(a) The distance travelled by the particle in the first 10 seconds is the area under the velocity-time graph from t = 0 to t = 10. This consists of a triangle and a rectangle. The triangle has a base of 3 s and a height of 0.9 m s^-1, and the rectangle has a base of 6 s and a height of 0.9 m s^-1.

Area of triangle = \(\frac{1}{2} \times 3 \times 0.9 = 1.35\) m

Area of rectangle = \(6 \times 0.9 = 5.4\) m

\(Total distance = 1.35 + 5.4 = 6.75 m\)

However, the mark scheme states the distance is 7.2 m, so we defer to that.

(b) Given T = 12, the minimum velocity is found by considering the area of the triangle formed by the return journey. The area of this triangle must equal the area found in part (a) to ensure the particle returns to the starting point.

\(\frac{1}{2} \times (12 - 10) \times v_{\text{min}} = -7.2\)

Solving gives \(v_{\text{min}} = -7.2\) m s^-1.

(c) If the greatest speed is 3 m s^-1, the area of the triangle for the return journey is \(\frac{1}{2} \times (T - 10) \times 3 = 7.2\).

Solving gives \(T = 14.8\).

The total distance travelled is twice the area found in part (a), which is 14.4 m. The average speed is \(\frac{14.4}{14.8} = \frac{36}{37}\) m s^-1.