A motorcyclist rides round a flat circular test track of radius 40 m…

Step 1. The two parts at A. Her speed falls steadily from $8.0\ \text{m/s}$ to zero in $10\ \text{s}$, so the tangential part is $a_t = 0.8\ \text{m/s}^2$, against the motion, all the time. At A the centripetal part is $\dfrac{v_0^2}{R} = \dfrac{(8.0\ \text{m/s})^2}{40\ \text{m}} = 1.6\ \text{m/s}^2$. The two are at right angles:

Step 2. Equal parts. The centripetal part equals $0.8\ \text{m/s}^2$ when $v^2 = a_t R = 32\ \text{m}^2/\text{s}^2$, that is $v = 5.66\ \text{m/s}$. Her speed falls by $8.0 - 5.66 = 2.34\ \text{m/s}$ at $0.8\ \text{m/s}^2$, which takes $t = 2.93\ \text{s}$.

Step 3. The angle turned. With a steady deceleration, the distance along the track is (average speed) $\times$ (time) $= \dfrac{8.0\ \text{m/s}}{2} \times 10\ \text{s} = 40\ \text{m}$. The angle is $\dfrac{s}{R} = 1.0\ \text{rad} = 57.3^\circ$.

Step 4. Chord and averages. The chord of a circle that subtends an angle $\phi$ at the centre is $2R\sin\dfrac{\phi}{2}$:

Average velocity $= \dfrac{38.35\ \text{m}}{10\ \text{s}} = 3.84\ \text{m/s}$, along AB. Average speed $= \dfrac{40\ \text{m}}{10\ \text{s}} = 4.0\ \text{m/s}$, a little more, because the path is longer than the chord.

Step 5. Average acceleration. Her velocity changes from $8.0\ \text{m/s}$ along the tangent at A to zero. So the change has the size $8.0\ \text{m/s}$, and the average acceleration is $\dfrac{8.0\ \text{m/s}}{10\ \text{s}} = 0.80\ \text{m/s}^2$, pointing backwards along the tangent at A.