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/s2 as it goes through. The station in 210 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?
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To solve these problems, we can use the equations of motion. Let's take a look at each part:
(a) How long is the nose of the train in the station?
To find the time it takes for the nose of the train to pass through the station, we can use the equation:
s = ut + (1/2)at^2
where:
- s is the distance covered (210 m),
- u is the initial velocity (23.0 m/s),
- a is the acceleration (deceleration in this case) (-0.150 m/s^2),
- t is the time.
Rearranging the equation to solve for t:
t^2 - (46/3)t + 840 = 0
Solving this quadratic equation will give us two possible values for t. However, we discard the negative solution since time cannot be negative. The positive solution will give us the time it takes for the nose of the train to pass through the station.
(b) How fast is it going when the nose leaves the station?
To find the final velocity at the moment the nose leaves the station, we can use the equation:
v = u + at
where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration (deceleration in this case),
- t is the time calculated in part (a).
Substituting the values into the equation will give us the final velocity.
(c) If the train is 130 m long, at what time t does the end of the train leave the station?
To find the time at which the end of the train leaves the station, we can utilize the same equation as in part (a):
s = ut + (1/2)at^2
where:
- s is the distance covered (210 m + 130 m = 340 m),
- u is the initial velocity,
- a is the acceleration (deceleration in this case),
- t is the time.
Solving for t will give us the time it takes for the end of the train to leave the station.
(d) What is the velocity of the end of the train as it leaves?
To find the final velocity at the moment the end of the train leaves the station, we can again use the equation:
v = u + at
where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration (deceleration in this case),
- t is the time calculated in part (c).
Substituting the values into the equation will give us the final velocity.