Path of an object at the vertex of an equilateral triangle

In summary: I'm sorry, but I can't help you with the math.Just so you are warned: I have no idea what comes out mathematically.Doing it numerically (this is excel ?:)) at least gives nice pictures :rolleyes:
  • #1
chaoseverlasting
1,050
3

Homework Statement



There are three objects at the Vertices of an equilateral triangle that start movin towards each other at the same time with a speed v.

Describe the path of the objects and the time taken for them to meet.

Homework Equations



V1=v3 - v2

Where all velocities are in vectors.

The Attempt at a Solution



The points moving towards each other will intersect at the Center of the triangle.

Velocity along the direction to the Center is vcos60 which will be he velocity at which he points move towards the Center.

The distance to be covered is d/sqrt(3).

This gives you the distance and time taken to cover it.

However, I'm struggling to formulate the problem in terms of vectors.

If point a moves towards point b by a distance dr (vector) in a time dt, I get stuck in trying to develop a vector equation for the path point a takes...

I've been struggling with this for about an hour now, but seem to be missing a connection.

Please help.

Thanks!
 
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  • #2
Hello chaos,

Is that all the information you have available ? I mean, they start moving, so I take it that theinitial velocity is zero. But what about the cause of the motion ? Mutual attraction, r does someone/somethig give them a kick ? Do they start moving towards each other all with the same speed ?
 
  • #3
Sorry, they all move with constant and same speed.

There is no force / acceleration at play.
 
  • #4
This the basic idea ? i.e. that e.g. a moves towards c with the same velocity as towards b (i.e. over the perpendicular in the middle of bc ?)

upload_2015-4-26_21-16-28.png


It should be clear that ##v_c \ne v_a - v_b##, so what is the meaning of the relevant equation ?
Where does the ##v\cos(60^\circ)## come in ?
"distance to be covered is ##d/\sqrt 3## " -- what is d ?
What is the complete problem statement ?
 
  • #5
So each of the points moves along the edge of the triangle towards the next point.

This, point a moves towards point b which moves towards point c.

As they always move towards the next point with the same speed, they always form the vertices of an ever smaller equilateral triangle and converge at the Center.

D would be the length of the side of the initial equilateral triangle.
 
  • #6
Ah, so we have a dog chasing hare(*) kind of thingy ! Like below ?
upload_2015-4-26_21-35-51.png


(*) actually, no: dog chasing hare let's te hare run straight -- yields a radiodrome, hefty mathematics
 
  • #7
Yes!

Its the math of it that I'm trying to figure out.

The thought process escapes me.

I think I'll just have to break it down a lot more.

Any approaches you'd suggest?
 
  • #8
You posted in introductory physics. What level of math are we at ? Do the differential equations in the link give you enough handle to set up such equations for the case at hand (hare path also curved ?)

upload_2015-4-26_22-11-6.png


Just so you are warned: I have no idea what comes out mathematically.
Doing it numerically (this is excel ?:)) at least gives nice pictures :rolleyes:
 

Attachments

  • upload_2015-4-26_22-2-57.png
    upload_2015-4-26_22-2-57.png
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  • #9
BvU said:
You posted in introductory physics. What level of math are we at ? Do the differential equations in the link give you enough handle to set up such equations for the case at hand (hare path also curved ?)

View attachment 82614

Just so you are warned: I have no idea what comes out mathematically.
Doing it numerically (this is excel ?:)) at least gives nice pictures :rolleyes:

So, the math itself in the links is alright.

I'd like to develop the equations themselves.

I was getting stuck with the velocity vs path vector stuff. What route of analysis should I take?

Also, how did you generate the graphs in excel?
 
  • #10
More or less by hand o0) .

a,b,c are x and y positions. point a is repeated to close the triangle ab bc ca

##\vec v_a = \left [ \left (\vec b_x -\vec a_x \over |\vec b -\vec a| \right ) |v| , \left (\vec b_y -\vec a_y \over |\vec b -\vec a| \right ) |v| \right ] ## etc.
and
##\vec a(t+\Delta t) = \vec a + \vec v_a \Delta t ##
 

Attachments

  • Chasers.xlsx
    15 KB · Views: 432
  • #11
chaoseverlasting said:
Velocity along the direction to the Center is vcos60 which will be he velocity at which he points move towards the Center.
Are you sure? But as you say it is a constant, so you can write down the distance from the centre at time t. Can you do something similar for tangential motion and obtain a path in polar coordinates?
 

1. What is the path of an object at the vertex of an equilateral triangle?

The path of an object at the vertex of an equilateral triangle is a straight line. The object will move in a straight line away from the vertex until it reaches the opposite side of the triangle.

2. Why is the path of an object at the vertex of an equilateral triangle a straight line?

This is due to the symmetry of an equilateral triangle. Since all three sides of the triangle are equal, the object will move away from the vertex at the same rate on each side, resulting in a straight line.

3. Does the angle of the equilateral triangle affect the path of the object at the vertex?

No, the angle of the equilateral triangle does not affect the path of the object at the vertex. As long as the triangle remains equilateral, the path will always be a straight line.

4. Can an object at the vertex of an equilateral triangle change direction?

No, the object will continue to move in a straight line away from the vertex until it reaches the opposite side of the triangle. It cannot change direction unless it encounters an external force.

5. What is the relationship between the length of the sides of the equilateral triangle and the path of the object at the vertex?

The length of the sides of the equilateral triangle does not have a direct relationship with the path of the object at the vertex. As long as the triangle remains equilateral, the path will always be a straight line.

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