(i) The displacement of P from O at t = 10 is the area under the velocity-time graph from t = 0 to t = 10. This area is a trapezium with bases 10 and 4, and height 6. The area is given by:

\(\text{Area} = \frac{1}{2} \times (10 + 4) \times 6 = 42 \text{ m}\)

(ii) To find the velocity of P at t = 12, we use the gradient of the line segment from t = 10 to t = 12. The change in velocity is from 6 to 0 over 4 seconds, so:

\(\frac{v}{2} = \frac{6}{4}\)

Solving gives \(v = 3 \text{ m s}^{-1}\).

(iii) The total distance covered by P is 49.5 m. The distance from t = 0 to t = 10 is 42 m. The remaining distance is covered from t = 10 to t = T at a constant velocity of 3 m s^-1:

\(42 + \frac{1}{2} (T - 10) \times 3 = 49.5\)

Solving for \(T\) gives \(T = 15 \text{ s}\).

(iv) The acceleration of Q from t = 2 to t = 6 is 1.75 m s^-2. The change in velocity over this time is:

\(V = 1.75 \times 4 = 7 \text{ m s}^{-1}\)

The distance traveled by Q is the area under its velocity-time graph:

\(\text{Distance} = \frac{1}{2} (13 + 6) \times 7 = 66.5 \text{ m}\)

The distance between P and Q when they both come to rest is:

\(66.5 + 42 - 7.5 = 101 \text{ m}\)