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An express train passes through a station. It enters with an initial velocity of 22.2 m/s and decelerates at a rate of 0.142 m/s2 as it goes through. The station is 206 m long. How long is the nose of train in the station?

Answer :

Kalahira
We can find the final velocity v at the end of the station. v^2 = (v0)^2 + 2ax v = sqrt{ (v0)^2 + 2ax } v = sqrt{ (22.2 m/s)^2 - (2)(0.142 m/s^2)(206 m) } v = 20.84 m/s We can find the time to decelerate to this velocity. v = v0 + at t = (v - v0) / a t = (20.84 m/s - 22.2 m/s) / -0.142 m/s^2 t = 9.58 seconds The nose of the train is in the station for 9.58 seconds.

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