$$ 1 \leftarrow $$\begin{tabular}{l} Two cars are racing around a circular track. The red car is traveling in a lane that is 79 ft from the center. The blue car is traveling in a lane that is 89 ft from the center. If both cars \ are traveling at 120 mph , how much later, in seconds, will the blue car finish if they race one lap around the track? (Hint: $$ 60 \mathrm{mph} \approx 88 \mathrm{fps} $$.) \ $$ \square $$ sec \ (Simplify your answer. Round to the nearest tenth as needed.) \ \hline\end{tabular}

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UpStudy AI · Accepted Answer

Let the distance each car travels be the circumference of its circular path.

1. For the red car, with radius $$ r = 79 $$ ft, the circumference is  
   $$
   C_{\text{red}} = 2\pi(79)
   $$  
   For the blue car, with radius $$ r = 89 $$ ft, the circumference is  
   $$
   C_{\text{blue}} = 2\pi(89)
   $$

2. Both cars are traveling at 120 mph. Given that 60 mph is approximately 88 fps, we have  
   $$
   120 \text{ mph} \approx 2 \times 88 = 176 \text{ fps}
   $$

3. The time taken by each car to complete one lap is given by the formula  
   $$
   \text{Time} = \frac{\text{Distance}}{\text{Speed}}
   $$
   Thus, for the red car:  
   $$
   T_{\text{red}} = \frac{2\pi(79)}{176}
   $$
   and for the blue car:  
   $$
   T_{\text{blue}} = \frac{2\pi(89)}{176}
   $$

4. The difference in time between the blue car and the red car is:  
   $$
   \Delta T = T_{\text{blue}} - T_{\text{red}} = \frac{2\pi(89) - 2\pi(79)}{176} = \frac{2\pi(89-79)}{176} = \frac{2\pi(10)}{176} = \frac{20\pi}{176}
   $$
   Simplify the fraction:  
   $$
   \frac{20\pi}{176} = \frac{5\pi}{44}
   $$

5. Approximating using $$\pi \approx 3.1416$$:  
   $$
   \Delta T \approx \frac{5 \times 3.1416}{44} \approx \frac{15.708}{44} \approx 0.357 \text{ seconds}
   $$  
   Rounding to the nearest tenth of a second, the blue car finishes approximately $$0.4$$ seconds later.

$$
\boxed{0.4 \text{ sec}}
$$
The blue car finishes approximately 0.4 seconds later than the red car.

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