(i) The velocity-time graph is a trapezium. The train accelerates to 25 m/s, travels at constant speed, then decelerates to rest. The right-hand slope is steeper than the left-hand slope because the deceleration is twice the acceleration.

(ii) Let the acceleration be a. Then, the deceleration is 2a. The time to decelerate from 25 m/s to 0 is T/2. The total time is 180 s, so the time at constant speed is:

\(180 - T - \frac{T}{2} = 180 - 1.5T\)

(iii) The total distance is 3300 m. Using the area under the velocity-time graph, the distance is:

\(0.5 \times [180 + (180 - 1.5T)] \times 25 = 3300\)

Solving for T:

\(0.5 \times (360 - 1.5T) \times 25 = 3300\)

\(360 - 1.5T = 264\)

\(1.5T = 96\)

\(T = 64\)

The time to decelerate is 32 s. The distance while decelerating is:

\(0.5 \times 32 \times 25 = 400 \text{ m}\)