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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

    tiny-tim

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    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

    tiny-tim

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    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

    tiny-tim

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    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

    tiny-tim

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    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?
     
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