An express subway train passes through an underground station. It enters at t =

Question

An express subway train passes through an underground station. It enters at t = 0 with an initial velocity of 23.5 m/s and decelerates at a rate of 0.150 m/s^2 as it goes through. The station in 210 m long. How long is the nose of the train in the station? Dollar How fast is it going when the nose leaves the station? m/s If the train is 130 m long, at what time t does the end of the train leave the station? dollar What is the velocity of the end of the train as it leaves? m/s Suppose you are looking down from a helicopter at three cars travelling in the same direction along a freeway. The positions of the three cars every 2 seconds are represented by dots on the diagram. The positive direction is to the right.

Explanation / Answer

Initial velocity u = 23.5 m/s

a = - 0.150 m/s2 and S = 210 m

use, v2 = u2 + 2aS

=> v2 = (23.5)2 + 2( - 0.150)(210)

=> v = 22.12 m/s

and v = u + at

=> 22.12 = 23.5 + ( -0.15)t

=> t = 9.2 seconds.

b] the speed of the train when its nose leaves the subway is, v = 22.12 m/s

c] The train needs to travel an additional distance of 130 m to clear the subway entirely.

so, u = 22.12 m/s and S' = 130 m

v2 = (22.12)2 + 2( - 0.15)(130)

=> v = 21.22 m/s

this is the velocity of the train as it entirely leaves the subway.

and so, t' = (22.12 - 21.22)/0.15 = 6 s

therefore, the total time taken to leave the subway = 9.2 + 6 = 15.2 seconds.

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