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Mechanics - Rectilinear Motion

  1. Aug 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A train travels between two stations 4.5km apart. If the maximum acceleration is 3m/s2 and the greatest deceleration is 4.5m/s2, find the shortest time between the two stations if the maximum speed allowed is 100km/hr.

    Max Acc. = 3ms-2
    Max Dec. = 4.5ms-2
    Max speed = 100kmhr-1 = 27.778ms-1
    Distance = 4.5km = 4500m

    2. Relevant equations
    v = ds/dt

    a = dv/dt

    vdv = ads


    3. The attempt at a solution
    I have done questions similar to this one. However, I have never seen a question where you are only given distance, max acceleration & deceleration and speed and you are told to find time. Thus, I have not been able to make an attempt at this question. I would appreciate any help to steer me in the right direction with this problem.

    Cheers
     
  2. jcsd
  3. Aug 9, 2009 #2
    The train is going to accelerate, hit max speed, then decelerate. Can you draw a speed-time graph? How would you relate that to distance?
     
  4. Aug 9, 2009 #3
    Thanks for the speedy reply. Um yeah, so that means the acceleration period of the v-t graph will have a gradient of 3, then it will hit the max speed at 27.778, then the deceleration period will have a gradient of -4.5. So that tells me that the question must be done in three parts?
     
  5. Aug 9, 2009 #4
    No, I'm asking something simpler: How do you find distance from a speed-time graph?
     
  6. Aug 9, 2009 #5
    Oh ok, that would be the area (integral) which is 4500m.
     
  7. Aug 9, 2009 #6
    Exactly: can't you now find the time using the graph?
     
  8. Aug 9, 2009 #7
    Can you? I'm not certain. I have to go, I'll be back online in about 2 hours. Thanks for your help so far.
     
  9. Aug 9, 2009 #8
    Yes you can, you need to find an expression for the total area. Hint: you can find the total time taken to accelerate / decelerate without using the graph
     
  10. Aug 9, 2009 #9
    Yes! I got it out.

    I used a = dv/dt, where a = 3, dv = 27.778 and dt = time taken to accelerate.
    Then did the same to find the time taken to decelerate. Then worked out the areas under the acc. dec. triangles, took these values away from the total distance. Then multiplied this by the velocity. And then finally added this time to the two other times. And most importantly the answer is correct.

    Cheers for all your help mate.
     
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