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

[Trooko]

• 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

 

[Astronuc]

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 x_B gets very large, the angle between A and B gets very small.

The 'speed' would be the magnitude of the velocity vector given by $\vec_A$ - $\vec_B$

 

[Trooko]

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?

 

[Astronuc]

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.