(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) it can accelerlerate half way, and deacceleralte halfway.
Vf^2=2ad solve for Vf, then avg velocity is half that, so time=1030/(Vf/2)
To solve these problems, we will use the equations of motion and principles of physics.
(a) To find the maximum speed the subway train can attain between stations, we need to use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (assumed to be zero as the train starts from rest between stations), a is the acceleration, and s is the distance between stations.
Plugging in the given values:
u = 0 m/s (initial velocity)
a = 1.52 m/s^2 (maximum acceleration)
s = 1030 m (distance between stations)
Rearranging the equation, we have:
v^2 = 0 + 2 * 1.52 * 1030
v^2 = 3145.6
v ≈ 56.02 m/s (rounded to two decimal places)
Therefore, the maximum speed a subway train can attain between stations is approximately 56.02 m/s.
(b) To find the travel time between stations, we can use the equation: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Since the train starts from rest, u = 0 m/s. Plugging in the given values:
s = 1030 m (distance between stations)
a = 1.52 m/s^2 (acceleration)
Rearranging the equation, we have:
1030 = 0 + (1/2) * 1.52 * t^2
2060 = 1.52 * t^2
t^2 ≈ 1355.26
t ≈ 36.82 s (rounded to two decimal places)
Therefore, the travel time between stations is approximately 36.82 seconds.
(c) To find the maximum average speed of the train, taking into account the 20-second stops at each station, we need to consider both the time spent accelerating and decelerating between stations and the time spent stationary at each station.
The time spent accelerating and decelerating is the same as the travel time between stations, which we found to be approximately 36.82 seconds in the previous step.
The time spent waiting at each station is given as 20 seconds.
Therefore, the total time for each trip between stations is 36.82 seconds + 20 seconds = 56.82 seconds.
The distance traveled between each start-up is the distance between stations, which is 1030 meters.
Using the formula for average speed (average speed = total distance / total time), we have:
Average speed = 1030 m / 56.82 s
Average speed ≈ 18.16 m/s (rounded to two decimal places)
Therefore, the maximum average speed of the train from one start-up to the next, taking into account the stops at each station, is approximately 18.16 m/s.