Question

An express subway train passes through an underground station. It enters at t = 0 with an initial velocity of 23.0 m/s and decelerates at a rate of 0.150 m/s^2 as it goes through. The station in 205 m long (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, at what time t does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?

Answer 1

a) Train's nose will be present 9.19 seconds in the station.

b) The nose leaves the station at 21.62 m/s.

c) Train's end will leave after 15.33 seconds from station.

d) The end leaves the station at 20.70 m/s.

Step-by-step explanation:

a) We have equation of motion s = ut + 0.5at²

Here u = 23 m/s, a = -0.15 m/s², s = 205 m

Substituting

205 = 23t + 0.5 x (-0.15) x t²

0.075t² -23 t +205 = 0

We will get t = 9.19 or t = 297.47

We have to consider the minimum time

So train's nose will be present 9.19 seconds in the station.

b) We have equation of motion v= u + at

Here u = 23 m/s, a = -0.15 m/s², t = 9.19

Substituting

v= 23 - 0.15 x 9.19 = 21.62 m/s

The nose leaves the station at 21.62 m/s.

c) We have equation of motion s = ut + 0.5at²

Here u = 23 m/s, a = -0.15 m/s², s = 205 + 130 = 335 m

Substituting

335 = 23t + 0.5 x (-0.15) x t²

0.075t² -23 t +335 = 0

We will get t = 15.33 or t = 291.33

We have to consider the minimum time

So train's end will leave after 15.33 seconds from station.

d) We have equation of motion v= u + at

Here u = 23 m/s, a = -0.15 m/s², t = 15.33

Substituting

v= 23 - 0.15 x 15.33 = 20.70 m/s

The end leaves the station at 20.70 m/s.