Sure, let's solve this problem step-by-step.
### 1. Rewrite the Formula as a Fraction Exponent
The given formula is:
[tex]\[ S = \sqrt{30 \cdot d \cdot f} \][/tex]
We know that a square root can be represented as a fractional exponent. Specifically, the square root of a variable [tex]\(x\)[/tex] can be written as [tex]\(x^{1/2}\)[/tex]. Therefore, rewriting the formula, we get:
[tex]\[ S = (30 \cdot d \cdot f)^{1/2} \][/tex]
### 2. Estimate the Speed of the Car Based on the Length of Skid Marks
We need to fill in the table for the estimated speed of the car based on the length of skid marks. We are given that it was a clear and dry day, so the coefficient of friction ([tex]\(f\)[/tex]) is 1.0.
#### Skid Mark Lengths and Corresponding Speeds:
1. 10 feet:
[tex]\[
S = (30 \cdot 10 \cdot 1.0)^{1/2}
\][/tex]
[tex]\[
S = (300)^{1/2}
\][/tex]
[tex]\[
S = \sqrt{300} \approx 17.32 \, \text{mph}
\][/tex]
2. 25 feet:
[tex]\[
S = (30 \cdot 25 \cdot 1.0)^{1/2}
\][/tex]
[tex]\[
S = (750)^{1/2}
\][/tex]
[tex]\[
S = \sqrt{750} \approx 27.39 \, \text{mph}
\][/tex]
3. 40 feet:
[tex]\[
S = (30 \cdot 40 \cdot 1.0)^{1/2}
\][/tex]
[tex]\[
S = (1200)^{1/2}
\][/tex]
[tex]\[
S = \sqrt{1200} \approx 34.64 \, \text{mph}
\][/tex]
4. 70 feet:
[tex]\[
S = (30 \cdot 70 \cdot 1.0)^{1/2}
\][/tex]
[tex]\[
S = (2100)^{1/2}
\][/tex]
[tex]\[
S = \sqrt{2100} \approx 45.83 \, \text{mph}
\][/tex]
5. 130 feet:
[tex]\[
S = (30 \cdot 130 \cdot 1.0)^{1/2}
\][/tex]
[tex]\[
S = (3900)^{1/2}
\][/tex]
[tex]\[
S = \sqrt{3900} \approx 62.45 \, \text{mph}
\][/tex]
Now, we can fill in the table with these calculated speeds:
\begin{tabular}{|l|l|}
\hline Skid Mark Lengths (feet) & Speed of Car (mph) \\
\hline 10 ft & 17.32 \\
\hline 25 ft & 27.39 \\
\hline 40 ft & 34.64 \\
\hline 70 ft & 45.83 \\
\hline 130 ft & 62.45 \\
\hline
\end{tabular}
