1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Subway train problem

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data
    (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 20.0 s at each station, what is the maximum average speed of the train, from one start-up to the next?

    2. Relevant equations
    velocity & time:
    v = v_0 + a t

    displacement & time:
    x = x_0 + v_0 t + (1/2) a t^2

    velocity & displacement:
    v^2 = v_0^2 + 2 a \\Delta x

    3. The attempt at a solution
    I tried the displacement and time equation for part A and it didn't work. I had no idea how to do the other two parts.
    How do you start this problem?
  2. jcsd
  3. Sep 1, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi mossfan563! :smile:

    It's difficult to say why it didn't work if you don't show us what you did! :wink:

    I'll guess for now :rolleyes: … did you remember to use only half the distance? :smile:
  4. Sep 1, 2008 #3
    Well I simply plugged in the distance between stations as the equation's displacement and the acceleration is given. I have to use half the distance?
  5. Sep 1, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    erm … depends on whether you want people to be able to get on and off without nets! :biggrin:
  6. Sep 1, 2008 #5
    Seriously, how do i start this problem?
  7. Sep 1, 2008 #6


    User Avatar
    Science Advisor
    Homework Helper

    You need to accelerate it for 408m at 1.54 m/s2, and then decelerate it for the other 408m. :smile:
  8. Sep 1, 2008 #7
    So the maximum speed would be the point between accelration and deceleration. So 408 would be the correct displacement instead of 816.
    Speed would be 35.449 m/s.
    Does that sound right?

    And for part B I would use the number I just got in the formula:
    V = V_0 + at?
  9. Sep 1, 2008 #8
    Ok I got part B wrong. How do you do part B?
  10. Sep 1, 2008 #9


    User Avatar
    Science Advisor
    Homework Helper

    Why do you make us guess? :confused:

    Well … I'm going to guess :rolleyes: … that you forgot to double the distance? :smile:
  11. Sep 1, 2008 #10
    Well I'm sorry if I'm making you guess. Trying to multitask. I know you use 816 as the displacement for this part. Acceleration is still 1.54. I'm just confused as to what formula to start with.

    Maybe this formula?:
    [tex] x = x_0 + v_0 t + (1/2) a t^2 [/tex]
  12. Sep 1, 2008 #11
    Can anyone point me towards the right formula?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Subway train problem
  1. Train problem (Replies: 2)

  2. A subway train (Replies: 1)

  3. Train Problem (Replies: 3)