# Tough kinematics question

#### joc

[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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it's probably d

#### StephenPrivitera

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.

#### Ambitwistor

The problem states that neither object is accelerating.

#### Guybrush Threepwood

both have constant speed, both chage direction.... seems like acceleration to me

#### Ambitwistor

They have constant velocity: neither one of them changes speed or direction.

#### physics247

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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#### Ambitwistor

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.

#### physics247

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.

#### Ambitwistor

Yes, so I thought it was an odd problem... your interpretation is probably correct.

#### turin

Homework Helper
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.

#### HallsofIvy

Homework Helper
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.

#### russ_watters

Mentor
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.

#### jcsd

Gold Member
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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#### krab

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.

#### joc

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...

#### Ambitwistor

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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#### turin

Homework Helper
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.

#### joc

turin: the magnitudes are constant and equal. i'm looking for the distance between the 2 at time=infinity.

#### turin

Homework Helper
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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#### turin

Homework Helper
I'm going to have to read through that webpage. I was thinking that this could not be solved analytically.

#### arcnets

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 &radic; (1+y'2)

Differentiate:
y' - y' - xy'' = &radic; (1+y'2)

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

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#### arcnets

Re: Re: tough kinematics question

Ah! It separates:

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

So we have

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

Edit: some corrections.

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#### arcnets

Re: Re: tough kinematics question

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 &radic; (1+y'2)

Differentiate:
y' - y' - xy'' = &radic; (1+y'2)

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

Thus, the arclength is
= 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.

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