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2 trains colliding, where am I wrong?

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data

    2 trains approach each other on same track with equal speed of 20 m/s. When they are 2km apart, they see each other and deaccelerate. If deaccelerations are uniform, how large the acc. would be to avoid collision.

    2. Relevant equations

    [tex]v^{2} - v_{0}^{2} = 2ax [/tex]

    3. The attempt at a solution

    The book says that the 2 trains will stop at 1000m and and they find out acceleration by usual method to be -0.2m/s2 with velocity 20m/s and distance 1000m

    But, I'm solving this problem by looking through a frame of reference going at -20m/s, so one of the trains appear to be stationary to me and the other moving with double velocity i.e. 40m/s. Then I calculate the acceleration which will stop this train in 2kms distance. Which comes out to be -0.4m/s2, twice the answer.

    Please guide me where I am wrong.

    Thanks.
     
  2. jcsd
  3. Oct 1, 2008 #2

    Mentallic

    User Avatar
    Homework Helper

    Using [tex]v^{2}=u^{2}+2as[/tex]

    Hence, [tex]a=\frac{v^{2}-u^{2}}{2s}[/tex]

    It can be seen that if the displacement (s) is twice, it will cause the acceleration to be halved; however, if the velocity (v) is 2x, it will cause the acceleration to be 22x=4x. So the final result is that the acceleration is 4/2=2x more than necessary.

    Without using formulas as the basis of understanding why using the method you chose is incorrect, think of it this way:

    The train travellling at 20m/s will slow to a stop in 1km by accelerating at -0.2m/s2. If the train speeds up to 40m/s and begins decelerating at -0.2m/s, it will not stop in 2km as you suspected, it will stop in 4km. Why is this? Well it's simply because if the train is travelling twice as fast, it covers the same distance in half the time and yes it will take twice as long to decelerate till stopping, but in that time it will have travelled twice further (4km instead of 2km).

    I think this is best illustrated with visual depictions, but I'm sorry, I can't provide those.
     
  4. Oct 1, 2008 #3
    But, accelerations don't change between inertial frames. I'm confused with this.

    Imagine a situation, 2 trains are approaching on a single track. And you are there beside the track in your car, moving with velocity 20m/s.

    Now the train coming from your front will appear to have 20 + 20 = 40m/s speed and the train beside you will appear to be standing still. Isn't it? Though the trains are moving with 20m/s each to a observer on ground.

    But then what happens to the positions of trains?

    Wait a second!

    What about the deacceleation of the train besides you? If 2 trains are deaccelerating together then the train beside you will appear to go backwards to you.

    Problem nailed!

    Thanks

    Relative velocity problems become really tricky sometimes
     
  5. Oct 1, 2008 #4

    Mentallic

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    Homework Helper

    I'm glad you nailed it, because I was barely able to follow how your mind was picturing the scenario. Once again, it is way better if these situations are represented with a model, rather than words :smile:
     
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