(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.52 m/s2 and subway stations are located 1030 m apart, what is the maximum speed a subway train can attain between stations?
b)what is the travel time between stations?
c)If a subway train stops for 20 s at each station, what is the maximum average speed of the train, from one start-up to the next?
a) To find the maximum speed between stations, we need to use the equation of motion that relates acceleration, velocity, and distance.
The equation is:
v^2 = u^2 + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
In this case, the initial velocity is 0 m/s (assuming the train starts from rest at each station), the acceleration is 1.52 m/s^2, and the displacement is 1030 m.
Plugging these values into the equation, we get:
v^2 = 0^2 + 2 * 1.52 * 1030
v^2 = 0 + 3145.6
v^2 = 3145.6
Taking the square root of both sides, we get:
v = sqrt(3145.6)
v ≈ 56.04 m/s
Therefore, the maximum speed a subway train can attain between stations is approximately 56.04 m/s.
b) To find the travel time between stations, we can use the equation of motion that relates velocity, time, and distance.
The equation is:
s = ut + (1/2)at^2
Where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity is 0 m/s (assuming the train starts from rest at each station), the acceleration is 0 m/s^2 (assuming the train maintains a constant speed between stations), and the displacement is 1030 m.
Plugging these values into the equation, we get:
1030 = 0 * t + (1/2) * 0 * t^2
1030 = 0 + 0
1030 = 0
Since the equation is not valid for this scenario, we cannot calculate the travel time between stations using the given information.
c) The maximum average speed of the train can be calculated by dividing the total distance traveled by the total time taken, considering the time spent at each station.
Since the distance between stations is 1030 m, and the train stops for 20 s at each station, the total time taken for each trip from one start-up to the next is the sum of the travel time and the time spent at each station.
Using the information from part b, the time spent traveling from one station to another is not available. Therefore, we cannot calculate the maximum average speed of the train.
To answer these questions, we can use the equations of motion and the given information.
(a) To determine the maximum speed a subway train can attain between stations, we can use the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (maximum speed),
- u is the initial velocity (zero since the train starts from rest),
- a is the acceleration (1.52 m/s^2),
- s is the distance between the stations (1030 m).
Rearranging the equation, we get:
v = sqrt(2as)
Substituting the given values, we have:
v = sqrt(2 * 1.52 * 1030) ≈ 50.36 m/s
Therefore, the maximum speed a subway train can attain between stations is approximately 50.36 m/s.
(b) To calculate the travel time between stations, we can use the equation:
t = s/v
where:
- t is the travel time,
- s is the distance between the stations (1030 m),
- v is the velocity (maximum speed) calculated in the previous step (50.36 m/s).
Substituting the values, we have:
t = 1030 / 50.36 ≈ 20.47 s
Therefore, the travel time between stations is approximately 20.47 seconds.
(c) To determine the maximum average speed of the train from one start-up to the next, we need to consider the time it takes for the train to accelerate to the maximum speed, travel the distance between stations, and decelerate to a full stop at the next station.
Since the train spends 20 seconds at each station, the total time for each trip cycle would be the travel time between stations plus twice the stop time:
Total time = 20.47 s + 20 s + 20 s ≈ 60.47 s
To calculate the maximum average speed, we can use the equation:
Average speed = total distance / total time
The total distance is twice the distance between stations (2 * 1030 m), which is 2060 m.
Substituting the values, we have:
Average speed = 2060 / 60.47 ≈ 34.02 m/s
Therefore, the maximum average speed of the train from one start-up to the next is approximately 34.02 m/s.
Divide the problem in half, accelerate and deaccelerate.
half distance = 515
515 = (1/2)(1.52) t^2
that gives you half the time
v = 0 + 1.52 t^2
total time = 2 t + 20
v average = 1030/total time