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Homework Help: Relative Motion of a train

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Two trains start from rest from two stations 5.0 km apart and travel, on two parallel railroad tracks, in opposite directions. One train accelerates uniformly at 2.5 m/s^2 and the other has a constant acceleration of 4.0 m/s^2.
    How far from the first station will they meet?

    2. Relevant equations

    1. d= at^2
    2. v^2 = u^2+ 2ad
    3. t= v-u/ a

    3. The attempt at a solution
    Train 1 relative to Train 2
    Initial velocity = 0 m/s
    Acceleration = 6.5 m/s^2
    Displacement = 5000m

    v^2 = u^2+ 2ad
    v = Square root of u^2 + 2ad
    v= Square root of 2ad
    v= Square root of 2(6.5 m/s^2) (5000m)
    v= Square root of 65000 m^2/s^2
    v= + or - 255 m/s

    t= v-u/ a
    t= 255ms^-1/6.5m/s^-2
    t= 39 seconds

    Train 1

    d= at^2 because initial velocity = 0
    d= (2.5 ms^-1) (39 seconds)
    d= 97.5m

    Therefore they will meet 97.5m from the first station.

    However this answer does not make sense to me because if I solve for distance with the other train it does not add to 5000m or 5.0 km. I do not know what I did wrong.

    Thanks In Advanced
  2. jcsd
  3. Nov 18, 2013 #2


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    Welcome to PF OmniNewton!

    This bit in red is wrong. The actual equation is:
    d = d0 + v0t + (1/2)at2

    Where d0 and v0 are the initial position and speed respectively. If these are both zero, it's still d = (1/2)at2. You are missing the factor of 1/2.

    There is a fundamental conceptual error here. You are assuming that the first train travels 5 km. It doesn't. The situation is that the two trains start out 5 km apart and head towards each other along parallel tracks. Since they are approaching each other, they will meet somewhere in the middle of the two stations, and you have to figure out where and when this occurs.

    In any case, you do not need this equation v2 = u2 + 2ad. The two trains will "meet" when their positions are the same. So, write down an expression for the position vs. time of train 1. Now write down an expression for the position vs. time of train 2 (hint: it does not have the same starting position, and the sign of its acceleration is opposite). Equate these two expressions, and solve for t. This gives you the time at which they meet. From this, you can get the position of each one.
  4. Nov 18, 2013 #3
    It actually all worked out once I used that equation d= at^2/2

    5000m worked because train 1 relative to train 2 equates to 5000m

    train 1's displacement = x

    train 2's displacement = 5000m - x

    so ultimately

    train 1's displacement relative to train 2 has to be 5000m.

    Correct me if I am wrong however.

    But thank you very much for the help you really helped me and it is very much appreciated :D

    Thank You Very Much For The Welcome.
  5. Nov 18, 2013 #4


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    If it worked out, then I guess your math must have been right, but what you are *saying* is totally wrong. First of all, train 1's *position* is x1. Its displacement is its change in position. Train 2's position is x2 relative to its station (or 5000 - x2 relative to the first station). Their separation (the distance between them) is NOT 5000 m at any time other than t = 0. E.g. at t = 0, x1 = 0 and x2 = 5000, so their separation is x2 - x1 = 5000 m. But suppose at some later time, the first train has moved 100 m along the track, and the second has moved 400 m along the track (totally made up numbers). So, train1's position (and displacement) is 100 m. Train 2's position (measured relative to the first station) is 5000 - 400 = 4600 m. Train 2's displacement is 400 m. The separation between the trains ("relative displacement") is 4600 m - 100 m = 4500 m. So, you can see that their relative distance is NOT 5000 m after t = 0, since they are moving *towards* each other.
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