(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.