# Question #f47c7

Apr 17, 2017

The speed of the particle is $\dot{s} \left(t\right) = \omega r = \sqrt{2} \omega \left(t\right)$

For angular speed $\omega$, we note that:

$\dot{\omega} \left(t\right) = \alpha \left(t\right) = \frac{\pi}{4}$

$\implies \omega \left(t\right) = \frac{\pi}{4} t + C$ and $\omega \left(0\right) = 0 \implies C = 0$

We also note that:

$\dot{\theta} \left(t\right) = \omega \left(t\right) = \frac{\pi}{4} t$

$\implies \theta \left(t\right) = \frac{\pi}{8} {t}^{2} + C$ and $\theta \left(0\right) = 0 \implies C = 0$

From that, for a quarter turn in time $t = \tau$, we have:

$\frac{\pi}{2} = \frac{\pi}{8} {\tau}^{2} \implies \tau = 2$

Returning to speed $\dot{s}$, we now have: $\dot{s} \left(t\right) = \sqrt{2} \frac{\pi}{4} t$

The average speed ${\dot{s}}_{a v e}$ over the time period $\tau$ is:

${\dot{s}}_{a v e} = \frac{{\int}_{0}^{\tau} \dot{s} \mathrm{dt}}{\tau}$

The numerator is just the distance travelled, ie a quarter turn along the circumference:

$= \frac{\frac{\pi}{2} \sqrt{2}}{2} = \frac{\pi}{2 \sqrt{2}}$