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Math problem. Solve it !

  1. May 18, 2004 #1
    Math problem. Solve it plz! :)

    Bill, Jack and Mike, each two of them are at the same distance (200 meters). At the same time Bill starts to chase Jack, Jack starts to chase Mike, and,.. Mike starts to chase Bill. Each has the velocity of 10 m/s, which remains unchainged till they meet in the center of the triangle. The question is what is the distance any of them passed (they all pass the same distance)? I could solve the problem if I knew the function of their movement (any of the 3). It looks something hyperbolic. I need the whole solution to the problems. Please reply.Tnx in advance

    netstrider
     
  2. jcsd
  3. May 18, 2004 #2

    matt grime

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    technically their velocities can't remain unchanged, or they'd go off in a straight line.

    it suffices to find the time elapsed, if that's any help.
     
  4. May 18, 2004 #3
    I meant speed, of course that direction chages. Sorry for bad english.
     
  5. May 19, 2004 #4
    The situation for each of them is identical, so they will all travel the same distance. Their speed also doesn't matter in this question.

    The three people are located on the sides of an equal sided triangle. The curve they travel on is a circle arc with as origin a point on an axis going through the starting point of the triangle and perpendicular to the side of the original triangle, as by this means the tangent at this circle arc will be the side of the triangle and hence point to the person on the point of the equal sided triangle. Now we now on what axis the circle's center is located, we also know one point on the circle is the starting point of the person and a second point in the center.

    This allows us to calculate the radius, which is sqrt(2)/3*(200m) and the angle traveled on the arc 60° or pi/3, making the distance traveled equal to pi*sqrt(2)/9*(200m). As the three people move along this circle arc, they will always be travelling in a direction towards each other and meet each other in the center, the size of the triangle they form will maintain its equal sided shape and gradually become smaller. I guess that answers it fully.
     
  6. May 19, 2004 #5

    uart

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    Ok if I take this as a starting point I can solve the problem, but can you give us any clues Simon as to how you deduced the above quoted material ?


    PS. Interesting problem netstrider. :)
     
  7. May 19, 2004 #6

    arildno

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    I cannot agree with this, Simon.
    Here's my solution:

    1. By symmetry, the three persons wil be equally distanced from the midpoint;
    let that distance be r(t) satisfying [tex]r(0)=r_{0}[/tex]

    2. Also by symmetry, each person wil have increased his position vector's angle to some fixed axis by the the same amount in the same time.

    3. Clearly, the angle can be related to the distance; in particular, we may write the angle as a function of position (we'll allow for possibly more than one cycle about the midpoint

    4. Hence, the position vector to i'th person may be written as:
    [tex]\vec{r}_{i}(t)=r(t)(\cos(\theta_{i}+\theta(r))\vec{i}+ \sin(\theta_{i}+\theta(r))\vec{j})=r(t)\vec{i}_{i,1}, \theta(r_{0})=0, [/tex]
    [tex]\vec{i}_{\theta,i}\cdot\vec{i}_{r,i}=0[/tex]

    [tex]\theta_{i}[/tex] has the interpretation of the initial angle the i'th person's position vector makes with the positive x-axis

    5. The velocity of person 1, for example, must satisfy the equation:
    [tex]\dot{r}\vec{i}_{r,1}+r\dot{r}\frac{d\theta}{dr}\vec{i}_{\theta,1}=V_{0}\frac{\vec{i}_{r,2}-\vec{i}_{r,1}}{2|\sin(\frac{\theta_{1}-\theta_{2}}{2})|}[/tex]
    where [tex]\dot{r}[/tex] is the temporal derivative of r(t) and [tex]V_{0}[/tex] the constant speed.

    6. Taking the dot product between the equation above and [tex]\vec{i}_{r,1}[/tex]
    yields the differential equation for r(t):
    [tex]\dot{r}=-V_{0}|\sin(\frac{\theta_{1}-\theta_{2}}{2})|[/tex]

    or:
    [tex]r(t)=r_{0}-V_{0}t|\sin(\frac{\theta_{1}-\theta_{2}}{2})|, r(T)=0\rightarrow[/tex]
    [tex]{T}=\frac{r_{0}}{V_{0}|\sin(\frac{\theta_{1}-\theta_{2}}{2})|}=\frac{2r_{0}}{V_{0}\sqrt{3}}[/tex]

    The last equality follows from the observation that the angle between 2 persons are initially 120 degrees on an equilateral triangle.

