Calculating Relative Speed in Circular Motion: Two Cyclists on a Track

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Two cyclists, A and B, travel at the same constant speed on a circular track, with B moving through the diameter and A along the circumference. The relative speed of A with respect to B depends on their velocity vectors and the angles between them. Initially, A moves away from B, only approaching after passing the quarter arc. The velocity of B is constant along the x-axis, while A's velocity is tangential to the circle. The calculation of their relative speed simplifies when considering the correct time formula, which is t = D/V.
Trooko
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  • Two cyclists( A, B) traveling with the same constant speed, v,
  • in a circular track.
  • They start at the same point on the circle.
  • cyclist B travel through the diameter of the circle, assume the x-axis on a xy
  • Cyclist A travel on the circumference of the circle

  • Find speed of A with respect to B.

Given: the constant speed, radius, angle between A and the x-axis through the centre, angle between A and B.

am I suppose to do something with the acceleration (normal/centripedal) and the two angles given.

thank you
 
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Trooko said:
Given: the constant speed, radius, angle between A and the x-axis through the centre, angle between A and B.

am I suppose to do something with the acceleration (normal/centripedal) and the two angles given.
It has to do with the velocity of each, and the velocity of A with respect to B is dependent on the angles between the velocity vectors.

When A and B start, A is moving away from B, and only starts moving toward B after A passes the quarter arc.

At time t = v/D = v/2R, B moves outside A's circular trajectory.

\vec_B is always v\,\hat{x}, whereas

\vec_A is always v\,\hat{\theta} where \hat{\theta} is the unit vector in the azimuthal direction (tangent to circumference of circle). As xB gets very large, the angle between A and B gets very small.

The 'speed' would be the magnitude of the velocity vector given by \vec_A - \vec_B
 
Thanks for the reply. It really does look a lot simpler now.

But when you wrote t= V/D, did you mean t = D/V?
 
Trooko said:
Thanks for the reply. It really does look a lot simpler now.

But when you wrote t= V/D, did you mean t = D/V?
Yes, t = D/V. My mistake.
 
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