- Rearrange the given formula to isolate the constant coefficient: \(f = \frac{2.15 v^2}{64.4 d}\).
- Calculate \(f\) for each provided pair of speed and distance data points.
- Average the resulting values of \(f\) to find a consistent estimate.
- Round the final result to the nearest hundredth, giving \(\boxed{0.35}\).
Explanation
- Understanding the Problem and Data
The problem provides a table of stopping distances \(d\) at different speeds \(v\) for a car on a wet road. The relationship is governed by the formula \[d(v) = \frac{2.15 v^2}{64.4 f}\] where \(f\) is a constant coefficient. Our objective is to determine the value of \(f\) based on these data points, rounded to the nearest hundredth.
The given data points are:
- For \(v = 20\), \(d = 38\)
- For \(v = 30\), \(d = 86\)
- For \(v = 40\), \(d = 153\)
- Calculating and Verifying the Coefficient f
To isolate the variable \(f\), we rearrange the given formula \(d = \frac{2.15 v^2}{64.4 f}\) by multiplying both sides by \(f\) and dividing by \(d\):
\[f = \frac{2.15 v^2}{64.4 d}\]
Now, we calculate the value of \(f\) using each of the provided data points to check for consistency.
- Using \(v = 20, d = 38\):
\[f = \frac{2.15 \times 20^2}{64.4 \times 38} = \frac{2.15 \times 400}{2447.2} = \frac{860}{2447.2} \approx 0.3514\]
- Using \(v = 30, d = 86\):
\[f = \frac{2.15 \times 30^2}{64.4 \times 86} = \frac{2.15 \times 900}{5538.4} = \frac{1935}{5538.4} \approx 0.3494\]
- Using \(v = 40, d = 153\):
\[f = \frac{2.15 \times 40^2}{64.4 \times 153} = \frac{2.15 \times 1600}{9853.2} = \frac{3440}{9853.2} \approx 0.3491\]
- Conclusion
The calculated values for \(f\) are approximately \(0.3514\), \(0.3494\), and \(0.3491\). These values are very close to each other. Taking the average of these values gives:
\[f_{avg} = \frac{0.3514 + 0.3494 + 0.3491}{3} \approx 0.3500\]
Rounding to the nearest hundredth, we get \(0.35\). Thus, the constant coefficient \(f\) is approximately \(\boxed{0.35}\).
Examples
Understanding the relationship between speed and stopping distance is crucial for road safety engineering. By using models like this, traffic engineers can calculate the safe speed limits for specific road conditions (such as wet pavement) to minimize accidents. This mathematical approach helps in designing safer infrastructure and setting appropriate driving regulations to ensure everyone reaches their destination safely.