Answer :
We are given the formula for the stopping distance:
[tex]$$
d(v)=\frac{2.15\,v^2}{64.4\,f},
$$[/tex]
where [tex]$v$[/tex] is the speed in mph, [tex]$d$[/tex] is the stopping distance in feet, and [tex]$f$[/tex] is the coefficient of friction on the road. We also have table data for a car traveling at 20 mph with a stopping distance of 31.8 ft. We can use this data to first find [tex]$f$[/tex].
Step 1. Determine the coefficient of friction [tex]$f$[/tex]:
For [tex]$v=20$[/tex] mph and [tex]$d=31.8$[/tex] ft, substitute into the formula:
[tex]$$
31.8 = \frac{2.15 \times (20)^2}{64.4\, f}.
$$[/tex]
Solve for [tex]$f$[/tex] as follows:
[tex]\[
f = \frac{2.15 \times 20^2}{64.4 \times 31.8}.
\][/tex]
First, compute [tex]$20^2$[/tex]:
[tex]$$
20^2 = 400.
$$[/tex]
Then the numerator is:
[tex]$$
2.15 \times 400 = 860.
$$[/tex]
And the denominator is:
[tex]$$
64.4 \times 31.8.
$$[/tex]
Thus,
[tex]$$
f = \frac{860}{64.4 \times 31.8}.
$$[/tex]
Evaluating this expression gives a value for [tex]$f \approx 0.42.$[/tex]
---
Step 2. Compute the stopping distance for [tex]$v = 35$[/tex] mph:
Now use the formula with [tex]$v=35$[/tex] mph:
[tex]$$
d(35)=\frac{2.15 \times (35)^2}{64.4\,f}.
$$[/tex]
First, calculate [tex]$35^2$[/tex]:
[tex]$$
35^2 = 1225.
$$[/tex]
Then the numerator becomes:
[tex]$$
2.15 \times 1225 = 2633.75.
$$[/tex]
Substitute [tex]$f\approx 0.42$[/tex] in the denominator:
[tex]$$
\text{Denominator} = 64.4 \times 0.42 \approx 27.04.
$$[/tex]
Thus, the stopping distance is:
[tex]$$
d(35)=\frac{2633.75}{27.04}\approx 97.4 \text{ ft}.
$$[/tex]
---
Final Answer:
The approximate stopping distance for a car traveling 35 mph on a wet road is about
[tex]$$\boxed{97.4\text{ ft}}.$$[/tex]
[tex]$$
d(v)=\frac{2.15\,v^2}{64.4\,f},
$$[/tex]
where [tex]$v$[/tex] is the speed in mph, [tex]$d$[/tex] is the stopping distance in feet, and [tex]$f$[/tex] is the coefficient of friction on the road. We also have table data for a car traveling at 20 mph with a stopping distance of 31.8 ft. We can use this data to first find [tex]$f$[/tex].
Step 1. Determine the coefficient of friction [tex]$f$[/tex]:
For [tex]$v=20$[/tex] mph and [tex]$d=31.8$[/tex] ft, substitute into the formula:
[tex]$$
31.8 = \frac{2.15 \times (20)^2}{64.4\, f}.
$$[/tex]
Solve for [tex]$f$[/tex] as follows:
[tex]\[
f = \frac{2.15 \times 20^2}{64.4 \times 31.8}.
\][/tex]
First, compute [tex]$20^2$[/tex]:
[tex]$$
20^2 = 400.
$$[/tex]
Then the numerator is:
[tex]$$
2.15 \times 400 = 860.
$$[/tex]
And the denominator is:
[tex]$$
64.4 \times 31.8.
$$[/tex]
Thus,
[tex]$$
f = \frac{860}{64.4 \times 31.8}.
$$[/tex]
Evaluating this expression gives a value for [tex]$f \approx 0.42.$[/tex]
---
Step 2. Compute the stopping distance for [tex]$v = 35$[/tex] mph:
Now use the formula with [tex]$v=35$[/tex] mph:
[tex]$$
d(35)=\frac{2.15 \times (35)^2}{64.4\,f}.
$$[/tex]
First, calculate [tex]$35^2$[/tex]:
[tex]$$
35^2 = 1225.
$$[/tex]
Then the numerator becomes:
[tex]$$
2.15 \times 1225 = 2633.75.
$$[/tex]
Substitute [tex]$f\approx 0.42$[/tex] in the denominator:
[tex]$$
\text{Denominator} = 64.4 \times 0.42 \approx 27.04.
$$[/tex]
Thus, the stopping distance is:
[tex]$$
d(35)=\frac{2633.75}{27.04}\approx 97.4 \text{ ft}.
$$[/tex]
---
Final Answer:
The approximate stopping distance for a car traveling 35 mph on a wet road is about
[tex]$$\boxed{97.4\text{ ft}}.$$[/tex]
