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We appreciate your visit to The table below shows the data for a car traveling at different speeds and their respective stopping distances Car Stopping Distances tex begin array c. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

The table below shows the data for a car traveling at different speeds and their respective stopping distances.

**Car Stopping Distances**

[tex]
\[
\begin{array}{|c|c|}
\hline
\text{Speed } (v \, \text{mph}) & \text{Stopping Distance } (d \, \text{ft}) \\
\hline
15 & 17.9 \\
\hline
20 & 31.8 \\
\hline
50 & 198.7 \\
\hline
\end{array}
\]
[/tex]

The formula for stopping distance is given by:

[tex]d(v) = \frac{2.15 v^2}{64.4}[/tex]

Calculate the stopping distance for a car traveling at 35 mph:

A. 41.7 ft
B. 49.7 ft
C. 97.4 ft
D. 115.3 ft

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]

Thanks for taking the time to read The table below shows the data for a car traveling at different speeds and their respective stopping distances Car Stopping Distances tex begin array c. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada