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Maximum acceleration homework

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    If the maximum acceleration that is tolerable for passengers in a subway is 1.70 m/s2, and the subway stations are located 699 m apart, what is the maximum speed that the train attains between stations?

    2. Relevant equations

    d= v(initial)t + .5at^2


    3. The attempt at a solution

    699m = 0t + .5(1.7)t^2

    Is this wrong, I don't know what to do. thanks
     
  2. jcsd
  3. Sep 3, 2009 #2

    kuruman

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    Re: subway

    First off, you are not calculating a speed. You are calculating how long it will take the train to reach the next station if it were accelerating all the way.

    You need to consider that the train will have to accelerate at the maximum acceleration for half the distance and then decelerate at the same maximum but opposite acceleration for the other half of the distance so that it is at rest when it reaches the next station. The maximum speed that problem is asking is attained at the half-way point.
     
  4. Sep 3, 2009 #3
    Re: subway

    so I would want to solve for t right? The time it will take? also should i cut the distance (d) in half to 349.5? Sorry, I'm just not very good at this but thank you very much for any help.
     
  5. Sep 4, 2009 #4

    rl.bhat

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    Re: subway

    It is also possible that the train attains maximum speed with maximum acceleration, moves certain distance and then again slows down with maximum retardation before coming stop.
     
  6. Sep 4, 2009 #5

    kuruman

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    Re: subway

    I don't think so. If the train accelerates at maximum acceleration, the final speed will be greater the longer time it accelerates. So it needs to accelerate for half the distance to maximize the acceleration time.
    Yes, solve for t required to reach the halfway mark at maximum acceleration. Then use the kinematic equation relating the final velocity and acceleration to find the speed at that time. That is the maximum speed because the train will have to start slowing down past that time.
     
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