(i) The velocity-time graph consists of three straight line segments: a positive gradient for 0 < t < 80, a zero gradient for 80 < t < 560, and a negative gradient for 560 < t < 600. The graph starts and ends on the t-axis at t = 0 and t = 600, respectively.

(ii) Using the area under the velocity-time graph for 0 < t < 80, we have:

\(\frac{1}{2} \times 80 \times v = 400\)

Solving for \(v\), we get \(v = 10 \text{ ms}^{-1}\).

(iii) The total distance is the area under the velocity-time graph. The area of the trapezium is:

\(D = \frac{1}{2} (600 + 480) \times 10 = 5400 \text{ m}\)

(iv) The acceleration for 0 < t < 80 is given by the gradient of the velocity-time graph:

\(v = 0 + at\)

\(10 = a \times 80\)

\(a = 0.125 \text{ ms}^{-2}\)