What is the Maximum Speed and Average Speed of a Subway Train Between Stations?

[tigerlili]

Homework Statement

(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.54 m/s2 and subway stations are located 816 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 21.0 s at each station, what is the maximum average speed of the train, from one start-up to the next?

Homework Equations

x= vot + 1/2 at^2 (i think? I'm not sure what else to use.. but time is not given)

The Attempt at a Solution

i don't know where to start :/

 

[berkeman]

Homework Statement

(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.54 m/s2 and subway stations are located 816 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 21.0 s at each station, what is the maximum average speed of the train, from one start-up to the next?

Homework Equations

x= vot + 1/2 at^2 (i think? I'm not sure what else to use.. but time is not given)

The Attempt at a Solution

i don't know where to start :/

Think about it. How will the train move from station to station? Will it accelerate all the way? Why not? How much of the way will it be accelerating versus decelerating?

 

[tomcue]

As a scientist, it is important to first understand the given information and identify any relevant equations that can be used to solve the problem. In this case, the maximum acceleration and distance between stations can be used to calculate the maximum speed that the train can attain. The equation v^2 = u^2 + 2as can be used, where v is the final velocity, u is the initial velocity (which can be assumed to be 0 in this case), a is the acceleration, and s is the distance between stations. Using this equation, the maximum speed can be calculated as √(2 * 1.54 * 816) = 40.44 m/s.

To calculate the travel time between stations, the equation s = ut + 1/2at^2 can be used, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. Since the initial velocity is 0, the equation simplifies to s = 1/2at^2. Plugging in the values, we get 816 = 1/2 * 1.54 * t^2, which simplifies to t = √(816/0.77) = 32.29 s.

To calculate the maximum average speed of the train, we can use the equation v = (u + v)/2, where u is the initial velocity and v is the final velocity. Since the train starts from rest and accelerates to the maximum speed, the average speed can be calculated as (0 + 40.44)/2 = 20.22 m/s. However, since the train also stops for 21.0 s at each station, the average speed will be slightly lower. We can calculate this by finding the average speed for each segment of the journey (acceleration phase and stopping phase) and taking the weighted average. This would result in a slightly lower maximum average speed of the train.