(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:

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