Math Problem: Solving a Triangular Chase with Constant Velocity

  • Thread starter netstrider
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In summary, three people - Bill, Jack, and Mike - are located at equal distances from each other (200 meters) and start chasing each other at the same time with a velocity of 10 m/s. They all travel the same distance until they meet at the center of an equilateral triangle. The distance traveled by each person can be calculated using the radius of a circle arc with its center located at a point on an axis perpendicular to the side of the original triangle. The angle traveled on the arc is 60 degrees or pi/3, resulting in a distance of pi*sqrt(2)/9*200m. The three people will always be traveling towards each other in a direction determined by the circle arc, and the size
  • #1
netstrider
3
0
Math problem. Solve it please! :)

Bill, Jack and Mike, each two of them are at the same distance (200 meters). At the same time Bill starts to chase Jack, Jack starts to chase Mike, and,.. Mike starts to chase Bill. Each has the velocity of 10 m/s, which remains unchainged till they meet in the center of the triangle. The question is what is the distance any of them passed (they all pass the same distance)? I could solve the problem if I knew the function of their movement (any of the 3). It looks something hyperbolic. I need the whole solution to the problems. Please reply.Tnx in advance

netstrider
 
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  • #2
technically their velocities can't remain unchanged, or they'd go off in a straight line.

it suffices to find the time elapsed, if that's any help.
 
  • #3
I meant speed, of course that direction chages. Sorry for bad english.
 
  • #4
The situation for each of them is identical, so they will all travel the same distance. Their speed also doesn't matter in this question.

The three people are located on the sides of an equal sided triangle. The curve they travel on is a circle arc with as origin a point on an axis going through the starting point of the triangle and perpendicular to the side of the original triangle, as by this means the tangent at this circle arc will be the side of the triangle and hence point to the person on the point of the equal sided triangle. Now we now on what axis the circle's center is located, we also know one point on the circle is the starting point of the person and a second point in the center.

This allows us to calculate the radius, which is sqrt(2)/3*(200m) and the angle traveled on the arc 60° or pi/3, making the distance traveled equal to pi*sqrt(2)/9*(200m). As the three people move along this circle arc, they will always be traveling in a direction towards each other and meet each other in the center, the size of the triangle they form will maintain its equal sided shape and gradually become smaller. I guess that answers it fully.
 
  • #5
Simon666 said:
The curve they travel on is a circle arc with as origin a point on an axis going through the starting point of the triangle and perpendicular to the side of the original triangle ...

Ok if I take this as a starting point I can solve the problem, but can you give us any clues Simon as to how you deduced the above quoted material ?


PS. Interesting problem netstrider. :)
 
  • #6
Simon666 said:
. The curve they travel on is a circle arc with as origin a point on an axis going through the starting point of the triangle and perpendicular to the side of the original triangle, as by this means the tangent at this circle arc will be the side of the triangle and hence point to the person on the point of the equal sided triangle. the center, the size of the triangle they form will maintain its equal sided shape and gradually become smaller. I guess that answers it fully.

I cannot agree with this, Simon.
Here's my solution:

1. By symmetry, the three persons wil be equally distanced from the midpoint;
let that distance be r(t) satisfying [tex]r(0)=r_{0}[/tex]

2. Also by symmetry, each person wil have increased his position vector's angle to some fixed axis by the the same amount in the same time.

