(i) The acceleration is calculated using the formula:

\(a = \frac{v - u}{t}\)

where \(v = 16 \text{ m/s}\), \(u = 0 \text{ m/s}\), and \(t = 40 \text{ s}\).

Thus, \(a = \frac{16 - 0}{40} = 0.4 \text{ m/s}^2\).

(ii) The total distance is the area under the velocity-time graph, which is a trapezium. The area \(A\) is given by:

\(A = \frac{1}{2} \times (600 + T) \times 16 = 9040\)

Solving for \(T\):

\(9040 = \frac{1}{2} \times (600 + T) \times 16\)

\(9040 = 8 \times (600 + T)\)

\(1130 = 600 + T\)

\(T = 530 \text{ s}\)

(iii) The distance travelled while decelerating is the area of the triangle at the end of the graph:

\(s = \frac{1}{2} \times (600 - 530 - 40) \times 16\)

\(s = \frac{1}{2} \times 30 \times 16\)

\(s = 240 \text{ m}\)