    7. It can be shown that each person's path is a logarithmic spiral into the origin.
     
    Last edited: May 19, 2004
  8. May 19, 2004 #7
    I took a guess and on second thought I am wrong. Posing it like that may however give a clue as how to solve the damn thing.
     
  9. May 19, 2004 #8
    Question, isn't in 4 your vector i depending on t since r is and hence your future derivation incorrect? Exactly how you get from 5) to 6) isn't clear to me either.
     
  10. May 19, 2004 #9

    arildno

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    I'm sorry: The vectors [tex] \vec{i},\vec{j}[/tex] are constant vectors, wheras in 4, I have a typo:
    The position vector should read: [tex]\vec{r}_{i}(t)=r(t)\vec{i}_{r,i}[/tex]

    The vector [tex]\vec{i}_{r,i}[/tex] is certainly a function of time; the derivative is:
    [tex]\frac{d\vec{i}_{r,i}}{dt}=\vec{i}_{\theta,i}\frac{d\theta}{dr}\frac{dr}{dt}[/tex]

    by the chain rule.
    ([tex]\vec{i}_{\theta,i}[/tex] is orthogonal on [tex]\vec{i}_{r,i}[/tex])
     
  11. May 19, 2004 #10

    arildno

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    As for step 5 to 6:

    The unit vector that person 1's velocity is directed along, is:
    [tex]\frac{\vec{i}_{r,2}-\vec{i}_{r,1}}{||\vec{i}_{r,2}-\vec{i}_{r,1}||}[/tex]

    [tex]||\vec{i}_{r,2}-\vec{i}_{r,1}||=\sqrt{(\vec{i}_{r,2}-\vec{i}_{r,1})\cdot(\vec{i}_{r,2}-\vec{i}_{r,1})}=\sqrt{2(1-\cos\gamma)}[/tex]

    [tex]\gamma=\theta_{1}+\theta(r)-(\theta_{2}+\theta(r))=\theta_{1}-\theta_{2}[/tex]

    [tex]1-\cos\gamma=2\sin^{2}(\frac{\gamma}{2})[/tex]
     
  12. May 19, 2004 #11
    A number of your steps are not quite clear to me and I also fail to see why the size of the velocity vector matters in any way to the form of the path.
     
  13. May 19, 2004 #12

    arildno

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    The reason why the velocity is important can be seen by considering the 2-person analogy:
    (That is each start at one end of a rod)
    Clearly, the time taken to the midpoint is: [tex]T=\frac{L}{2V_{0}}[/tex]

    where L is the length of the rod.
    The n-person case, where all starts at the periphery of a circle of radius r, and where the angles between them is constant,
    [tex]\delta\theta=\frac{360}{n}[/tex]
    satisfies the relation
    [tex]T=\frac{r}{V_{0}|\sin(\frac{180}{n})|}[/tex]

    The 3-person case is merely a special case
     
    Last edited: May 19, 2004
  14. May 19, 2004 #13
    The time it takes to get to the midpoint isn't asked. And I see neither why it matter in order to obtain the curve of the path, only when the time it takes them to get to the center was asked it would seem necessary to me, and with your transition from step 5 to step 6 I meant how making that cross product results in the equation in 6.
     
  15. May 19, 2004 #14

    arildno

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    However, you are absolutely correct in stating that the size of the speed is irrelevent for the FORM of the curve (it is a logarithmic spiral!), but the time to traverse the curve is dependent on the speed.
     
  16. May 19, 2004 #15

    arildno

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    I have no cross products anywhere!!
     
  17. May 19, 2004 #16
    I believe this is a case I have encountered once in an exam, hastily making all kinds of calculations that seem very beautiful and probably correct but that have little to do with the answer asked, you gave the time needed to get to the center while the question explicitly specified the distance was asked.
     
  18. May 19, 2004 #17

    arildno

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    What's the difference between the midpoint and the centre of the triangle?
     
  19. May 19, 2004 #18
    My mistake, dot product, getting late here, getting sleepy.
     
  20. May 19, 2004 #19

    arildno

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    Multiply speed with time taken. Period.
     
  21. May 19, 2004 #20
    Aren't both the same, don't lookt at me, I'm not native English but a dumb Belgian. :confused:
     
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