# Is displacement in circular motion a chord or an arc?

When taking the displacement between two points along a circular path to calculate its velocity, do you take the length of a chord connecting the two points or do you take the length of the arc connecting them?

When deriving the formula $a_c=\frac{v^2}{r}$, you presume that the displacement is a chord ($d$): $$\frac{\Delta v}{v}=\frac{d}{r}$$ $$a_c=\frac{\Delta v}{t}=\frac{v}{r}\frac{d}{t}=\frac{v}{r}v=\frac{v^2}{r}$$

However, other equations seem to suggest that displacement is an arc ($s$). For example, the equation $v=\omega r$, where $s=\theta r$ and $\omega=\frac{\theta}{t}$, suggests that $v=\frac{s}{t}$ rather than $\frac{d}{t}$.

Moreover, common equations like $a_c=4π^2rf^2$ are derived from the equation $v=\frac{2πr}{T}$, which suggests also that $v=\frac{s}{t}$ rather than $\frac{d}{t}$ (since $d=0$ in the reference frame of a full revolution).

If the formula $v=\omega r$ describes only speed, not velocity, and is based on distance instead of displacement (which would then be unequivocally a chord), then I’d understand. But the second equation, $a_c=4π^2rf^2$ involves acceleration, which definitely is the change in velocity, not speed. How can you derive an equation for acceleration based on one for speed if speed does not represent the magnitude of velocity?

• Check in the treatment that you are using the distinction between "displacement" and "distance". After that all should be clear. Aug 30 '15 at 5:36
• @dmckee So there is a distinction, then, between speed and velocity in the context of circular motion? Speed isn’t simply the magnitude of the velocity? Aug 30 '15 at 6:05
• Instantaneous speed is the magnitude of instantaneous velocity, but that does not mean that average speed is the magnitude of average velocity. Aug 30 '15 at 6:17
• @dmckee Why is that so? Aug 30 '15 at 6:29
• Because taking the magnitude is a nonlinear operation (it involves the Pythagorean sum of vector components) and so it is not left invariant by a linear operation like taking the mean. (This is kind of a circular answer, because that's how we define "nonlinearity"). Aug 30 '15 at 7:11