3. Clearly, the angle can be related to the distance; in particular, we may write the angle as a function of position (we'll allow for possibly more than one cycle about the midpoint

4. Hence, the position vector to i'th person may be written as:
[tex]\vec{r}_{i}(t)=r(t)(\cos(\theta_{i}+\theta(r))\vec{i}+ \sin(\theta_{i}+\theta(r))\vec{j})=r(t)\vec{i}_{i,1}, \theta(r_{0})=0, [/tex]
[tex]\vec{i}_{\theta,i}\cdot\vec{i}_{r,i}=0[/tex]

[tex]\theta_{i}[/tex] has the interpretation of the initial angle the i'th person's position vector makes with the positive x-axis

5. The velocity of person 1, for example, must satisfy the equation:
[tex]\dot{r}\vec{i}_{r,1}+r\dot{r}\frac{d\theta}{dr}\vec{i}_{\theta,1}=V_{0}\frac{\vec{i}_{r,2}-\vec{i}_{r,1}}{2|\sin(\frac{\theta_{1}-\theta_{2}}{2})|}[/tex]
where [tex]\dot{r}[/tex] is the temporal derivative of r(t) and [tex]V_{0}[/tex] the constant speed.

6. Taking the dot product between the equation above and [tex]\vec{i}_{r,1}[/tex]
yields the differential equation for r(t):
[tex]\dot{r}=-V_{0}|\sin(\frac{\theta_{1}-\theta_{2}}{2})|[/tex]

or:
[tex]r(t)=r_{0}-V_{0}t|\sin(\frac{\theta_{1}-\theta_{2}}{2})|, r(T)=0\rightarrow[/tex]
[tex]{T}=\frac{r_{0}}{V_{0}|\sin(\frac{\theta_{1}-\theta_{2}}{2})|}=\frac{2r_{0}}{V_{0}\sqrt{3}}[/tex]

The last equality follows from the observation that the angle between 2 persons are initially 120 degrees on an equilateral triangle.

7. It can be shown that each person's path is a logarithmic spiral into the origin.
 
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  • #7
uart said:
Ok if I take this as a starting point I can solve the problem, but can you give us any clues Simon as to how you deduced the above quoted material ?
I took a guess and on second thought I am wrong. Posing it like that may however give a clue as how to solve the damn thing.
 
  • #8
Question, isn't in 4 your vector i depending on t since r is and hence your future derivation incorrect? Exactly how you get from 5) to 6) isn't clear to me either.
 
  • #9
I'm sorry: The vectors [tex] \vec{i},\vec{j}[/tex] are constant vectors, wheras in 4, I have a typo:
The position vector should read: [tex]\vec{r}_{i}(t)=r(t)\vec{i}_{r,i}[/tex]

The vector [tex]\vec{i}_{r,i}[/tex] is certainly a function of time; the derivative is:
[tex]\frac{d\vec{i}_{r,i}}{dt}=\vec{i}_{\theta,i}\frac{d\theta}{dr}\frac{dr}{dt}[/tex]

by the chain rule.
([tex]\vec{i}_{\theta,i}[/tex] is orthogonal on [tex]\vec{i}_{r,i}[/tex])
 
  • #10
As for step 5 to 6:

The unit vector that person 1's velocity is directed along, is:
[tex]\frac{\vec{i}_{r,2}-\vec{i}_{r,1}}{||\vec{i}_{r,2}-\vec{i}_{r,1}||}[/tex]

[tex]||\vec{i}_{r,2}-\vec{i}_{r,1}||=\sqrt{(\vec{i}_{r,2}-\vec{i}_{r,1})\cdot(\vec{i}_{r,2}-\vec{i}_{r,1})}=\sqrt{2(1-\cos\gamma)}[/tex]

[tex]\gamma=\theta_{1}+\theta(r)-(\theta_{2}+\theta(r))=\theta_{1}-\theta_{2}[/tex]

[tex]1-\cos\gamma=2\sin^{2}(\frac{\gamma}{2})[/tex]
 
  • #11
A number of your steps are not quite clear to me and I also fail to see why the size of the velocity vector matters in any way to the form of the path.
 
