(i) The train accelerates from rest with an acceleration of 0.025 m s^-2 for 600 s. The velocity at the end of this period is given by:

\(v(600) = 0.025 \times 600 = 15 \text{ m s}^{-1}\)

The train then travels at a constant speed of 15 m s^-1 for 2600 s. Finally, it decelerates to rest with a deceleration of 0.0375 m s^-2. The time taken to decelerate to rest is given by:

\(0 = 15 - 0.0375 t_3\)

\(t_3 = \frac{15}{0.0375} = 400 \text{ s}\)

The total time is:

\(600 + 2600 + 400 = 3600 \text{ s}\)

(ii) The velocity-time graph consists of three segments: an upward slope for 600 s, a horizontal line for 2600 s, and a downward slope for 400 s. The distance is the area under the graph:

\(d = \frac{1}{2} \times 600 \times 15 + 2600 \times 15 + \frac{1}{2} \times 400 \times 15\)

\(d = 4500 + 39000 + 3000 = 46500 \text{ m}\)

(iii) The speed of the train is 7.5 m s^-1 at two points: during acceleration and deceleration. During acceleration:

\(7.5 = 0.025 t\)

\(t = \frac{7.5}{0.025} = 300 \text{ s}\)

During deceleration:

\(7.5 = 15 - 0.0375 (t - 3200)\)

\(0.0375 (t - 3200) = 7.5\)

\(t - 3200 = \frac{7.5}{0.0375} = 200\)

\(t = 3400 \text{ s}\)