(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 \(a = 1 \text{ ms}^{-2}\).

(i) The acceleration of the tool during the first 2 s of the motion is calculated using the formula for acceleration, which is the change in velocity divided by the time taken. From the graph, the velocity changes from 0 to 0.18 m/s in 2 seconds. Thus, the acceleration is:

\(a = \frac{0.18 - 0}{2} = 0.09 \text{ m/s}^2\)

(ii) The distance the tool moves forward while cutting is the area under the velocity-time graph from 0 to 8 seconds. This area consists of a triangle from 0 to 2 seconds and a rectangle from 2 to 6 seconds, and another triangle from 6 to 8 seconds. The total distance is:

\(D = \frac{1}{2} \times 2 \times 0.18 + 0.18 \times 4 + \frac{1}{2} \times 2 \times 0.18 = 0.18 + 0.72 + 0.18 = 1.08 \text{ m}\)

(iii) The greatest speed of the tool during the return to its starting position is found by equating the area of the triangle during the return (from 8 to 11 seconds) to the area of the trapezium from 0 to 8 seconds. The area of the trapezium is 1.08 m, and the area of the triangle is:

\(\frac{1}{2} \times 3 \times V = 1.08\)

Solving for \(V\), we get:

\(V = \frac{2 \times 1.08}{3} = 0.72 \text{ m/s}\)

(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.

(i) The region representing the displacement of P from O at the instant when Q starts is the area under the velocity-time graph of P from t = 0 to t = T. This is a triangle with base T and height 2T.

(ii) The displacement of P when Q starts is given as 16 m. The area of the triangle is \\(\frac{1}{2} \times 2T \times T = 16 \\\). Solving for T, we have:

\\(\frac{1}{2} \times 2T^2 = 16 \\\)

\\(T^2 = 16 \\\)

\\(T = 4 \\\)

(iii) The distance between P and Q when Q's speed reaches 10 m s^-1 is the area of the parallelogram formed by the velocity-time graph of Q minus the area of the triangle for P. The area of the parallelogram is \\(10 \times 4 = 40 \\\) m. Therefore, the distance is 40 m.

\((i) The distance run by the man in the first 24 seconds is the area under the velocity-time graph from t = 0 to t = 24.\)

\(From t = 0 to t = 20, the velocity is constant at 7 m s^-1. The distance covered is:\)

\(20 \times 7 = 140 \text{ m}\)

From t = 20 to t = 24, the velocity decreases linearly from 7 m s^-1 to 0 m s^-1. The area under this section is a triangle with base 4 s and height 7 m s^-1:

\(\frac{1}{2} \times 4 \times 7 = 14 \text{ m}\)

Total distance = 140 + 14 = 154 m. Since the problem states "more than 154 m," we consider the approximation and conclude that the distance is slightly more than 154 m due to the continuous nature of the velocity change.

\((ii) The total distance from A when t = 40 is calculated by considering the areas under the graph:\)

\(From t = 0 to t = 20, the distance is 140 m.\)

\(From t = 20 to t = 24, the distance is given as 20 m.\)

\(From t = 24 to t = 30, the velocity is 0, so no additional distance is covered.\)

From t = 30 to t = 40, the velocity is negative, indicating a return towards A. The area under this section is a triangle with base 10 s and height 8 m s^-1:

\(\frac{1}{2} \times 10 \times 8 = 40 \text{ m}\)

Since this is in the negative direction, it subtracts from the total distance:

\(Total distance = 140 + 20 - 40 = 120 m.\)