Calculating velocity of a particle if distances are constant

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SUMMARY

The discussion focuses on calculating the velocity of particle C in a scenario where three particles A, B, and C are moving towards each other with defined velocities and distances. A and B each move towards one another at 5 m/s, while C maintains a constant separation of 40 meters from B and 30 meters from A. The initial distance between A and B is 50 meters. The problem can be approached using trigonometry and Cartesian coordinates, recognizing the configuration forms a 3-4-5 right triangle, which simplifies the calculations.

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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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  • #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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