The velocity of train (in ms-1) at t=4s (Given , e=2.7)is