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Collision( kinematics)

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data
    The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. The freight train is traveling at 15.0 m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of 0.100 m/s^2, while the freight train continues with constant speed. Take x=0 at the location of the front of the passenger train when the engineer applies the brakes.

    Where will the collision take place?




    2. Relevant equations

    y=yo+vo*t+1/2*a*t^(2)

    -b+- Sqrt(b^(2)-4ac)/2a

    3. The attempt at a solution

    yp-passanger train and yf=freight train

    Yp=yo+vo*t+1/2*a*t^(2)
    yp=25m/s*t - 0.05m/s^(2)*t^(2)

    Yf=yo + vo*t +1/2*a*t^(2)
    yf=200m +15m/s*t

    yp=yf

    -0.05m/s^(2)*t^(2) + 10m/s*t -200m

    I get t= 177.4596 seconds and t=22.5403 seconds

    How do I know which time is correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 22, 2013 #2

    Dick

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    I haven't checked your numbers. But sure you get two solutions. The later one assumes the trailing train survives the first collision and passes through the leading train and then collides again as the trailing train catches up with it again. Which do you think is the physically plausible one?
     
  4. Jan 22, 2013 #3
    The 22.5403 seconds one!
     
  5. Jan 22, 2013 #4

    Dick

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    I'd agree with that. The first collision is usually the last.
     
    Last edited: Jan 22, 2013
  6. Jan 23, 2013 #5
    Thank you!
     
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