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Linear Motion

  1. Aug 29, 2006 #1

    I have a question about the following problem:

    A train is traveling at 100 m/s. The engineer applies the breaks because he sees the Batmobile stuck on the tracks ahead. The breaks cause an acceleration of -2.2 m/s^2. The engineer applies the breaks when the train is 960 m from the Batmobile. How long does it take the train to reach the Batmobile?

    I thought I would use the linear motion equation of:

    X = Xo + Vo(t) + 1/2at^2

    But I became confused because I didn't know what to do about the distance "X" Is 960 the initial or final distance? A hint would be greatly appreciated.

  2. jcsd
  3. Aug 29, 2006 #2


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    It depends on the coordinate system you choose to treat the problem with!

    You basically have you two choices: 1) Put the origin on the Batmobile, or 2) Put the origin 960m away from the batmombile. It would be instructive for you to try to do the problem both ways and see if you can get the same answer. Don't hesistate to post if further confusion occurs. :smile:
  4. Aug 30, 2006 #3
    When you say origin what do you mean? Thanks!
  5. Aug 30, 2006 #4


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    A point which you define as zero displacement; as in the point (0,0) on a graph.
  6. Aug 30, 2006 #5


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    You can do it either way: You can take the trains initial position to be 0 (i.e. you are setting up your "coordinate system" so that the "origin" is at the train) and then the train must go 960 m to reach the batmobile:
    X= 960, X0= 0, V0= 100, A= -2.2.
    Solve the equation 960= 0+ 100t+ (1/2)(-2.2)t2.

    Or: take the trains initial position to be X0= 960 m away from the batmobile so that when it hits the batmobile its distance is X= 0. Then V0= -100 (since the train is going from 960 to 0 the speed is negative) and A= +2.2 (since it's speed is increasing from -100 toward 0).
    Solve the equation 0= 960- 100t+ (1/2)(2.2)t2.

    You should be able to see that those two equations are really the same.
  7. Aug 30, 2006 #6
    LOL, I get it now. I have to use the quadratic equation to solve the problem. I understand now. Thanks for your help!
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