Question: Two window washers start at the heights shown. One is rising, the other is descending. How long does it take for the two window washers to reach the same height? Explain.
21 ft high
rising 8 in./s
50 ft high
descending 11 in./s
After t seconds, the height of the rising window washer is exactly $21 + \frac{2}{3} t$ feet, and the height of the descending window washer is exactly $50 - \frac{2}{3} t$ feet. Setting these expressions equal to each other and solving for $t$ yields the exact value $t = \frac{348}{19}$. Rounding to one decimal place, the window washers reach the same height after about 18.3 seconds.
(Simplify your answers.)
1. **State the problem:** Two window washers start at different heights: one at 21 ft rising at 8 in./s, the other at 50 ft descending at 11 in./s. We want to find the time $t$ when they reach the same height.
2. **Convert speeds to feet per second:** Since 12 inches = 1 foot,
- Rising speed: $8 \text{ in./s} = \frac{8}{12} = \frac{2}{3}$ ft/s
- Descending speed: $11 \text{ in./s} = \frac{11}{12}$ ft/s
3. **Write height expressions after $t$ seconds:**
- Rising washer height: $21 + \frac{2}{3} t$
- Descending washer height: $50 - \frac{11}{12} t$
4. **Set heights equal to find $t$: $$21 + \frac{2}{3} t = 50 - \frac{11}{12} t$$**
5. **Solve for $t$: Move terms to one side:**
$$\frac{2}{3} t + \frac{11}{12} t = 50 - 21$$
6. **Find common denominator for left side:**
$$\frac{8}{12} t + \frac{11}{12} t = 29$$
7. **Combine terms:**
$$\frac{19}{12} t = 29$$
8. **Isolate $t$: $$t = 29 \times \frac{12}{19} = \frac{348}{19}$$**
9. **Simplify and approximate:**
$$\frac{348}{19} \approx 18.3 \text{ seconds}$$
**Answer:** The window washers reach the same height after about 18.3 seconds.