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Motion problem

  1. Oct 13, 2009 #1
    Two trains heading straight for each other on the same track are 850m apart when their engineers see each other and hit the brakes, giving both trains a constant deceleration (a). The Express is heading west at a speed of 15.0 m/s, while the east bound Flyer is traveling at a speed of 25 m/s. Calculate the minimum deceleration (a) required for a collision to be avoided.

    V1 speed of first train = 15.0 m/s
    V2 speed of second train = 25 m/s
    x=850 m

    Can someone please help me solve this problem. I have been staring at it for about an hour and cannot even begin to go anywhere.
  2. jcsd
  3. Oct 13, 2009 #2


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    What is the relevant kinematic equation which relates vi, vf, a and x?
    In this problem what is the final velocities of the trains?
    If west bound train moves a distance x m before it stops, what is the distance traveled by east bound train?
  4. Oct 13, 2009 #3
    The kinematic equation would be..

    V^2 = V0^2 + 2a (X-X0)

    The final velocity would have to be 0 since they have to come to a complete stop to avoid a collision.

    I am not really sure about the distance.. since the total distance is 850..possibly 850-x?
  5. Oct 13, 2009 #4


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    Yes. You are right. Now wright down two equations for two trains. And solve for x. From that you can find a.
  6. Oct 13, 2009 #5
    ok let me try this

    15^2 = 2a(850-x)


    25^2 = 2a(850-x)

    a = 25^2 / (1700 - 2x) plug back into eqn. 1? to solve for x?
  7. Oct 13, 2009 #6


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    First one should be
    15^2 = 2ax.
    Now 15^2/25^2 = 2ax/2a(850-x)
    Solve for x.
  8. Oct 13, 2009 #7
    ok I get x = 225m is this correct? It was a little difficult to solve for x. The acceleration should cancel correct?
  9. Oct 13, 2009 #8
    Also plugging that number back in I get the acceleration for both of the equations to be

    a= -0.5 m/s^2 minimum
  10. Oct 13, 2009 #9


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    Yes. You are right.
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