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Homework Help: Tough kinematics question

  1. Oct 31, 2003 #1

    joc

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    [POST CORRECTED]

    would greatly appreciate it if someone could help me with the following problem:

    2 point particles, A and B, are a distance d apart in a vacuum.

    from time t=0 onwards, A has a constant nonzero velocity with magnitude v and a direction perpendicular to AoBo, where Ao and Bo are the initial positions of A and B.

    from time t=0 onwards, B has a velocity of constant magnitude v which is always directed in direction BA (i.e. towards A).

    find the distance AB at time=infinity.

    (there are apparently relatively short ways to do this problem.)
     
    Last edited: Nov 2, 2003
  2. jcsd
  3. Oct 31, 2003 #2
    it's probably d
     
  4. Oct 31, 2003 #3
    Hi joc! Welcome to the Forums.

    You didn't show any work! Go to the homework section and read the Sticky post by Tom.

    Since you're new here, I'll give you a hint, but next time show your work.
    HINT:
    It helps to know that both object are accelerating.
     
  5. Oct 31, 2003 #4
    The problem states that neither object is accelerating.
     
  6. Oct 31, 2003 #5
    both have constant speed, both chage direction.... seems like acceleration to me
     
  7. Oct 31, 2003 #6
    They have constant velocity: neither one of them changes speed or direction.
     
  8. Oct 31, 2003 #7
    Apparently the speed of B is constant and its direction is changing as A moves. This is indeed acceleration.

    Similarly, the magnitude of velocity of A is constant (ie speed is constant) but its direction is changing.
     
    Last edited: Oct 31, 2003
  9. Oct 31, 2003 #8
    Re: Re: tough kinematics question

    Hmm, perhaps I misinterpreted the statement of the problem. I took "the line AB" to be the original line between the two particles.
     
  10. Oct 31, 2003 #9
    In that case, they are both moving in a straight line, and at t = infinity they would be an infinite distance apart. Not much of a problem.
     
  11. Oct 31, 2003 #10
    Yes, so I thought it was an odd problem... your interpretation is probably correct.
     
  12. Oct 31, 2003 #11

    turin

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    If AB is considered to be the line that seperates the two particles, and A only moves perpendicular to this line (as stated in the original post), then the motion of B is the only thing that changes the distance between particles (because the motion of A is perpendicular to their separation, and therefore cannot effect the separation).

    AB0 is the initial separation of A and B.

    If B moves with a constant speed v directly towards A, then, at time t = AB0/v, the separation between the particles will be zero. If both AB0 and v are finite, then the particles will have zero seperation at time t = infinity (assuming that B always moves TOWARDS A).

    I think the problem needs to be reworded.
     
  13. Nov 1, 2003 #12

    HallsofIvy

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    Yes that's true: "B has a velocity of constant magnitude v which is always directed in direction BA (i.e. towards A)."

    The problem says "A has a constant nonzero velocity with magnitude v and a direction perpendicular to the line AB." Since it specifically says "constant nonzero velocity" I would interpret that to mean A travels along a straight line perpendicular to the line between the ORIGINAL positions of A and B.
     
  14. Nov 1, 2003 #13

    russ_watters

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    I think its poorly worded (maybe on purpose) but here's what appears to be going on:

    A is moving at t=0 perpendicular to the line AB and continues in that direction with constant velocity (no change in speed or direction).

    B is doing a "tail chase" which takes it on a curved (hyperbolic I think) path which by time infinity approaches being directly behind A.

    You could construct an equation for the velocity component on B perpendicular to the original line AB, integrate for distance and subtract the distance A has traveled to find out how far apart they are.
     
  15. Nov 1, 2003 #14

    jcsd

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    The line AA0 (A is A's current postion, A0 is A's intial psotion) is always vt, B's sepration from A (AB) is equal to [A0B0x2 + (AA0 - y)2]1/2 (x is B's component of separation from B0 in the direction parallel to A0B0 and y is B's component of separation from B0 in the direction paralell to AA0.

    Due to the nature of the curve at t = infinity x = A0B0 and AB = y => (from the equations in the last paragraph)

    AB = vt/2, which is an infinte separation.

    Anyway that's my first attempt, I think though that I've either made a mistake or a faulty assumption (partly becasue I was expecting the answer to be finite and because I haven't really taken into account the fact that B's magnitude of velocity is always v along AB).
     
    Last edited: Nov 1, 2003
  16. Nov 2, 2003 #15

    krab

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    OK. I'll generously take the interpretation of this sloppily-worded problem that will make it the most interesting: both particles have velocity of fixed magnitude, but varying direction.

    Particles A and B are initially side-by-side. Take a small time step. B has moved a little toward A, A has moved a little up, thus slightly tilting the line joining the two. So tilt your perspective a bit so the line is no longer tilted. So we are still in the same condition as at t=0, except the particles are a little closer together. Obviously then, at t=infinity, the distance AB is zero.
     
  17. Nov 2, 2003 #16

    joc

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    my apologies to all. made a mistake in the wording; the question is actually as russ_watters and HallsofIvy have explained it:

    A has a constant velocity (no change in magnitude or direction) which is directed perpendicular to the AoBo (the line segment AB at t=0).

    B has a varying velocity with constant magnitude that is directed along line segment BA, B and A representing, for lack of a simpler phrasing, the current positions of the particles.

    my attempts at working haven't come close to yielding any insights or answers so i didn't post them.

    anyway, the answer is finite. btw russ, i've tried doing as you suggested but i have to involve an angle in order to obtain the components of velocity; i can't seem to get rid of the angle by expressing it in terms of time. seems like differential equations hold the answer...
     
  18. Nov 2, 2003 #17
    Okay, the question as stated is a standard pursuit curve problem. MathWorld has a derivation,

    http://mathworld.wolfram.com/PursuitCurve.html

    -- the trick I didn't think of is eliminating the explicit time dependence by substituting the arc length integral -- but it looks to me that there is an error in their derivation. You can also try here:

    http://www.math.rutgers.edu/pub/nwkmath/dupre/pursuit.ps [Broken]
     
    Last edited by a moderator: May 1, 2017
  19. Nov 2, 2003 #18

    turin

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    If the magnitude of the velocity of A is greater than or equal to the magnitude of the velocity of B, then B will never catch A.
     
  20. Nov 3, 2003 #19

    joc

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    turin: the magnitudes are constant and equal. i'm looking for the distance between the 2 at time=infinity.
     
  21. Nov 3, 2003 #20

    turin

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    OK, I think you will need to solve the differential equation.

    Let the initial position of A be (xAo,yAo) = (0,0).

    Let the initial position of B be (xBo,yBo) = -(AoBo,0).


    The equation of motion for A is simply: (xA,yA) = (0,vt), where v is the magnitude of the velocity.

    B is a little bit more complicated.
    There are two equations that come to my mind: (dxB/dt)^2 + (dyB/dt)^2 = v^2,
    and
    (dxB/dt)/(dyB/dt) = (yA - yB)/(xA - xB)
     
    Last edited: Nov 3, 2003
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