(i) The area under the velocity-time graph represents the distance traveled. For the first stage, the area is a triangle with base 5 s and height equal to the greatest speed, which we denote as \(v_{\text{max}}\). The area is \(\frac{1}{2} \times 5 \times v_{\text{max}} = 10\). Solving for \(v_{\text{max}}\), we get \(v_{\text{max}} = 4\) m s^-1.

(ii) During the third stage, the lift accelerates at 2 m s^-2 for 3 s. Using \(v = u + at\), where \(u = 0\), \(a = 2\), and \(t = 3\), we find \(V = 0 + 2 \times 3 = 6\) m s^-1.

(iii) The lift travels 34.5 m in the third stage. The area under the graph for this stage is a trapezium with parallel sides of 9.5 s and \(T\) s, and height 6 m s^-1. The area is \(\frac{1}{2} \times (T + 9.5) \times 6 = 34.5\). Solving for \(T\), we find \(T = 2\) s.

(iv) The deceleration is the negative of the gradient of the final part of the third stage. The change in velocity is from 6 m s^-1 to 0 m s^-1 over a time of \(24.5 - (18 + 2) = 4.5\) s. The deceleration is \(\frac{6}{4.5} = \frac{4}{3}\) m s^-2.