(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.