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Two trains and straight line movement

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    The train travels from place A to place B for one hour. Forty minutes after the departure, first train meets with second train, which departed from place B ten minutes after first train, and it’s traveling with an average speed of 40 km / h.

    2. Relevant equations

    What is the distance between points A and B? Assuming that the movement of trains at all times can be described as consistent evenly.

    3. The attempt at a solution

    I really need some help because I don not know where to start.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 9, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi mmoadi! Welcome to PF! :wink:

    Start by calling the distance x km … then the speed of the first train is x km/hour.

    Show us what you get. :smile:
     
  4. Oct 10, 2009 #3
    I know that the it takes for the first train to travel the distance between A and B in 1 hour.
    Does this mean that the v1 (velocity of the first train) is v1= d/t which gives us that the v1 = d after I plug in the time which is 1 hour?
    I also know that the two trains meet after the first train has traveled for 40 min (t1) and that the average speed of the second train is 40 km/h (v2).
    So, in my opinion the whole distance is s= v1t1 + v2t2. Am I right?
    So how do I continue?
     
  5. Oct 10, 2009 #4
    I did some more thinking.
    So, I know that the speed(v1) of the first train is v1=d. And the two trains meet after 40 min(2/3h) and the second train leaves 10 min after the first train, so does this mean that it travels for 40min-10min=30 min (1/2h) before it meets the first train. If I am right than it comes to:
    s= v1t1 + v2t2 = d(2/3h) + 40km/h(1/2h)= 2/3d + 20km
    d - 2/3d = 20 km
    d(1-2/3) = 20 km
    1/3d=20 km
    d= 60 km

    Is my solution right?
     
  6. Oct 10, 2009 #5

    tiny-tim

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    :biggrin: Woohoo! :biggrin:

    Yes, that's exactly the way to do it …

    and we can always check the answer (which I did :wink:) by putting 60 back into the original question, and confirming that it works! :smile:
     
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