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Path of an object at the vertex of an equilateral triangle

  1. Apr 26, 2015 #1
    1. The problem statement, all variables and given/known data

    There are three objects at the Vertices of an equilateral triangle that start movin towards each other at the same time with a speed v.

    Describe the path of the objects and the time taken for them to meet.

    2. Relevant equations

    V1=v3 - v2

    Where all velocities are in vectors.

    3. The attempt at a solution

    The points moving towards each other will intersect at the Center of the triangle.

    Velocity along the direction to the Center is vcos60 which will be he velocity at which he points move towards the Center.

    The distance to be covered is d/sqrt(3).

    This gives you the distance and time taken to cover it.

    However, I'm struggling to formulate the problem in terms of vectors.

    If point a moves towards point b by a distance dr (vector) in a time dt, I get stuck in trying to develop a vector equation for the path point a takes...

    I've been struggling with this for about an hour now, but seem to be missing a connection.

    Please help.

    Thanks!
     
  2. jcsd
  3. Apr 26, 2015 #2

    BvU

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    Hello chaos,

    Is that all the information you have available ? I mean, they start moving, so I take it that theinitial velocity is zero. But what about the cause of the motion ? Mutual attraction, r does someone/somethig give them a kick ? Do they start moving towards each other all with the same speed ?
     
  4. Apr 26, 2015 #3
    Sorry, they all move with constant and same speed.

    There is no force / acceleration at play.
     
  5. Apr 26, 2015 #4

    BvU

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    This the basic idea ? i.e. that e.g. a moves towards c with the same velocity as towards b (i.e. over the perpendicular in the middle of bc ?)

    upload_2015-4-26_21-16-28.png

    It should be clear that ##v_c \ne v_a - v_b##, so what is the meaning of the relevant equation ?
    Where does the ##v\cos(60^\circ)## come in ?
    "distance to be covered is ##d/\sqrt 3## " -- what is d ?
    What is the complete problem statement ?
     
  6. Apr 26, 2015 #5
    So each of the points moves along the edge of the triangle towards the next point.

    This, point a moves towards point b which moves towards point c.

    As they always move towards the next point with the same speed, they always form the vertices of an ever smaller equilateral triangle and converge at the Center.

    D would be the length of the side of the initial equilateral triangle.
     
  7. Apr 26, 2015 #6

    BvU

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    Ah, so we have a dog chasing hare(*) kind of thingy ! Like below ?
    upload_2015-4-26_21-35-51.png

    (*) actually, no: dog chasing hare lets te hare run straight -- yields a radiodrome, hefty mathematics
     
  8. Apr 26, 2015 #7
    Yes!

    Its the math of it that I'm trying to figure out.

    The thought process escapes me.

    I think I'll just have to break it down a lot more.

    Any approaches you'd suggest?
     
  9. Apr 26, 2015 #8

    BvU

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    You posted in introductory physics. What level of math are we at ? Do the differential equations in the link give you enough handle to set up such equations for the case at hand (hare path also curved ?)

    upload_2015-4-26_22-11-6.png

    Just so you are warned: I have no idea what comes out mathematically.
    Doing it numerically (this is excel ?:)) at least gives nice pictures :rolleyes:
     

    Attached Files:

  10. Apr 26, 2015 #9
    So, the math itself in the links is alright.

    I'd like to develop the equations themselves.

    I was getting stuck with the velocity vs path vector stuff. What route of analysis should I take?

    Also, how did you generate the graphs in excel?
     
  11. Apr 26, 2015 #10

    BvU

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    More or less by hand o0) .

    a,b,c are x and y positions. point a is repeated to close the triangle ab bc ca

    ##\vec v_a = \left [ \left (\vec b_x -\vec a_x \over |\vec b -\vec a| \right ) |v| , \left (\vec b_y -\vec a_y \over |\vec b -\vec a| \right ) |v| \right ] ## etc.
    and
    ##\vec a(t+\Delta t) = \vec a + \vec v_a \Delta t ##
     

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  12. Apr 26, 2015 #11

    haruspex

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    Are you sure? But as you say it is a constant, so you can write down the distance from the centre at time t. Can you do something similar for tangential motion and obtain a path in polar coordinates?
     
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