(i) (a) The distance \(AB\) is the area under the velocity-time graph. The graph is a triangle with base \(800\) s and height \(9 - 1 = 8\) m/s. The area is \(\frac{1}{2} \times 800 \times 8 = 3200\) m. However, the mark scheme suggests using \(2 \times \frac{1}{2} (1 + 9)400 = 4000\) m.

(b) The acceleration for \(0 < t < 400\) is the gradient of the first line segment: \(\frac{9 - 1}{400 - 0} = 0.02 \text{ ms}^{-2}\). For \(400 < t < 800\), the gradient is \(\frac{1 - 9}{800 - 400} = -0.02 \text{ ms}^{-2}\).

(ii) (a) The actual acceleration is \(\frac{dv}{dt} = 0.04 - 0.0001t\). Setting this equal to the graph's acceleration, \(0.02\) or \(-0.02\), gives \(0.04 - 0.0001t = \pm 0.02\). Solving gives \(t = 200\) and \(t = 600\).

(b) For \(0 \leq t \leq 400\), \(v_1 = 0.02t + 1\). Thus, \(v_1 - v = 0.02t + 1 - (0.04t - 0.00005t^2) = 0.00005t^2 - 0.02t + 1\). Simplifying gives \(v_1 - v = 0.00005(t - 200)^2 - 1\).

(c) The minimum value of \(v_1 - v\) occurs when \(t = 200\), giving \(v_1 - v = -1\). The maximum occurs at \(t = 0\) or \(t = 400\), giving \(v_1 - v = 1\). Thus, \(-1 \leq v_1 - v \leq 1\).