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Train Kinematics Problem

  1. Jul 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Two trains, one traveling 60.0mph east and the other traveling 90.0mph west, are on the same track. When they are 10000 ft apart both train's brakes are applied. If they both have the same deceleration, what is the minimum deceleration to prevent a collision?

    2. Relevant equations

    x = xo + vot + 0.5at2

    3. The attempt at a solution
    I converted velocities to ft/s, 60mph = 88ft/s , 90mph = 132ft/s

    I chose the location of the eastbound train to be the origin of the x-axis, then wrote out the positions of both trains as:
    x1 = 88t - 0.5at2
    x2 = 10000 - 132t + 0.5at2

    I supposed that the minimum acceleration to prevent a collision implies that x1 approaches x2. So
    x1 = x2
    88t - 0.5at2 = 10000 - 132t + 0.5at2

    Solving for a gives:
    a = (88+132)/t - 10000/t2

    I am currently stuck here, and really puzzled as to why a is dependent on t. Acceleration must be a constant, so I think I must have some error somewhere that I cannot find. Any help is greatly appreciated.
     
  2. jcsd
  3. Jul 28, 2014 #2
    Welcome to PF.

    Use the another equation of kinematics of motion without time.

    And Use the relative velocity concept. Here two trains going in opposite direction. What is the velocity of first train relative to second.?
     
  4. Jul 28, 2014 #3

    Nathanael

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    a is dependent on t simply because that's how you chose to set up the problem. That is, you chose to use time as an intermidiate step ("the middle man")

    You have one more constraint (equation) that you're overlooking.

    You've set [itex]x_1=x_2[/itex] because you want to find the minimum
    (and the minimum is where they just barely don't crash, meaning, [itex]x_1\approx x_2[/itex])

    But [itex]x_1=x_2[/itex] is true even if they do crash, isn't it? So what else must be true?
     
  5. Jul 28, 2014 #4

    Orodruin

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    Hint: If they have the same deceleration, they will not stop at the same time.
     
  6. Jul 28, 2014 #5
    Okay, I think that I have got the solution.

    Using x = xo + (v2 - vo2) / (2a) :

    x1 = -(88)2 / (-2a)
    x2 = 10000 - (132)2 / (2a)

    x1 ≈ x2

    I was then able to solve for a and obtain a = 1.26 FT/s2.

    I see now that I failed to notice some crucial bits of information during my first attempts at this problem.

    Thanks everyone for the help!
     
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