  • #12
The reason why the velocity is important can be seen by considering the 2-person analogy:
(That is each start at one end of a rod)
Clearly, the time taken to the midpoint is: [tex]T=\frac{L}{2V_{0}}[/tex]

where L is the length of the rod.
The n-person case, where all starts at the periphery of a circle of radius r, and where the angles between them is constant,
[tex]\delta\theta=\frac{360}{n}[/tex]
satisfies the relation
[tex]T=\frac{r}{V_{0}|\sin(\frac{180}{n})|}[/tex]

The 3-person case is merely a special case
 
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  • #13
The time it takes to get to the midpoint isn't asked. And I see neither why it matter in order to obtain the curve of the path, only when the time it takes them to get to the center was asked it would seem necessary to me, and with your transition from step 5 to step 6 I meant how making that cross product results in the equation in 6.
 
  • #14
However, you are absolutely correct in stating that the size of the speed is irrelevant for the FORM of the curve (it is a logarithmic spiral!), but the time to traverse the curve is dependent on the speed.
 
  • #15
I have no cross products anywhere!
 
  • #16
arildno said:
However, you are absolutely correct in stating that the size of the speed is irrelevant for the FORM of the curve (it is a logarithmic spiral!), but the time to traverse the curve is dependent on the speed.
I believe this is a case I have encountered once in an exam, hastily making all kinds of calculations that seem very beautiful and probably correct but that have little to do with the answer asked, you gave the time needed to get to the center while the question explicitly specified the distance was asked.
 
  • #17
What's the difference between the midpoint and the centre of the triangle?
 
  • #18
arildno said:
I have no cross products anywhere!
My mistake, dot product, getting late here, getting sleepy.
 
  • #19
Multiply speed with time taken. Period.
 
  • #20
arildno said:
What's the difference between the midpoint and the centre of the triangle?
Aren't both the same, don't lookt at me, I'm not native English but a dumb Belgian. :confused:
 
  • #21
arildno said:
Multiply speed with time taken. Period.
I think I should have kept my mouth instead of sounding increasingly stupid, anyway, could you help me a hand with my last unsolved question?
 
  • #22
I've glanced at it earlier; now, it's getting a bit late, so I'll check it tomorrow..
 
  • #23
:smile: Thanks. Could you also explain how the dot product yields the result in 6, still don't understand that?
 
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  • #24
All right, dot product:
We have:
[tex]\vec{i}_{r,1}=\cos(\theta_{1}+\theta(r))\vec{i}+ \sin(\theta_{1}+\theta(r))\vec{j}[/tex]
[tex]\vec{i}_{r,2}=\cos(\theta_{2}+\theta(r))\vec{i}+ \sin(\theta_{2}+\theta(r))\vec{j}[/tex]

The dot product:
[tex]\vec{i}_{r,1}\cdot\vec{i}_{r,2}=\cos(\theta_{1}+\theta(r))\cos(\theta_{2}+\theta(r))+\sin(\theta_{1}+\theta(r))\sin(\theta_{2}+\theta(r))=[/tex]
[tex]\cos(\theta_{1}+\theta(r)-(\theta_{2}+\theta(r)))=\cos(\theta_{1}-\theta_{2})[/tex]

By the half-angle formula, we have:
[tex]1-\cos(\theta_{1}-\theta_{2})=2\sin^{2}(\frac{\theta_{1}-\theta_{2}}{2})[/tex]

Combining these expressions yields the transition from 5) to 6)

I'll answer to your own thread where it is posted
 
  • #25
If you take arildno's solution for T and put in r0 = a/sqrt(3), you find that the distance traveled is 2a/3 where a is your 200 m. Of course, the solution is independent of v0.

Here's another way to solve this problem - using relative velocities.
Imagine yourself as one of the guys, say Bill, who is chasing Jack. You approach Jack with a velocity of vo.
Also Jack's velocity has a component in your direction given by v0*cos60 or v0/2.
So, his velocity relative to you in your direction is the sum of these, namely 3v0/2.
The relative distance he travels is 'a'. So the time taken is 2a/3v0.
Therefore, the distance traveled is 2a/3.
 
