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Train collision problem

  1. Oct 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Train A travels with the velocity v= 30m/s, suddenly the driver sees another train ahead, Train B which travels with the velocity v=10m/s. Immediately Train A's driver starts the brakes ( a=1m/s²) while Train B still travels with its velocity. The moment A starts the brakes the distance between A's front and B's back is 200m.

    Will both trains collide? If yes, where?

    2. Relevant equations

    3. The attempt at a solution
    i searched the internet and found an equation:

    s= v² / 2a

    I was going to try it this way but ended up being confused. cause i wouldn't get metres out of it, instead seconds

    unknown person's attempt

    my attempt:

    s= 30m/s / 2*1m/s² = 30m/s * 1 / 2m/s²

    I don't know how to get a lenght out of this equation.
  2. jcsd
  3. Oct 28, 2014 #2
    Ok I tried another equation

    s=1/2 * (deltaV)² / a

    s= 1/2 * (30m/s) ² / 1m/s²

    s=1/2 * 900m²/s² / 1m/s²

    s=1/2 * 900m/ 1


    is this correct? I would be so glad and happy if it was, usually I feel so dumb since I started university
  4. Oct 28, 2014 #3
    I calculated that they do collide but i don't know how to calculate where
    How do I find out where they collide?

    Train A travels 30seconds 450m
    during that time Train B travels 150m (+200m) it is less , so they collide.
  5. Oct 28, 2014 #4
    Maybe you should start by trying to write the kinematic equations for both Train 1 and Train 2 given their velocities in terms of time. Then calculate the difference in distance as

    ##x_{A} - x_{B}##

    The introductory physics equation you are looking for is:

    ##x(t) = v_{0}t + \frac{1}{2}at^{2}##

    where ##x(t)## is the distance, ##t## is the time, ##v_{0}## is the initial velocity, and ##a## is the acceleration.

    As an extra hint, you should have two separate equations of the form:

    ##x_{A}(t) = v_{0,A}t + \frac{1}{2}a_{A}t^{2}##
    ##x_{B}(t) = v_{0,B}t + \frac{1}{2}a_{B}t^{2}##

    ##x_{A} - x_{B} = v_{0,A}t + \frac{1}{2}a_{A}t^{2} - v_{0,B}t - \frac{1}{2}a_{B}t^{2}##

    But several terms in these equations equal zero (which ones?). It's important to note that you do not know the time of the collision (if there is one), but you can always calculate it using the quadratic formula. The important thing is that if one or both of the two solutions for time may be negative.
  6. Oct 28, 2014 #5

    thanks a lot for helping me

    i calculated both and got x_{A} - x_{B} = 600m

    is this correct and even possible? cause how are they supposed to collide after 600m if Train A stops completely after 450m?
  7. Oct 28, 2014 #6
    Ahh...I apologize, I forgot to add one more term to the equations:

    ##x_{A}(t) = x_{0,A} + v_{0,A}t + \frac{1}{2}a_{A}t^{2}##
    ##x_{B}(t) = x_{0,B} + v_{0,B}t + \frac{1}{2}a_{B}t^{2}##

    We need to take into account the initial displacement between the two trains.

    You should get that the trains collide when ##x_{A} - x_{B} = 0## and your time should be an integer number.
  8. Oct 29, 2014 #7
    I need a bit more help, i don't know what to use as xoa and xob :-(
  9. Oct 29, 2014 #8


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    Staff: Mentor

    You choose any convenient point on the railway line to be the point you'll call x = 0. Each train can then be located relative to this point.

    You'll then be looking to find whether there is ever a time, t, when the trains are identical distances from that origin.

    Xoa and Xob are therefore the distances from the origin of those trains at the start of observations, i.e., at t = 0.
    Last edited: Oct 29, 2014
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