Question
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The wind speed near the center of a tornado is represented by the equation \( S=93 \log d+65 \), where \( d \) is the distance, in miles, that the tornado travels and \( S \) is the wind speed, in miles per hour. If the tornado traveled 5.2 miles, what was its wind speed in miles per hour? Round your answer to one decimal place. You do not need to include units.

Ask by Gordon Santiago. in the United States
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The wind speed is approximately 131.6 miles per hour.

Solution

Substitute \( d=5.2 \) into the equation \( S=93 \log d+65 \). Evaluate the expression by following steps: - step0: Evaluate: \(S=93\log_{10}{\left(d\right)}+65\) - step1: Substitute: \(S=93\log_{10}{\left(5.2\right)}+65\) - step2: Convert the expressions: \(S=93\log_{10}{\left(\frac{26}{5}\right)}+65\) - step3: Simplify: \(S=93\log_{10}{\left(26\right)}-93\log_{10}{\left(5\right)}+65\) - step4: Check if the solution is in the defined range: \(S=93\log_{10}{\left(26\right)}-93\log_{10}{\left(5\right)}+65,d>0\) - step5: Find the intersection: \(\left\{ \begin{array}{l}S=93\log_{10}{\left(26\right)}-93\log_{10}{\left(5\right)}+65\\d>0\end{array}\right.\) The wind speed near the center of the tornado when it traveled 5.2 miles is approximately 131.588311 miles per hour. Rounding this to one decimal place, the wind speed is approximately 131.6 miles per hour.

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Beyond the Answer

To find the wind speed \( S \) when the tornado traveled 5.2 miles, we can substitute \( d = 5.2 \) into the equation: \[ S = 93 \log(5.2) + 65 \] First, calculate \( \log(5.2) \), which is approximately 0.715. Now plug this value into the equation: \[ S = 93 \times 0.715 + 65 \approx 66.495 + 65 \approx 131.495 \] Rounding to one decimal place, the wind speed is approximately 131.5.

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