How do I find the average speed when an airplane is flying in circles?

Jun 2017
Lima, Peru
The problem is as follows:

At an air exhibition a biplane is flying in circles as shown in the figure from below. It is known that type of motion of this airplane is a rotation with constant angular acceleration. The radius of the circle is $2\,m$. Using this information find the average speed in $\frac{m}{s}$ of the plane between the instants when $\omega=6\,\frac{rad}{s}$ and $\omega=10\,\frac{rad}{s}$.​

The given alternatives in my book are:


I'm not sure if my attempt is correct in the solution of this problem but I thought that to get the average speed is given by:


Since it mentions that it is a rotation with constant angular acceleration, then what I need is the acceleration. Using the information from the picture I'm getting this:

Considering $r=2\,m$



Therefore the acceleration can be found:





Since what they request is the average speed then what is needed is the displacement for the given speeds.



But the time elapsed for that displacement is also required for getting the average speed:


$\omega_{f}=\omega_{o}+\alpha t$



So it is only $1\,s$ elapsed in the given interval.

Therefore the average speed would be:

$\overline{v}=\frac{s}{t}=\frac{8\times 2}{1}=16\,\frac{m}{s}$

Which is what appears in alternative number $3$. But is my solution correct?. :)
Last edited by a moderator:


Forum Staff
Dec 2006
That's okay, but for constant angular acceleration, \(\displaystyle \Delta\theta = \frac{\omega_1 + \omega_2}{2}\Delta t\),
so \(\displaystyle \overline\omega = \frac{\Delta\theta}{\Delta t} = \frac{\omega_1 + \omega_2}{2} = \frac{6 + 10}{2}\text{rad/s} = 8\text{ rad/s}\).
Hence (as the radius is $\text{2 m}$) $\overline v = 8\times2\text{ m/s} = 16\text{ m/s}$.