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Homework Help: Tips for problem

  1. Apr 19, 2004 #1
    I really need help solving this problem!
    Give me some hints and tips on how can it be solved.

    Two trains are leaving from two points, A and B , one towards each other ,at the same time.The distance between A and B is 200Km.
    The train leaving from A has a constant speed of 40 Km/h the other one is running at 60Km/h.An eagle starts flying back and forth inbetween the trains until the locomotives meet.( It means that the eagles flyes from A towards B, meets the locomotive B and return back to A and so on.) The egale's flight speed is 80km/h.
    find out:
    -What is the distance the eagle flyes
    -How many times the eagles goes back and forth

  2. jcsd
  3. Apr 19, 2004 #2


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    Do NOT try to calculate the total distance the eagle flies by finding the distance from one train to another each time the eagle takes off! Find the time it takes until the trains meet (I do hope they are not on the same track. I would hate for the eagle to get squashed between them!). Now multiply that time by 80km/h to find how far the eagle flew.
  4. Apr 19, 2004 #3
    ok, but, how many times does the eagles goes back and forth
  5. Apr 20, 2004 #4
    I know this problem as the fly and bicycle problem. I'm afraid I will not be able to help you much with the second part of your question; you'll have to get one of the Math Mentors in here. However, it is my guess that the second part is a trick question as the eagle's body is not infinitesimally small. If it was, then it goes back and forth and infinite number of times. But if you have to factor in the eagle's true body size, it would be a different story.

    Someone get a professional in here!!!
  6. Apr 20, 2004 #5
    Come to think to it, even the first part of the problem cannot be solved using the special trick HallsofIvy suggested. There's a reason as to why the fly and bicycle analogy is more commonly used.
  7. Apr 20, 2004 #6


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    The trains are "closing" on one another at 100 km/hr. It will take them exactly 2 hours to cover the 200 km between them and that is the time the eagle flies.

    As others have said, unless you give a specific size for the eagle, there is no way to answer the second part- there is always some smaller interval between the trains for the eagle to fly. The point of the "bicycle and fly" is that we take the fly to be a point and declare (ala Zeno) that he must make an infinite number of loops.

    Okay, here goes: doing exactly what I warned about!!

    Assume that, at the instant the trains are 200km apart, the eagle takes off from train A to train B at 80km/hr. Let t be the time it takes the eagle to arrive at train B. In that time, train B will have moved 60t km closer so the eagle actually flies 200-60t km. We must have 80t= 200- 60t or 140t= 200 so t= 200/140= 10/7 hrs. The two trains are closing on one another at 40+ 60= 100 km/hr so during that same time the two trains have moved 1000/7 km closer together. When the eagle "turns around" the two trains are 200- 1000/7= 400/7 km apart.

    Again, let t be the time back to train A. Since train A has moved 40t in that time, the eagle must fly 400/7- 40t km or 400/7- 40t= 80t so 120t= 400/7 and t= 10/21 hours. In that time, the trains have moved a total of 1000/21 km closer: they are now 400/7- 1000/21= 200/21 km apart.
    (So far the eagle has flown for 10/7+ 10/21 hr.)

    Let t be time to fly back to B. In that time, B moves 60t km so the eagle must fly 200/21- 60t km: 80t= 200/21- 60t so 140t= 200/21 or t= 10/147 hr. In that time the trains have moved 1000/147 km closer so they are now 200/21- 1000/147= 400/147 km apart.
    (So far the eagle has flown 10/7+ 10/21+ 10/147 hrs.)

    Let t be the time back to A. In that time A moves 40t km so the eagle must fly
    400/147- 40t km: 80t= 400/147- 40t so 120t= 400/147 or t= 10/441 hrs. (That's actually about 1 min 22 sec.!). The total time the eagle has flown now is
    10/7+ 10/21+ 10/147+ 10/441 hr= 1.99 hrs.

    You see what's happening- we are getting an infinite series that will, eventually, sum to 2 hrs. If you don't want to say that the eagle will have to fly an infinite number of "legs", you will have to declare, at some point, that the distance between the trains is smaller that the eagle itself, "declare victory", and annouce the answer!
  8. Apr 20, 2004 #7
    This may be of some help to you:


    This situation is really beginning to remind me of that opening scene with the French soldiers in Monty Python and the Holy Grail, where they debate the coconut-swallow migration. :)
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