(i) The velocity-time graph is a trapezium with three straight lines: a positive gradient for the first 3 seconds, a zero gradient for the next 6 seconds, and a negative gradient for the last 4 seconds. The time points 0, 3, 9, and 13 are shown on the time axis. The maximum velocity reached is 2.7 m/s.
(ii) To find the total distance, we calculate the area under the velocity-time graph, which is a trapezium. The formula for the area of a trapezium is:
\(\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}\)
Here, the bases are the time intervals: 6 s and 13 s, and the height is the maximum velocity 2.7 m/s. Thus, the total distance is:
\(\text{Total distance} = \frac{1}{2} \times (6 + 13) \times 2.7 = 25.65 \text{ m}\)
Alternatively, using the constant acceleration equations for each stage:
- Stage 1: \(s_1 = 0.5 \times 0.9 \times 3^2 = 4.05 \text{ m}\)
- Stage 2: \(s_2 = 2.7 \times 6 = 16.2 \text{ m}\)
- Stage 3: \(s_3 = 0.5 \times (2.7 + 0) \times 4 = 5.4 \text{ m}\)
Total distance = \(s_1 + s_2 + s_3 = 4.05 + 16.2 + 5.4 = 25.65 \text{ m}\)
