(i) The velocity-time graph consists of three straight line segments:

• From \(t = 0\) to \(t = 8\), velocity increases from 0 to 8 m/s.
• From \(t = 8\) to \(t = 20\), velocity decreases from 8 m/s to 2 m/s.
• From \(t = 20\) to \(t = 26\), velocity increases from 2 m/s to 6.5 m/s.

(ii) The shaded region for \(20 \leq t \leq 26\) is the area under the velocity-time graph between these times.

(iii) To show \(s = 0.375t^2 - 13t + 202\):

1. Calculate \(s(20)\):

\(s(20) = \frac{1}{2} \times 8 \times 8 + \frac{1}{2} \times (8 + 2) \times 12 = 92 \text{ meters}.\)

2. Calculate acceleration in the third phase:

\(a = \frac{6.5 - 2}{6} = 0.75 \text{ m/s}^2.\)

3. Use the equation for displacement:

\(s(t) = 92 + 2(t - 20) + 0.375(t - 20)^2.\)

4. Simplify to get:

\(s(t) = 0.375t^2 - 13t + 202.\)