Tough kinematics question

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  • #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.)
 
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Answers and Replies

  • #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.
 
  • #4
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The problem states that neither object is accelerating.
 
  • #5
both have constant speed, both chage direction.... seems like acceleration to me
 
  • #6
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They have constant velocity: neither one of them changes speed or direction.
 
  • #7
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Originally posted by joc

from time t=0 onwards, B has a velocity of constant magnitude v which is always directed in direction BA (i.e. towards A).
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.
 
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  • #8
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Hmm, perhaps I misinterpreted the statement of the problem. I took "the line AB" to be the original line between the two particles.
 
  • #9
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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.
 
  • #10
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Yes, so I thought it was an odd problem... your interpretation is probably correct.
 
  • #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.
 
  • #12
HallsofIvy
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Apparently the speed of B is constant and its direction is changing as A moves. This is indeed acceleration.
Yes that's true: "B has a velocity of constant magnitude v which is always directed in direction BA (i.e. towards A)."

Similarly, the magnitude of velocity of A is constant (ie speed is constant) but its direction is changing.
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.
 
  • #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.
 
  • #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).
 
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  • #15
krab
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Originally posted by joc
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 the line AB.

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.)
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.
 
  • #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...
 
  • #17
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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]
 
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  • #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.
 
  • #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.
 
  • #20
turin
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Originally posted by joc
turin: the magnitudes are constant and equal. i'm looking for the distance between the 2 at time=infinity.
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)
 
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  • #22
turin
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I'm going to have to read through that webpage. I was thinking that this could not be solved analytically.
 
  • #23
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Originally posted by joc
(there are apparently relatively short ways to do this problem.)
Yeah...one should think so, shouldn't one? After all, it's a very special version of a pursuit curve.

Let me try from scratch.
If we let particle A start at the origin, and let it move up the y-axis (with velocity v), then particle B starts on the x-axis (say, at x = -d). Now B moves to the topright in a curve, and the curve's tangent is directed straight toward A's position, right?

Let B = (x,y), then this means
y - xy' = vt.
Now, since B has the same speed as A, vt is also the arclength of the curve:
y - xy' = integral √ (1+y'2)

Differentiate:
y' - y' - xy'' = √ (1+y'2)

y'' = -[√ (1+y'2)]/x.

Now that's at least a DEQ, but I don't know how to continue. Any DEQ experts?

(Edit: √ sign)
 
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  • #24
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Ah! It separates:
y''/√ (1+y'2) = -1/x.

I think this integrates to:
√ (1+y'2) - 1 = -ln(x/-d)

So we have
y' = √ [(1-ln(x/-d))2-1]

Anyone can integrate that? I can't. Or is there some error?

Edit: some corrections.
 
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  • #25
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Today I found some time to solve this.

Originally posted by arcnets
If we let particle A start at the origin, and let it move up the y-axis (with velocity v), then particle B starts on the x-axis (say, at x = -d). Now B moves to the topright in a curve, and the curve's tangent is directed straight toward A's position, right?

Let B = (x,y), then this means
y - xy' = vt.
Now, since B has the same speed as A, vt is also the arclength of the curve:
y - xy' = integral √ (1+y'2)

Differentiate:
y' - y' - xy'' = √ (1+y'2)

y'' = -[√ (1+y'2)]/x.
Or, y''/√ (1+y'2) = -1/x.

In the other post I made a mistake. Because the integral of this is:
ln(y' + √(1+y'2)) = ln(d/x),
thus
√(1+y'2) = d/x - y',
so
y' = (d/x - x/d)/2.

Thus, the pursuit curve's equation is
y = 1/2*ln(x/d) - x2/4d + d/4.

We also see that
√(1+y'2) = d/x - y' = (d/x + x/d)2.

Thus, the arclength is
L = integral √(1+y'2) dx
= integral (d/x + x/d)/2 dx
= 1/2*ln(x/d) + x2/4d - d/4.

Thus, the vertical distance between A and B is
y - vt = y - L = d/2 - x2/2d,
which approaches d/2 as x approaches zero.

So the answer is d/2.
 

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