(i) To find the acceleration, we use the formula for acceleration as the gradient of the velocity-time graph.
For 0 < t < 30, the velocity changes from 1.5 m s-1 to 2.1 m s-1 over 30 seconds. Thus, the acceleration is:
\(a = \frac{2.1 - 1.5}{30} = 0.02 \text{ m s}^{-2}\)
For 30 < t < 40, the velocity changes from 2.1 m s-1 to 0 m s-1 over 10 seconds. Thus, the acceleration is:
\(a = \frac{0 - 2.1}{10} = -0.21 \text{ m s}^{-2}\)
\((ii) The distance AB is the area under the velocity-time graph from t = 0 to t = 60.\)
Calculate the area of each segment:
- From t = 0 to t = 30: Trapezium with bases 1.5 and 2.1, height 30.
- Area = \(\frac{1}{2} (1.5 + 2.1) \times 30 = 54 \text{ m}\)
- From t = 30 to t = 40: Triangle with base 10, height 2.1.
- Area = \(\frac{1}{2} \times 2.1 \times 10 = 10.5 \text{ m}\)
- From t = 40 to t = 52: Triangle with base 12, height 2.2.
- Area = \(\frac{1}{2} \times 2.2 \times 12 = 13.2 \text{ m}\)
- From t = 52 to t = 60: Triangle with base 8, height 2.2.
- Area = \(\frac{1}{2} \times 2.2 \times 8 = 8.8 \text{ m}\)
\(Total displacement = 54 + 10.5 - 13.2 - 8.8 = 42.5 m\)
(iii) The total distance walked is the sum of the absolute values of the areas:
\(Total distance = 54 + 10.5 + 13.2 + 8.8 = 86.5 m\)
