High School

The table below shows the data for a car stopping on a wet road. What is the approximate stopping distance for a car traveling at 35 mph?

**Car Stopping Distances**

[tex]
\[
\begin{tabular}{|c|c|}
\hline
v & \begin{tabular}{c}
d \\
(ft)
\end{tabular} \\
\hline
15 & 17.9 \\
\hline
20 & 31.8 \\
\hline
50 & 198.7 \\
\hline
\end{tabular}
\]
[/tex]

Possible stopping distances:
- 41.7 ft
- 49.7 ft
- 97.4 ft
- 115.3 ft

Use the formula:

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

Calculate the stopping distance for 35 mph and choose the closest answer.

Answer :

To find the approximate stopping distance for a car traveling at 35 mph on a wet road, we use the formula provided:

[tex]\[ d = \frac{2.15 \times v^2}{64.4 \times f} \][/tex]

Where:
- [tex]\( d \)[/tex] is the stopping distance in feet.
- [tex]\( v \)[/tex] is the velocity of the car in mph.
- [tex]\( f \)[/tex] is the friction coefficient (which is often assumed to be 0.7 for wet roads).

Let's go through the solution step-by-step:

1. Identify the given values:
- Velocity ([tex]\( v \)[/tex]) = 35 mph
- Friction coefficient ([tex]\( f \)[/tex]) = 0.7

2. Understand the formula:
- This formula calculates the stopping distance using a combination of velocity ([tex]\( v \)[/tex]) and the friction coefficient ([tex]\( f \)[/tex]). The term [tex]\( 2.15 \times v^2 \)[/tex] represents a factor based on speed, while the denominator [tex]\( 64.4 \times f \)[/tex] accounts for road conditions.

3. Substitute the values into the formula:
[tex]\[
d = \frac{2.15 \times (35)^2}{64.4 \times 0.7}
\][/tex]

4. Calculate the stopping distance:
- First, calculate [tex]\( v^2 \)[/tex]: [tex]\( 35 \times 35 = 1225 \)[/tex]
- Then, multiply by 2.15: [tex]\( 2.15 \times 1225 = 2637.75 \)[/tex]
- Calculate the denominator: [tex]\( 64.4 \times 0.7 = 45.08 \)[/tex]
- Divide the results to find the stopping distance:
[tex]\[
d = \frac{2637.75}{45.08} \approx 58.42
\][/tex]

The approximate stopping distance for the car traveling at 35 mph on a wet road is approximately 58.4 feet.