- Identify the variables: \(v = 47\), \(f = 0.34\).
- Substitute these into the formula \(d(v) = \frac{2.15 v^2}{58.4 f}\).
- Calculate the numerator: \(2.15 \times 47^2 = 4749.35\).
- Calculate the denominator: \(58.4 \times 0.34 = 19.856\).
- Divide the numerator by the denominator to get \(d \approx 239\) feet.
The final answer is \(\boxed{239 \text{ feet}}\).
Explanation
- Analysis of the Problem
The problem asks to calculate the stopping distance of a car using the formula \(d(v) = \frac{2.15 v^2}{58.4 f}\), where \(v\) is the initial velocity in mph and \(f\) is a constant related to friction. Given the values \(v = 47\) mph and \(f = 0.34\), we need to substitute these into the equation to find the value of \(d\).
- Step-by-Step Calculation
We begin by substituting the given variables into the formula:
\[d = \frac{2.15 \times (47)^2}{58.4 \times 0.34}\]
First, compute the square of the initial velocity:
\[47^2 = 2209\]
Next, calculate the numerator:
\[2.15 \times 2209 = 4749.35\]
Then, calculate the denominator:
\[58.4 \times 0.34 = 19.856\]
Finally, divide the numerator by the denominator:
\[d = \frac{4749.35}{19.856} \approx 239.19 \text{ feet}\]
The approximate stopping distance is \(239\) feet. Comparing this to the given options, we find it corresponds to option (a).
Examples
Stopping distance is a critical concept in road safety and traffic engineering. When designing roads, traffic engineers use such formulas to determine appropriate speed limits and safe distances between vehicles. For instance, if a driver sees an obstacle on the road, knowing how long it takes for the car to stop at a given speed helps them make better decisions while driving, contributing to safer traffic environments. By calculating these distances, we can better understand how different factors like velocity and road friction influence safety.