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19. The tower at Philadelphia City Hall contains four clocks that have a radius of about 3.96 meters. Find how far the minute hand travels after each number of rotations around the clock face. Round to the nearest hundredth. \( \begin{array}{lll}\text { a. } 2 \text { rotations } & \text { b. } \frac{1}{2} \text { rotation } & \text { c. } 5 \frac{3}{4} \text { rotations }\end{array} \)

Ask by Graham Perkins. in the United States
Mar 11,2025

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Answer

- **a.** 49.74 meters - **b.** 12.44 meters - **c.** 143.00 meters

Solution

The first step is to find the circumference of the circle traced by the tip of the minute hand. The circumference is given by \[ C = 2\pi r, \] where \(r = 3.96\) meters. Thus, \[ C = 2\pi (3.96) \approx 2 \times 3.14159 \times 3.96 \approx 24.87 \text{ meters}. \] Now we calculate the distance traveled for each case. **a. 2 rotations** The distance traveled is \[ \text{Distance} = 2 \times C \approx 2 \times 24.87 \approx 49.74 \text{ meters}. \] **b. \(\frac{1}{2}\) rotation** The distance traveled is \[ \text{Distance} = \frac{1}{2} \times C \approx 0.5 \times 24.87 \approx 12.44 \text{ meters}. \] **c. \(5\frac{3}{4}\) rotations** First, convert \(5\frac{3}{4}\) to a decimal: \[ 5\frac{3}{4} = 5.75. \] Then the distance traveled is \[ \text{Distance} = 5.75 \times C \approx 5.75 \times 24.87 \approx 143.00 \text{ meters}. \] Thus, the answers rounded to the nearest hundredth are: - **a.** 49.74 meters - **b.** 12.44 meters - **c.** 143.00 meters

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To find how far the minute hand travels after a certain number of rotations, we first need to calculate the circumference of the clock face using the formula \( C = 2\pi r \), where \( r \) is the radius. In this case, \( r \) is approximately 3.96 meters. Calculating the circumference: \( C = 2\pi(3.96) \approx 24.86 \) meters. Now we can use this to find the distance traveled for each specific number of rotations. For two rotations: Distance = \( 2 \times C = 2 \times 24.86 \approx 49.72 \) meters. For half a rotation: Distance = \( \frac{1}{2} \times C \approx \frac{1}{2} \times 24.86 \approx 12.43 \) meters. For \( 5 \frac{3}{4} \) rotations: Distance = \( 5.75 \times C \approx 5.75 \times 24.86 \approx 143.38 \) meters. So the distances are approximately: a. 49.72 meters b. 12.43 meters c. 143.38 meters

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