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Train motion

  1. Jan 25, 2014 #1
    1. The problem statement, all variables and given/known data
    You are an engineer working for a new interstate bullet train start-up company. Your manager, Saul Enderman, is charged with designing the first 1500-meter-long segment of a new train line. (Another team is in charge of designing the remaining segments of the line.) During this initial segment of track, the trains must meet the following criteria:
    The train must start at the station at rest (velocity = 0).
    The train’s speed must not exceed 24 m/s.
    The train’s acceleration must not exceed 1 m/s2.
    The train must cover the 1500-meter-long segment in less than 2 minutes (120 seconds).
    Your manager has decided that the train’s motion should be described by the function x(t)=0.0011 m/s^3 t^3, where x is the train’s position along the track (with x=0 being the starting position at the station) and t is the amount of time since the train has left the station (with t=0 being the starting time at which the train leaves the station).
    Your job is to review your boss’s proposed function and determine which of the above criteria it successfully meets and which it fails to meet. (Hint: It fails at least one.) If your boss’s proposed function fails any of the criteria, propose another function that meets all four criteria. Explain how you reviewed your boss’s function and your function and determined their successes or failures.


    2. Relevant equations
    x(t)=0.0011 m/s^3 t^3
    or maybe 1500msin^2(pi*t/240s)


    3. The attempt at a solution
    I have tried to do this using higher and lower power terms and have failed both times. I am so lost at this point
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 25, 2014 #2

    Dick

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    The question is about being about to answer questions about the speed and acceleration of the train knowing x(t). It's not complicated. The velocity is the derivative of x(t) and the acceleration is the derivative of the velocity. You know that much I hope. You should at least be able to start. Handling the last question doesn't even require knowing that. Just try?
     
  4. Jan 26, 2014 #3
    I have been working on this for a few days. The issue is not that it is hard to take the derivatives of the function i was given. The issue is that whenever i find a new equation to satisfy one criteria, it fails another.
     
  5. Jan 26, 2014 #4

    Dick

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    Not sure I see what the problem is. Start with the first question. Is the velocity zero at t=0? What's your expression for the velocity v(t)?
     
  6. Jan 26, 2014 #5
    Velocity is 0 at t =0. my function is V(t) = .0033m/s^3 t^2. which then fails the velocity at 120 seconds. And when i start to change any of the coefficients it just gets worse. I have even tried it in this format v(t) = 3at^2-2bt
     
  7. Jan 26, 2014 #6

    Dick

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    Ah, I see what's making it hard. It is a little tricky to do it that way. How about taking a simple approach. Just accelerate at 1m/s^2 for 24 seconds. Then stop accelerating.
     
  8. Jan 26, 2014 #7
    Well someone helped me come up with a piecewise function. which reads

    x(t) = (0m/s)t+1/2(1m/s^2)t^2 if 0≤t≤24
    (24m/s)(t-24) if t≥24

    However they could not really explain to me how they got that. They said something about equation of motion and constant acceleration. but we havent learned any equations of motion yet. we are only in our second week.
     
  9. Jan 26, 2014 #8

    Dick

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    That's pretty much what I was suggesting. For the first 24 seconds v(t)=(1m/s^2)t and a(t)=(1m/s^2). You get the first piece by just starting from a(t)=(1m/s^2) and working backwards. And the second piece isn't quite right. x(24) should be the same number on both pieces, shouldn't it?
     
  10. Jan 26, 2014 #9
    I dont think so. it should be x(120). so x(24) for first part and then x(120-24) for the second piece right?
     
  11. Jan 26, 2014 #10

    Dick

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    I'm not sure I see why. I'm just saying that if you graph the first piece for 0<=x<=24 and the second part for 24<=x<=120, they should have the same value at x=24. The train can't suddenly jump from one to another.
     
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