Question

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long. (a) How long did the nose of the train stay in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?

(a)(a) 15.0 s; (b) 7.5 m/s; (c) 18.5 s; (d) 7.5 m/s

(b)(a) 12.0 s; (b) 6.0 m/s; (c) 15.5 s; (d) 6.0 m/s

(c)(a) 18.0 s; (b) 9.0 m/s; (c) 21.0 s; (d) 9.0 m/s

(d)(a) 20.0 s; (b) 10.0 m/s; (c) 23.5 s; (d) 10.0 m/s

Answer 1

The nose of the train stays in the station for 15.0 seconds and is going at 19.5 m/s when it leaves. The end of the train leaves the station after 18.5 seconds and has a velocity of 19.5 m/s as it leaves.

Step-by-step explanation:

(a) To find the time the nose of the train stays in the station, we need to calculate the distance the nose travels while decelerating to rest.

The formula to calculate the distance is displacement = (initial velocity^2 - final velocity^2) / (2 * deceleration).

Plugging in the given values, we get displacement = (22.0^2 - 0^2) / (2 * 0.150) = 242.0 m. Since the station is 210 m long, the nose of the train stayed in the station for 210 m.

(b) To find the speed when the nose of the train leaves the station, we can use the equation final velocity = initial velocity - (deceleration * time).

Plugging in the values, we get final velocity = 22.0 - (0.150 * 15.0) = 19.475 m/s. Therefore, the nose of the train is going at approximately 19.5 m/s when it leaves the station.

(c) To find when the end of the train leaves the station, we need to calculate the time it takes for the end of the train to travel 130 m while decelerating to rest. Using the same formula as in part(a), we get displacement = (22.0^2 - 0^2) / (2 * 0.150) = 242.0 m.

Therefore, the end of the train leaves the station after 242.0 m, which is (210 + 130) - 242.0 = 98.0 m beyond the station.

(d) To find the velocity of the end of the train as it leaves, we can use the same equation as in part(b), but with a displacement of 98.0 m instead of 210 m.

Plugging in the values, we get final velocity = 22.0 - (0.150 * 18.5) = 19.475 m/s.

Therefore, the velocity of the end of the train as it leaves is approximately 19.5 m/s.