1. **State the problem:** Flaumback drives to a job interview in 1.5 hours at 60 km/h. We want to find how many minutes longer the drive takes if he drives at 50 km/h instead. 2. **Write down the known values:** - Speed 1: $v_1 = 60$ km/h - Time 1: $t_1 = 1.5$ hours - Speed 2: $v_2 = 50$ km/h 3. **Find the distance using the first speed and time:** $$d = v_1 \times t_1 = 60 \times 1.5 = 90 \text{ km}$$ 4. **Find the time taken at the second speed:** $$t_2 = \frac{d}{v_2} = \frac{90}{50}$$ 5. **Simplify the fraction:** $$t_2 = \frac{\cancel{90}}{\cancel{50}} = \frac{9}{5} = 1.8 \text{ hours}$$ 6. **Calculate the difference in time:** $$\Delta t = t_2 - t_1 = 1.8 - 1.5 = 0.3 \text{ hours}$$ 7. **Convert the time difference to minutes:** $$0.3 \times 60 = 18 \text{ minutes}$$ **Final answer:** The drive takes 18 minutes longer at 50 km/h.