(i) The area under the velocity-time graph from t = 0 to t = 2.5 represents the distance XA. The graph forms a triangle with base 2.5 s and height equal to the maximum speed. Using the formula for the area of a triangle, \(\frac{1}{2} \times 2.5 \times \text{speed}_{\text{max}} = 4\). Solving for \(\text{speed}_{\text{max}}\), we get \(\text{speed}_{\text{max}} = 3.2 \text{ ms}^{-1}\).

(ii) For the first 2 s of the second stage, the particle accelerates at 3 m s^-2. Using \(V = u + at\), where \(u = 0\), \(a = 3 \text{ ms}^{-2}\), and \(t = 2 \text{ s}\), we find \(V = 0 + 3 \times 2 = 6 \text{ ms}^{-1}\).

(iii) The total distance from A to B is 48 m. The area under the graph from t = 2.5 to t = 14.5 must equal 48 m. The graph forms a trapezium with a height of 6 ms^-1 and bases of 12 s and T s. Using the area of a trapezium, \(\frac{1}{2} \times 6 \times (12 + T) = 48\). Solving for T, we find T = 8.5 s.

(iv) The particle decelerates from V to 0 over the remaining time. Using \(a = \frac{0 - V}{14.5 - 8.5}\), where V = 6 ms^-1, we find the deceleration \(a = 1 \text{ ms}^{-2}\).