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Solving for Constant acceleration

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data
    When a high-speed passenger train traveling at vP = 55 m/s rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D = 1000 m ahead). The locomotive is moving at vL = 12 m/s. The engineer of the passenger train immediately applies the brakes. Assume that an x axis extends in the direction of motion. What must be the constant acceleration along that axis if a collision is to be just avoided?

    2. Relevant equations
    Kinematic Equations?

    3. The attempt at a solution

    I found the time with respect to acceleration which is (-43/a)
    The distance the 12m/s train moves, d=12(-43/a)
    The distance the passenger train moves (55m/s). d=1000-516/a
    I'm having issues piecing this all together.
    Any help to clarify this would be greatly appreciated.
  2. jcsd
  3. Sep 29, 2010 #2
    The first thing I need to say here is that the question doesn't make it clear which way the loco is moving. Towards the express or in the same direction. The only clue is that both velocities are given as positive, which would imply that the loco is moving in the same direction. If it isn't then you will need to change the solution to take that into account.
    A diagram always helps.
    The loco moves a distance x in the time it takes the express to stop. (time = t)
    The express moves a distance s.
    We know s = x +1000 [ignoring the length of the loco]
    The loco is moving at constant speed so its equation is simple.
    The express is decelerating so its motion will be described by the standard equations for uniform acceleration.
    v²=u² +2as
    s=ut + ½at²


    To solve this you need to write down two equations, one for the loco (with x and t) and one for the express.
    From these you can find the answer.
  4. Oct 1, 2010 #3
    Thanks much!
    I ended up solving the equation 0^2=43^2+2(a)(1000) for a, and got -0.9245 m/s^2.
    Last edited: Oct 1, 2010
  5. Oct 1, 2010 #4
    -0.9245 what? kilogrammes? miles per hour per hour?
  6. Oct 1, 2010 #5
    Well it's acceleration so m/s^2 :)
  7. Oct 3, 2010 #6
    Yes, that's a good way of solving this. (My way used two steps and is not really necessary)
    Just be careful with your method that you make it clear where the 43 comes from.
    You need to say you are using relative velocity of approach and calculating the time for the separation to reach zero. At this point the express will be just behind the loco and travelling at the same speed.
    My method was actually unnecessarily complicated. The suvat equation for the express would require its speed to drop to 12 m/s (not zero) in the distance s to avoid the loco.
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