A cyclist rides round a circular track of radius 5.0 m at a steady speed…

Step 1. Path and displacement. The path is $\dfrac{2\pi \times 5.0\ \text{m}}{4} = 7.85\ \text{m}$. The displacement goes from $(5, 0)$ to $(0, 5)$: its size is $\sqrt{(5\ \text{m})^2 + (5\ \text{m})^2} = 7.07\ \text{m}$.

Step 2. The averages. Average speed: $\dfrac{7.85\ \text{m}}{2.0\ \text{s}} = 3.93\ \text{m/s}$. Average velocity: $\dfrac{7.07\ \text{m}}{2.0\ \text{s}} = 3.54\ \text{m/s}$, towards the north-west.

Step 3. The two velocities. Her speed is $3.93\ \text{m/s}$ all the way. At the east end she moves north, and at the north end she moves west.

Step 4. Average acceleration. $\Delta\vec{v}$ is $3.93\ \text{m/s}$ west minus $3.93\ \text{m/s}$ north. Its size is $\sqrt{2} \times 3.93\ \text{m/s} = 5.55\ \text{m/s}$, so $a_{\text{av}} = \dfrac{5.55\ \text{m/s}}{2.0\ \text{s}} = 2.78\ \text{m/s}^2$.