1. **Problem statement:** A vehicle is traveling at 35 mph (56.32 kph). The driver has a perception and reaction time of 1.6 seconds. The deceleration rate during braking is 22.54 feet per second squared (fps²). We need to find the total stopping distance. 2. **Understanding the problem:** The total stopping distance consists of two parts: - The distance traveled during the driver's perception and reaction time (reaction distance). - The distance traveled while the vehicle is braking to a stop (braking distance). 3. **Formulas used:** - Reaction distance: $$d_{reaction} = v \times t$$ where $v$ is velocity and $t$ is reaction time. - Braking distance: $$d_{braking} = \frac{v^2}{2a}$$ where $a$ is the deceleration rate. 4. **Convert velocity to consistent units:** Given velocity $v = 35$ mph. Convert mph to feet per second (fps): $$v = 35 \times \frac{5280}{3600} = 35 \times 1.4667 = 51.33\, \text{fps}$$ 5. **Calculate reaction distance:** $$d_{reaction} = v \times t = 51.33 \times 1.6 = 82.13\, \text{feet}$$ 6. **Calculate braking distance:** $$d_{braking} = \frac{v^2}{2a} = \frac{(51.33)^2}{2 \times 22.54} = \frac{2635.7}{45.08} = 58.47\, \text{feet}$$ 7. **Calculate total stopping distance:** $$d_{total} = d_{reaction} + d_{braking} = 82.13 + 58.47 = 140.6\, \text{feet}$$ **Final answer:** The total stopping distance is approximately **140.6 feet**.