Answer :
We are given a formula for the stopping distance of a car moving at a speed [tex]$v$[/tex]:
[tex]$$
d(v)=\frac{2.15\,v^2}{64.4\,f},
$$[/tex]
where [tex]$f$[/tex] is the friction coefficient. The table provides a data point that will allow us to determine [tex]$f$[/tex]. In particular, when [tex]$v=15\text{ mph}$[/tex] the stopping distance is [tex]$d(15)=17.9\text{ ft}$[/tex].
### Step 1. Solve for the Friction Coefficient [tex]$f$[/tex]
Using [tex]$v=15\text{ mph}$[/tex] and [tex]$d(15)=17.9\text{ ft}$[/tex] in the formula, we have:
[tex]$$
17.9=\frac{2.15 \cdot 15^2}{64.4\,f}.
$$[/tex]
To solve for [tex]$f$[/tex], multiply both sides by [tex]$64.4\,f$[/tex]:
[tex]$$
17.9 \times 64.4\,f = 2.15 \cdot 15^2.
$$[/tex]
Now, isolate [tex]$f$[/tex] by dividing both sides by [tex]$17.9 \times 64.4$[/tex]:
[tex]$$
f = \frac{2.15 \cdot 15^2}{64.4 \times 17.9}.
$$[/tex]
Evaluating the numbers, we obtain a friction coefficient of approximately
[tex]$$
f \approx 0.42.
$$[/tex]
### Step 2. Compute the Stopping Distance for 35 mph
Now that we have determined [tex]$f \approx 0.42$[/tex], we can compute the stopping distance when [tex]$v = 35\text{ mph}$[/tex]. Substitute [tex]$v=35$[/tex] and the computed friction coefficient into the formula:
[tex]$$
d(35)=\frac{2.15 \cdot 35^2}{64.4\,f}.
$$[/tex]
Substitute [tex]$f \approx 0.42$[/tex] into the equation:
[tex]$$
d(35) \approx \frac{2.15 \cdot 35^2}{64.4 \times 0.42}.
$$[/tex]
After performing the calculation, the stopping distance comes out to be approximately
[tex]$$
d(35) \approx 97.4\text{ ft}.
$$[/tex]
### Final Answer
The stopping distance for a car traveling at 35 mph is approximately
[tex]$$
\boxed{97.4\text{ ft}}.
$$[/tex]
[tex]$$
d(v)=\frac{2.15\,v^2}{64.4\,f},
$$[/tex]
where [tex]$f$[/tex] is the friction coefficient. The table provides a data point that will allow us to determine [tex]$f$[/tex]. In particular, when [tex]$v=15\text{ mph}$[/tex] the stopping distance is [tex]$d(15)=17.9\text{ ft}$[/tex].
### Step 1. Solve for the Friction Coefficient [tex]$f$[/tex]
Using [tex]$v=15\text{ mph}$[/tex] and [tex]$d(15)=17.9\text{ ft}$[/tex] in the formula, we have:
[tex]$$
17.9=\frac{2.15 \cdot 15^2}{64.4\,f}.
$$[/tex]
To solve for [tex]$f$[/tex], multiply both sides by [tex]$64.4\,f$[/tex]:
[tex]$$
17.9 \times 64.4\,f = 2.15 \cdot 15^2.
$$[/tex]
Now, isolate [tex]$f$[/tex] by dividing both sides by [tex]$17.9 \times 64.4$[/tex]:
[tex]$$
f = \frac{2.15 \cdot 15^2}{64.4 \times 17.9}.
$$[/tex]
Evaluating the numbers, we obtain a friction coefficient of approximately
[tex]$$
f \approx 0.42.
$$[/tex]
### Step 2. Compute the Stopping Distance for 35 mph
Now that we have determined [tex]$f \approx 0.42$[/tex], we can compute the stopping distance when [tex]$v = 35\text{ mph}$[/tex]. Substitute [tex]$v=35$[/tex] and the computed friction coefficient into the formula:
[tex]$$
d(35)=\frac{2.15 \cdot 35^2}{64.4\,f}.
$$[/tex]
Substitute [tex]$f \approx 0.42$[/tex] into the equation:
[tex]$$
d(35) \approx \frac{2.15 \cdot 35^2}{64.4 \times 0.42}.
$$[/tex]
After performing the calculation, the stopping distance comes out to be approximately
[tex]$$
d(35) \approx 97.4\text{ ft}.
$$[/tex]
### Final Answer
The stopping distance for a car traveling at 35 mph is approximately
[tex]$$
\boxed{97.4\text{ ft}}.
$$[/tex]
