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Linear Motion - Minimum Retardation to Avoid Crash?

  1. Jan 30, 2012 #1
    First post! I would be grateful if anyone could give me any advice on this particular type of problem (i.e min retardation to avoid a crash). I'm not necessarily looking for the answer to this specific question, I would rather if someone could point me in the right direction as to how to go about these questions. Thanks!

    1. A passenger train, which is travelling at 80 m s[itex]^{-1}[/itex] is 1500 m behind a goods train which is travelling at 30 m s[itex]^{-1}[/itex] in the same direction on the same track. At what rate must the passenger train decelerate to avoid a crash? (Ignore the lengths of the trains)



    2. s = ut + [itex]\frac{1}{2}[/itex]at[itex]^{2}[/itex]
    v = u + at
    s = ([itex]\frac{u + v}{2}[/itex])t
    v[itex]^{2}[/itex] = u[itex]^{2}[/itex] + 2as



    3. I worked out that at current speeds, the trains would collide after 30 seconds, using a velocity-time graph, letting the distance travelled by the passenger train = distance travelled by the goods train + 1500. (30T + 1500 = 80V). From there I summized that for the trains to never crash, the passenger train should decelerate to at least the same speed as the goods train ( 30 m s[itex]^{-1}[/itex]). Then I used v = u + at for the passenger train, and had 30 = 80 + a(30), where a comes out at - [itex]\frac{5}{3}[/itex] m s[itex]^{-2}[/itex]. I've compared this answer to the correct one.

    Thanks,
    project_ILE
     
  2. jcsd
  3. Jan 30, 2012 #2
    You wrote,

    "I've compared this answer to the correct one."


    %^) So did you get the right answer? If you did fine, if not then someone might care to find the error.
     
  4. Jan 30, 2012 #3
    Sorry, I thought I had finished my post! This question is from an exam paper, I have the numeric answer, but not the method. My own answer (above) is not correct.
     
  5. Jan 31, 2012 #4
    The answer might help potential helpers. What was it 8^)
     
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