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I'm stuck help someone

  1. Aug 25, 2005 #1
    I'm stuck.... help someone!!

    here is the problem:
    Two trains are travelling on the same track. One is going West at 25 km/h and the other is going East at 30 km/h. When they are 100 km apart an eagle takes off from the first train and flies at 50 km/h towards the other train. On reaching the other train it reverses direction and goes off to the first train, and keeps on flying between the two.

    Assuming that the eagle takes no time in reversing direction and that its speed remains constant calculate :

    i) the distance travelled by the eagle until the two trains collide.

    ii) the number of time the eagle touches the two trains

    p.s:I tried my hand on the problem but didn't succeed.All I can conclude from my calculations is that the process is never ending and a an infinite series is formed by the distance travelled by the eagle everytime it flies from one train to another.I know the answer to the first part will be finite but I can not really find a method to sum all the terms of the sequence bcoz it's not a normal AP or GP.Someone help me please!!!
  2. jcsd
  3. Aug 25, 2005 #2
    For question (i), by collision I assume the trains are traveling towards each other. Anyway,

    i) When the trains are 100 km apart, they will collide in

    [tex] t = \frac{{100km}}{{25\frac{{km}}{{hr}} + 30\frac{{km}}{{hr}}}} = \frac{{20}}{{11}}hr [/tex]

    In [itex] \left( 20/11 \right) [/itex] hr, the distance the eagle travels will be

    [tex] 50\frac{{km}}{{hr}} \cdot \frac{{20}}{{11}}hr = \frac{{1000}}
    {{11}}km \approx \boxed{91 km} [/tex]

    But significant figures say that your answer should be 90 km :smile:
    Last edited: Aug 25, 2005
  4. Aug 25, 2005 #3


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    Don't forget the finite size of the eagle! :)
  5. Aug 25, 2005 #4


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    As pointed out- the simple way to do this problem is calculate the time until the two trains collide and divide by the speed of the eagle! Of course, if one were really determined to do it the hard way, one might calculate lengths of the individual flights back and forth and add- getting an infinite geometric series.

    There is a story that someone posed this question to VonNeumann who thought for about 15 seconds and then gave the correct answer. The person laughed and said "You know, some people try to do that as an infinite series". VonNeumann looked puzzled ans said "But I did it as an infinite series!"
  6. Aug 25, 2005 #5


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    I've always thought von Neumann was pulling the guy's leg!
  7. Aug 25, 2005 #6
    Shouldn't it be a little over 10 times that distance? :wink:
  8. Aug 25, 2005 #7


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    He probably was. He was well known for that!

    (Though not as quite as well known for treating everyone around him as a git as Feynman!)
  9. Aug 25, 2005 #8
    Yes, and so I updated the post :wink:

    I have a long procedure for the second problem, but it concludes that:
    *If [itex] t_n [/itex] represents the time spent on [itex] n [/itex] quantity of flights, then

    [tex] t_n = \frac{{100km + t_{n - 1} \left( {50\tfrac{{km}}{{hr}} - v_n } \right)}}{{50\tfrac{{km}}{{hr}} + v_{n - 1} }} [/tex]
    [tex] t_0 = 0 \, {hr} , \;v_n = \frac{5}{2}\left[ {11\tfrac{{km}}{{hr}} + \left( { - 1} \right)^n } \right] [/tex]

    *(Feel free to omit the units for mathematical clarity :smile:)
    I haven't bothered to simplify this (sorry:frown:), but the number of flights will be the value of [itex] n [/itex] for which
    [tex] t _ n = \frac{20}{11} \, {hr} [/tex]
  10. Aug 26, 2005 #9
    thanx people for all ur help but @ bomba923 the sequence u have given here is a recurrence relation(if I'm not wrong) but I don't know how to solve these sort of relations!(bcoz I haven't learnt it at school :cry: )If u could please solve for [tex]n[/tex] I would be really thankful to u!
    Best regards,
  11. Aug 26, 2005 #10


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    It's infinity times. Why? The eagle flies faster than both trains. Say it starts from train 1, flies to train 2. Train 1 and train 2 is [tex]\Delta s_1[/tex](m) apart. Because the eagle flies faster than train 1, it will meet train 2 before train 1 does. That means, when it reaches train 2, train 1 has not reached train 2 yet. And they are [tex]\Delta s_2[/tex](m) apart. And the flies flies back to train 1, it reaches train 1 before train 2. At the time it reaches train 1, the 2 trains are [tex]\Delta s_3[/tex](m) apart, and so on... The distance between the two trains decreases, but the eagle will never touch one of the two train at the exact time the other train touches it (the eagle is faster than 2 trains).
    So you notice:[tex]\Delta s_1 > \Delta s_2 > \Delta s_3 > \Delta s_4 > ... > \Delta s_n[/tex](m)
    The distance between two trains gets smaller and smaller, the eagle completes its loop faster.
    We have:
    [tex]\lim_{n \to \infty} \Delta s_n = 0[/tex]
    So the eagle touches the two train 'infinity' times.
    Viet Dao,
    Last edited: Aug 26, 2005
  12. Aug 26, 2005 #11
    Only if you assume the eagle has no size
    :biggrin: As Tide said :biggrin:
    :cool: Please don't squish the bird between the trains!
    Anyway, if you provide some specifications as to the eagle's size, you CAN set a finite number of flights in between.....that is, until the trains are so close as to squish the eagle, and he flies away.......
  13. Aug 27, 2005 #12


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    I, for one, suspect that he was speaking the truth. I know that when I was a hot shot grad student I could work out the infinite series and sum it in my head in about 30 seconds because I tried it. But I should mention that it was for a problem that was somewhat simpler. That is, the two trains went at the same speed. I remember it as 50mph trains, 100mph crow and a 100 mile track.

    One calculates how far the eagle flies to its first point of collision by dividing the distance by the combined speeds. This gives a new problem, which is simply the reduced distance with the eagle flying in the other direction. It really isn't that complicated.

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