1. **State the problem:** We have a motorist's journey represented on a distance-time graph from 07:30 to 11:00 am covering 70 km. The motorist stopped for breakfast between points A and B. 2. **Find the distance traveled before stopping for breakfast:** Point A is at (08:00, 5 km), so the motorist traveled 5 km before stopping. 3. **Calculate the slope (gradient) of each line segment:** The slope formula is $$\text{slope} = \frac{\text{change in distance}}{\text{change in time}} = \frac{\Delta y}{\Delta x}$$ - Convert time to hours from 07:30 (start time): - 07:30 = 0 hours - 08:00 = 0.5 hours - 09:30 = 2 hours - 11:00 = 3.5 hours (i) Slope of OA: $$\frac{5 - 0}{0.5 - 0} = \frac{5}{0.5} = 10$$ (ii) Slope of AB: $$\frac{10 - 5}{2 - 0.5} = \frac{5}{1.5} \approx 3.33$$ (iii) Slope of BC: $$\frac{70 - 10}{3.5 - 2} = \frac{60}{1.5} = 40$$ 4. **Interpretation:** - The motorist traveled 5 km before stopping. - The slope represents speed in km/h for each segment. - OA slope = 10 km/h, AB slope = 3.33 km/h (slow, indicating stop), BC slope = 40 km/h (fast after breakfast). **Final answers:** (a) Distance before breakfast = 5 km (b) Slopes: (i) OA = 10.00 (ii) AB = 3.33 (iii) BC = 40.00