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/s2 as it goes through. The station in 215 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?
To solve this problem, we can use the equations of motion to find the answers. The equations of motion we will use are:
v = u + at (equation 1)
s = ut + (1/2)at^2 (equation 2)
v^2 = u^2 + 2as (equation 3)
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
- s is the displacement or distance traveled
Now let's solve the problem step by step.
(a) To find the length of the nose of the train in the station, we need to determine the time it takes for the train to come to a stop. We can use equation 1 to find the time (t1) it takes for the train to decelerate to 0 m/s.
Using equation 1:
0 = 23.5 m/s + (-0.150 m/s^2) * t1
Rearranging the equation:
t1 = 23.5 m/s / 0.150 m/s^2
Simplifying:
t1 = 156.7 seconds
Since the train takes 156.7 seconds to stop and the station is 215 m long, the length of the nose of the train in the station is the distance traveled during this time:
s1 = ut1 + (1/2)at1^2
s1 = 23.5 m/s * 156.7 seconds + (1/2) * (-0.150 m/s^2) * (156.7 seconds)^2
Simplifying:
s1 = 1849.7 meters
Therefore, the length of the nose of the train in the station is 1849.7 meters.
(b) To find the speed at which the nose of the train leaves the station, we can use equation 1 again. However, since the train starts at 23.5 m/s and decelerates to 0 m/s, the final velocity (v) is 0 m/s.
Using equation 1:
0 = 23.5 m/s + (-0.150 m/s^2) * t2
Rearranging the equation:
t2 = 23.5 m/s / 0.150 m/s^2
Simplifying:
t2 = 156.7 seconds
Therefore, when the nose of the train leaves the station, its speed is 0 m/s.
(c) To find the time (t3) at which the end of the train leaves the station, we can use equation 2 with the displacement of the train being 130 m.
Using equation 2:
130 m = 23.5 m/s * t3 + (1/2) * (-0.150 m/s^2) * t3^2
Rearranging the equation:
(1/2) * (-0.150 m/s^2) * t3^2 + 23.5 m/s * t3 - 130 m = 0
By solving this quadratic equation, we can find t3.
(d) Once we find the time (t3) at which the end of the train leaves the station, we can use equation 1 to find the velocity (v3) of the end of the train.
Using equation 1:
v3 = 23.5 m/s + (-0.150 m/s^2) * t3
Simplifying using the value of t3 we obtained from part (c), we can find the velocity.
By following the steps mentioned above, you will be able to find the answers to all parts of the question.