1. **State the problem:** We want to find the number of seconds in which the number of clicks heard equals the speed of the train in miles per hour. 2. **Understand the problem:** Each click corresponds to the train passing over a joint between two rails. Each rail is 30 feet long. 3. **Convert units:** Speed in miles per hour (mph) needs to be related to clicks per second. 4. **Calculate the distance per click:** Each click corresponds to 30 feet traveled. 5. **Convert speed from mph to feet per second (fps):** $$\text{Speed (fps)} = \text{Speed (mph)} \times \frac{5280 \text{ feet}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} = \text{Speed (mph)} \times \frac{5280}{3600} = \text{Speed (mph)} \times 1.4667$$ 6. **Relate clicks per second to speed:** Clicks per second = $$\frac{\text{Speed (fps)}}{30} = \frac{\text{Speed (mph)} \times 1.4667}{30} = \text{Speed (mph)} \times 0.0489$$ 7. **Find the number of seconds for clicks to equal speed:** Let $t$ be the number of seconds. Clicks in $t$ seconds = speed in mph $$\text{Clicks per second} \times t = \text{Speed (mph)}$$ Substitute clicks per second: $$\text{Speed (mph)} \times 0.0489 \times t = \text{Speed (mph)}$$ Divide both sides by $\text{Speed (mph)}$ (assuming speed > 0): $$\cancel{\text{Speed (mph)}} \times 0.0489 \times t = \cancel{\text{Speed (mph)}}$$ $$0.0489 \times t = 1$$ 8. **Solve for $t$:** $$t = \frac{1}{0.0489} \approx 20.45$$ **Final answer:** The number of seconds in which the number of clicks heard equals the speed in miles per hour is approximately **20 seconds**.