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Three points at the vertices of equilateral triangle

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  1. Jul 12, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-7-12_23-44-19.png
    2. Relevant equations


    3. The attempt at a solution
    The direction of motion of all the three particles are changing w.r.t. lab frame.
    But , their relative velocity remains constant.
    relative velocity = vb - va

    And the distance between the two particles decreases from a to 0.

    So, t = a/|vb - va| I am not clear why I am taking this step. I just feel tempted to take it.
    |vb - va| = (√3)v

    t = a/(√3)v

    W.r.t. A's frame of reference , B is always moving at an angle 60° to the line joining A and B. Then, how can the two ever meet?


    After seeing the answer (which is 2a/3v)in the book,

    Due to the symmetry of the problem, the three will converge at the centre of the triangle.

    The distance between the center and the vertices is a/√3.

    This is the magnitude of the displacement traveled by each point.

    So, taking t = (a/√3)/ [|vb - va| = (√3)v]
    = a/3v

    Even now, I am missing a factor of 2.

    I know that the relative velocity is constant and the displacement of each particle but I can't connect the two information.
     
  2. jcsd
  3. Jul 12, 2017 #2
    For each unit you travel along your path, how far towards the center do you travel?
    That ratio needs to be multiplied by your a/sqrt(3) to get the total path length.
     
  4. Jul 12, 2017 #3
    Each point is spiraling in towards the center. And you need to know the length of that spiral path.
    At every moment in time, the rate of descent towards the center is proportional to the rate along the spiral. That ratio can be determined by comparing its path to a path that simply orbits the center in a circular motion. The spiral and the circle will meet at a particular angle. What is that angle?
    Since the rate of descent is always proportional to the distance traveled, the length of that spiral is proportional to the initial radius r=a/√3.
     
  5. Jul 12, 2017 #4

    haruspex

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    Right. Consider two particles. Think about components of their relative velocity. Which component is of interest?
    No, it's easier than that.
     
  6. Jul 12, 2017 #5
    W.r.t. A's frame of reference ,

    B is always moving at an angle 60° to the line joining A and B.
    For meeting A, B travels along BA a distance 'a' with speed 3v/2.
    Hence, t = 2a/3v

    So, the problem is solved.
    But, I did't understand how B can meet A as B is also moving perpendicular to BA with speed (√3)v/2. Please explain this.
     
  7. Jul 13, 2017 #6

    haruspex

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    in each short interval of time, dt, the movement that B makes perpendicular to AB does not alter the distance between them. A cancels the effect of that movement merely by changing direction slightly.
     
  8. Jul 21, 2017 #7
    This I didn't understand.

    W.r.t. A's frame,
    B moves towards A along BA and away from A along perpendicular to AB. (Please, correct it if this sentence is not proper grammatically.)
    Thus, the motion in perpendicular direction always remains away from A. So, how could B meet A?
     
  9. Jul 21, 2017 #8

    haruspex

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    A very small movement perpendicular to AB is not away from A. It is neutral.
    If you stand still at A and person B moves always perpendicular to AB then that person goes around in a circle, always at the same radius from you.
     
  10. Jul 21, 2017 #9
    O.K.
    Now I understood that the perpendicular motion doesn't change the distance and the motion towards A along BA reduces the distance. This is how B meets A.

    Thank you, haruspex. Thanks a lot.
     
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