Question
A 3-ft wide circular track for a camera dolly is set up for a movie scene. The two rails of the track form concentric circles. The radius of the inner circle is 52 ft. How much farther does a wheel on the outer rail travel than a wheel on the inner rail of the track in one turn?
52 ft
3 ft
Question content area bottom
Part 1
The wheel on the outer rail travels
enter your response here feet farther than the wheel on the inner rail of the track in one turn.
(Round to the nearest tenth as needed.)
52 ft
3 ft
Question content area bottom
Part 1
The wheel on the outer rail travels
enter your response here feet farther than the wheel on the inner rail of the track in one turn.
(Round to the nearest tenth as needed.)
Answers
GPT-4o mini
To determine how much farther a wheel on the outer rail travels compared to a wheel on the inner rail in one turn, we first need to calculate the circumferences of both circles.
1. **Inner circle's radius**: \( r_{inner} = 52 \) ft
The circumference \( C_{inner} \) of the inner circle is given by the formula:
\[
C_{inner} = 2 \pi r_{inner} = 2 \pi (52)
\]
2. **Outer circle's radius**: Since the track is 3 ft wide, the radius of the outer circle is:
\[
r_{outer} = r_{inner} + 3 = 52 + 3 = 55 \text{ ft}
\]
The circumference \( C_{outer} \) of the outer circle is given by:
\[
C_{outer} = 2 \pi r_{outer} = 2 \pi (55)
\]
3. **Calculate the difference in circumferences**:
The distance that the wheel on the outer rail travels farther than the wheel on the inner rail is:
\[
\text{Distance} = C_{outer} - C_{inner}
\]
Let's calculate \( C_{inner} \) and \( C_{outer} \):
- For the inner circle:
\[
C_{inner} = 2 \pi (52) \approx 2 \times 3.14159 \times 52 \approx 326.72 \text{ ft}
\]
- For the outer circle:
\[
C_{outer} = 2 \pi (55) \approx 2 \times 3.14159 \times 55 \approx 345.58 \text{ ft}
\]
4. **Now subtract the two results**:
\[
\text{Distance} = C_{outer} - C_{inner} \approx 345.58 - 326.72 \approx 18.86 \text{ ft}
\]
5. **Round to the nearest tenth**:
\[
\text{Distance} \approx 18.9 \text{ ft}
\]
Thus, the wheel on the outer rail travels approximately **18.9 feet** farther than the wheel on the inner rail of the track in one turn.
1. **Inner circle's radius**: \( r_{inner} = 52 \) ft
The circumference \( C_{inner} \) of the inner circle is given by the formula:
\[
C_{inner} = 2 \pi r_{inner} = 2 \pi (52)
\]
2. **Outer circle's radius**: Since the track is 3 ft wide, the radius of the outer circle is:
\[
r_{outer} = r_{inner} + 3 = 52 + 3 = 55 \text{ ft}
\]
The circumference \( C_{outer} \) of the outer circle is given by:
\[
C_{outer} = 2 \pi r_{outer} = 2 \pi (55)
\]
3. **Calculate the difference in circumferences**:
The distance that the wheel on the outer rail travels farther than the wheel on the inner rail is:
\[
\text{Distance} = C_{outer} - C_{inner}
\]
Let's calculate \( C_{inner} \) and \( C_{outer} \):
- For the inner circle:
\[
C_{inner} = 2 \pi (52) \approx 2 \times 3.14159 \times 52 \approx 326.72 \text{ ft}
\]
- For the outer circle:
\[
C_{outer} = 2 \pi (55) \approx 2 \times 3.14159 \times 55 \approx 345.58 \text{ ft}
\]
4. **Now subtract the two results**:
\[
\text{Distance} = C_{outer} - C_{inner} \approx 345.58 - 326.72 \approx 18.86 \text{ ft}
\]
5. **Round to the nearest tenth**:
\[
\text{Distance} \approx 18.9 \text{ ft}
\]
Thus, the wheel on the outer rail travels approximately **18.9 feet** farther than the wheel on the inner rail of the track in one turn.