  • #26
Great solution Gokul!
Much simpler than mine :smile:
 
  • #27
arildno said:
7. It can be shown that each person's path is a logarithmic spiral into the origin.

Hi there. Sorry to trawl this post up after, what, three years I think? I came across this problem recently and found it very interesting. Martin Gardner has an elegant solution (in an "Aha!" book) for what he calls the "Four Turtle Problem". I've read various other geometrical arguments that use careful considerations of symmetry etc. to avoid using calculus.

Anyway, those solutions are fine, but now I'm pulling my hair out trying to parameterise a turtle's/person's motion w.r.t time. It's getting me really frustrated because it seems to me it should be a really simple problem, yet I can't for the life of me seem to do this! Well, polar or Cartesian I'm not too fussed...either way I'd really appreciate it if someone could help me draw up the necessary D.E's (e.g. I suppose dr(t)/dt and then a relation to dtheta(t)/dt?) and then from there I'll be able to solve the problem...

Thank you!
 
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  • #28
  • #30
Actually it's interesting to see how easy it is to generalise when we change reference frames, especially seeing how hard the problem is even for a triangle if we don't.

Suppose we have n ligers on the sides of a regular n-gon, and each moves towards the one in front of it with constant speed v.

Each angle in a regular n-gon is (n-2)*pi/n=pi-2*pi/n.
From the point of view of one of the ligers the liger in front and the liger behind is coming at constant speed.
This speed is the speed v the liger is traveling plus v*cos(pi-2*pi/n), the parallel component of the liger in front.

So if they are initially a distance a apart, they meet after
[tex]t=\frac{a}{v(1-\cos(\frac{2\pi}{n}))}[/tex]
So they each travel a distance
[tex]\frac{a}{1-\cos(\frac{2\pi}{n})}[/tex]

In fact we can generalise slightly more - if the speed is a function of separation between the ligers (when they get the smell of another liger in their nostrils they start sprinting) and time (accelerating), then if l is the seperation
[tex]\frac{dl}{dt}=\frac{a}{v(l,t)(1-\cos(\frac{2\pi}{n}))}[/tex]
So solving this differential equation yields the time, and can be subsequently used to find the distance traveled.

If you use an asymmetric arrangement (e.g. irregular polygon) it is not as apparent, since the shape can change as a function of time.
 
  • #31
Thanks for the reply! just what i was looking for
 

1. How do you solve a triangular chase with constant velocity?

To solve a triangular chase with constant velocity, you first need to determine the initial positions and velocities of all the objects involved. Then, you can use the equations of motion to calculate the time it takes for the objects to reach a certain point and the distance they travel during that time. Finally, you can use trigonometry to determine the angles and distances between the objects at any given time.

2. What are the equations of motion used in solving a triangular chase?

The equations of motion used in solving a triangular chase are the equations for constant velocity, which are:
- s = ut + 1/2at^2 (where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time)
- v = u + at (where v is the final velocity)
- s = 1/2(u+v)t (where s is the distance traveled, u is the initial velocity, v is the final velocity, and t is the time)

3. How does trigonometry help in solving a triangular chase?

Trigonometry helps in solving a triangular chase by allowing us to calculate the angles and distances between the objects at any given time. This is important in determining the relative positions and movements of the objects in the chase. Trigonometric functions such as sine, cosine, and tangent are used to calculate these values.

4. What are some real-life applications of solving a triangular chase with constant velocity?

Solving a triangular chase with constant velocity has many real-life applications, such as in sports where players are chasing after a ball or in navigation where objects are moving towards a specific destination. It is also used in video games and simulations to create realistic movements and interactions between objects.

5. What are some common challenges in solving a triangular chase with constant velocity?

Some common challenges in solving a triangular chase with constant velocity include accurately determining the initial positions and velocities of the objects, accounting for any external factors that may affect the motion, and dealing with more complex scenarios where the objects may have varying velocities or accelerations. It is also important to double-check the calculations and equations used to avoid errors in the final solution.

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