# Average velocity in uniform circular motion

1. Feb 18, 2015

Suppose a particle undergoing uniform circular motion completes a cycle, then $<\vec{v}> = \frac{Δ\vec{r}}{Δt} = \vec{0}$. But then again, isn't it also the mean of the two velocity vectors at that point? And aren't those two velocity vectors equal? So if the velocity at that point were, say, $\vec{v}_1$, then this would also be the average velocity. So which is it?

2. Feb 18, 2015

### jbriggs444

The average velocity (over time) is always equal to displacement (end position minus start position) divided by elapsed time (end time minus start time). In the case of constant acceleration, this will be equal to the mean of the starting and ending velocities. But in uniform circular motion, acceleration is not constant. Its magnitude is constant, but its direction is changing.

3. Feb 18, 2015

Thanks. I'm curious though, why does acceleration have to be constant for this relation to hold?

4. Feb 18, 2015

### jbriggs444

An average (over time) of any continuously varying quantity is taken as the integral of that quantity over the relevant time interval.

$v_{avg} = \frac{\int_{t_i}^{t_f} v(t)\,dt}{t_f-t_i}$

Intuitively [and restricting our attention to one dimension] this amounts to graphing velocity on the y axis versus time on the x axis and finding the average height of the graph. You take the area under the curve and divide by the length of the time interval. If acceleration is constant, that means that the graph will be a straight line. You can compute the area under the curve by averaging the heights (velocities) at the end points.

If acceleration is not constant, you cannot compute the area this way. The graph might bulge up above the line between the end points. Or it might sag under that line. Or both. In such a case, the area under the curve has to be determined in some other way.

5. Feb 18, 2015

I tried to compute the integral you mentioned, and as you said, the relation only holds for constant acceleration [$v = v_0 + at$]