1. **Problem Statement:** We have a motorist's journey graph showing distance traveled over time from 7:30 am to 11:00 am. (a) Find the distance traveled before stopping for breakfast. (b) Find the gradient (slope) of line segments OA, AB, and BC. 2. **Understanding the graph points:** - O = (7:30 am, 0 km) which we take as (0 hours, 0 km) - A = (8:00 am, 15 km) which is (0.5 hours, 15 km) - B = (9:00 am, 20 km) which is (1.5 hours, 20 km) - C = (11:00 am, 70 km) which is (3.5 hours, 70 km) 3. **(a) Distance before breakfast:** The motorist stopped for breakfast after point B, so distance before breakfast is distance at B. $$\text{Distance before breakfast} = 20 \text{ km}$$ 4. **(b) Gradient (slope) formula:** The gradient between two points $(t_1, d_1)$ and $(t_2, d_2)$ is: $$\text{slope} = \frac{d_2 - d_1}{t_2 - t_1}$$ 5. **Calculate slopes:** (i) OA: from O(0,0) to A(0.5,15) $$\text{slope}_{OA} = \frac{15 - 0}{0.5 - 0} = \frac{15}{0.5} = 30$$ (ii) AB: from A(0.5,15) to B(1.5,20) $$\text{slope}_{AB} = \frac{20 - 15}{1.5 - 0.5} = \frac{5}{1} = 5$$ (iii) BC: from B(1.5,20) to C(3.5,70) $$\text{slope}_{BC} = \frac{70 - 20}{3.5 - 1.5} = \frac{50}{2} = 25$$ 6. **Final answers:** (a) Distance before breakfast = 20 km (b) Slopes: - OA = 30.00 - AB = 5.00 - BC = 25.00