Question

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and accelerates opposite to the motion at a rate of 0.150 m/s² as it goes through. The station is 210.0 m long. (a) How fast is it going when the nose leaves the station? (b) How long is the nose of the train in the station? (c) If the train is 130 m long, what is the velocity of the end of the train as it leaves? (d) When does the end of the train leave the station?

a) (a) 20.9 m/s; (b) 1.96 s; (c) 21.3 m/s; (d) 1.96 s
b) (a) 23.4 m/s; (b) 1.40 s; (c) 22.0 m/s; (d) 1.40 s
c) (a) 19.6 m/s; (b) 2.80 s; (c) 19.0 m/s; (d) 2.80 s
d) (a) 24.1 m/s; (b) 1.60 s; (c) 23.5 m/s; (d) 1.60 s

Answer 1

The nose of the train stays in the station for approximately 1.96 seconds. The speed of the train when the nose leaves the station is approximately 20.9 m/s. The velocity of the end of the train as it leaves the station is approximately 21.3 m/s.

Step-by-step explanation:

To solve this problem, we can use the equations of motion to analyze the train's motion through the station.

(a) To find the speed of the train when the nose leaves the station, we can use the equation:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Given:

Initial velocity (u): 22.0 m/s (opposite to the motion)

Acceleration (a): -0.150 m/s² (opposite to the motion)

Distance (s): 210.0 m

Plugging in the values, we have:

v^2 = (22.0)^2 + 2(-0.150)(210.0)

Solving for v, we get:

v = 20.898 m/s

Therefore, the speed of the train when the nose leaves the station is approximately 20.9 m/s.

(b) To find the duration the nose of the train stays in the station, we can use the equation:

v = u + at

Where t is the time taken.

Plugging in the values, we have:

20.898 = 22.0 + (-0.150)t

Solving for t, we find:

t = 1.96 s

Therefore, the nose of the train stays in the station for approximately 1.96 seconds.

(c) To find the velocity of the end of the train as it leaves the station, we can use the same equation as in part (a), but with a distance of 130.0 m:

v^2 = (22.0)^2 + 2(-0.150)(130.0)

Solving for v, we get:

v = 21.346 m/s

Therefore, the velocity of the end of the train as it leaves the station is approximately 21.3 m/s.

(d) To find the time when the end of the train leaves the station, we can use the following equation:

s = ut + 0.5at^2

Where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken.

Plugging in the values, we have:

210.0 = 22.0t + 0.5(-0.150)t^2

Solving for t, we find:

t = 1.96 s

Therefore, the end of the train leaves the station at approximately the same time as the nose, which is 1.96 seconds.