# Difference between velocity vectors, relative speed

Consider two particles moving in the same direction instantaneously but one of them follows a circular path and one follows a straight path. If I adopt a rotating frame of reference in which the particle $A$ is stationary, math tells me that the relative speed between $A$ and $B$ is $V_{rel}=V_B-V_A-\frac{V_B h}{R}$, where $h$ is the distance between the particles and $R$ is the radius of the circular path.

I understand that this is right, I can infer this from the formulas of rotating reference systems. But can someone please explain why it doesn't work to simply take $V_A-V_B$. The relative velocity should be the difference between two velocity vectors, I think that's very intuitive. But in this case it isn't? Why? When I see a physical situation like this (e.g. they could be airplanes, one doing a loop), how can I know if I can simply take the difference between the velocity vectors to find the relative velocity or not?

• What is $V_B$ exactly? – David Z May 21 '13 at 19:33
• @DavidZaslavsky $V_B$ and $V_A$ are velocities. Imagine for example two airplanes. One flying in a loop, how would that pilot perceive his relative velocity to the pilot in the other airplane? (This is the question which I worked on when I got stuck with this conceptual problem) – user714 May 21 '13 at 19:35
• Yes, it's clear that they are velocities, but how exactly is $V_B$ defined? Velocity of what (presumably object B) with respect to what? – David Z May 21 '13 at 19:37
• @DavidZaslavsky Sorry, I didn't understand the question. Velocities of objects, for example airplanes, with respect to an inertial frame. – user714 May 21 '13 at 19:49

In general, $A$ will see $B$ going at a different velocity than $B$ sees $A$.
Imagine that, in your example, $\mathbf{V_B}=\mathbf{V_A}$ (zero relative velocity) in the inertial frame. In the rotating frame, $B$ will lag a bit behind, because it is further than $A$ from the rotation center. This shows that the relative velocity is not zero in that frame.