Calculating velocity of a particle if distances are constant

AI Thread Summary
The discussion revolves around calculating the velocity of particle C based on the movements of particles A and B towards each other at 5 m/s. The initial distances between the particles are specified, with A and B starting 50 m apart, while maintaining fixed separations from C. Participants clarify the angles involved and the use of trigonometry, with suggestions to use Cartesian coordinates for simplification. The configuration of the particles is likened to a 3-4-5 right triangle, leading to insights about their trajectories and eventual collinearity. The conversation highlights the complexities of the problem, including the implications of the particles' movements on their distances over time.
anthraxiom
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Homework Statement
There are 3 particles a, b and c. A moves towards B with velocity 5 m/s and B moves towards A with same speed as A. C moves such that separation between B and C is always 40 and separation between A and C is always 30. initial distance between A and B was 50m. Find velocity of C as a function of time.
Relevant Equations
dr/dt = v
I tried to solve using trigonometry.
Let angle between line CA ad BA be theta and angle between BA and AC be alpha.
30sin theta= 40 sin alpha
30 cos theta + 40 cos alpha + 2vt=50
I have no idea how to proceed after this.
 
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A moves towards a? B moves toward A with the same speed as B? What are the statements saying!
 
its just saying that A moves to B with 5m/s and B moves towards A at 5m/s
 
anthraxiom said:
Homework Statement: There are 3 particles a, b and c. A moves towards a
You mean A moves towards B.
anthraxiom said:
with velocity 5 m/s and B moves towards A with same speed as B.
You mean same speed as A.
anthraxiom said:
C moves such that separation between B and C is always 40 and separation between A and C is always 30. initial distance between A and B was 50m. Find velocity of C as a function of time.
Relevant Equations: dr/dt = v

I tried to solve using trigonometry.
Let angle between line CA ad BA be theta and angle between BA and AC be alpha.
You mean angle between BA and BC is alpha.
anthraxiom said:
30sin theta= 40 sin alpha
30 cos theta + 40 cos alpha + 2vt=50
Where v = 5m/s.

I would approach it using Cartesian coordinates. What would a convenient X axis and origin be, using a certain symmetry in the question?
 
haruspex said:
You mean A moves towards B.

You mean same speed as A.

You mean angle between BA and BC is alpha.
My bad...
 
haruspex said:
I would approach it using Cartesian coordinates. What would a convenient X axis and origin be, using a certain symmetry in the question?
I got it thanks
 
Setting up this problem becomes easier if one recognizes that the starting configuration is a 3-4-5 right triangle.
 
anthraxiom said:
C moves such that separation between B and C is always 40 and separation between A and C is always 30.
Don't these conditions make the three trajectories parallel?
 
Lnewqban said:
Don't these conditions make the three trajectories parallel?
No (assuming I understand the question correctly).

At t = 0, the configuration is this (not to scale):
1747589251392.webp


Consider the positions some time later. A and B are closer (A and B are heading towards each other with speed-of-approach = 10 m/s) and the configuration is, for example, this:

1747589281511.webp
 
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Likes jbriggs444 and Lnewqban
  • #10
In that case, A and B are satellites of C, and when the three particles become collinear (distance between A and B is 10 m), A stops moving "towards B with velocity 5 m/s and B" stops moving "towards A with same speed as A".
 
  • #11
Lnewqban said:
In that case, A and B are satellites of C, and when the three particles become collinear (distance between A and B is 10 m), A stops moving "towards B with velocity 5 m/s and B" stops moving "towards A with same speed as A".

Not sure about that. In post #1 it says:
anthraxiom said:
... A moves towards B with velocity 5 m/s and B moves towards A with same speed as A. ...
I interpret that as A and B moving towards each other along the same straight line. It would then be implicit that the question is asking for v(t) during the first four seconds (so we needn't consider the infeasible case where |AB| is less the 10 m).
 
  • #12
Lnewqban said:
In that case, A and B are satellites of C, and when the three particles become collinear (distance between A and B is 10 m), A stops moving "towards B with velocity 5 m/s and B" stops moving "towards A with same speed as A".
Indeed. The equations will lead to the y coordinate (distance from C to the line AB) becoming imaginary.
 
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