(i) The area representing the displacement of P from X at the instant when Q starts is a composite figure consisting of two rectangles. The first rectangle has boundaries at t = 0 and t = 20, with v = 0 and v = 2.5. The second rectangle has boundaries at t = 20 and t = T, with v = 0 and v = 4.

(ii) The displacement of P when Q starts is given by:

\(50 + 4(T - 20) = 70\)

Simplifying gives:

\(4T - 30 = 70\)

\(4T = 100\)

\(T = 25\)

(iii) The distance between P and Q when Q's speed reaches 4 m s^-1 is calculated by:

\(\text{Distance} = 70 + (4 - 2.5) \times 20\)

\(\text{Distance} = 70 + 30 = 100 \text{ m}\)

(iv) For the displacement-time graph, P is represented by two straight line segments: the first with a positive slope and the second with a steeper slope, with t = 20 indicated. For Q, the first and second segments are parallel to P's and displaced to the right, with t = 25 and t = 45 indicated.

(i) To find the distance AC, calculate the area under the velocity-time graph from t = 0 to t = 10 s. The velocity is 7 m/s for 10 seconds, so:

\(AC = 7 \times 10 = 70 \text{ m}\)

To find the distance AB, calculate the net displacement by subtracting the area from t = 15 to t = 30 s (where velocity is -4 m/s) from the distance AC:

\(BC = 4 \times 15 = 60 \text{ m}\)

\(AB = AC - BC = 70 - 60 = 10 \text{ m}\)

(ii) The graph of x against t consists of three connected straight line segments:

1. From t = 0 to t = 10, x increases linearly from 0 to 70 m.
2. From t = 10 to t = 15, x remains constant at 70 m.
3. From t = 15 to t = 30, x decreases linearly from 70 m to 10 m.

The graph should show:

• At t = 10, x = 70 m.
• At t = 15, x = 70 m.
• At t = 30, x = 10 m.