The distance it takes a truck to stop can be modeled by the function:

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

where
- [tex]\(d\)[/tex] = stopping distance in feet
- [tex]\(v\)[/tex] = initial velocity in miles per hour
- [tex]\(f\)[/tex] = a constant related to friction

When the truck's initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft. Determine the value of [tex]\(f\)[/tex], rounded to the nearest hundredth.

[tex]\[ f \approx \square \][/tex]



Answer :

Sure! Let's break down the problem step by step to determine the value of [tex]\( f \)[/tex] given the stopping distance function:

The given function for stopping distance [tex]\( d \)[/tex] is:

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

We know from the problem statement:
- The initial velocity [tex]\( v \)[/tex] is [tex]\( 40 \)[/tex] mph.
- The stopping distance [tex]\( d \)[/tex] is [tex]\( 138 \)[/tex] feet.

Our goal is to determine the value of [tex]\( f \)[/tex], rounded to the nearest hundredth.

First, we substitute the known values into the stopping distance formula:

[tex]\[ 138 = \frac{2.15 \cdot 40^2}{64.4 f} \][/tex]

To isolate [tex]\( f \)[/tex], we'll carry out the following steps:

1. Compute [tex]\( 40^2 \)[/tex]:
[tex]\[ 40^2 = 1600 \][/tex]

2. Substitute [tex]\( 1600 \)[/tex] for [tex]\( 40^2 \)[/tex] in the equation:
[tex]\[ 138 = \frac{2.15 \cdot 1600}{64.4 f} \][/tex]

3. Calculate the product [tex]\( 2.15 \cdot 1600 \)[/tex]:
[tex]\[ 2.15 \cdot 1600 = 3440 \][/tex]

The equation now reads:
[tex]\[ 138 = \frac{3440}{64.4 f} \][/tex]

4. To solve for [tex]\( f \)[/tex], we need to isolate [tex]\( f \)[/tex] on one side of the equation. Multiply both sides of the equation by [tex]\( 64.4 f \)[/tex] to eliminate the denominator on the right-hand side:
[tex]\[ 138 \cdot 64.4 f = 3440 \][/tex]

5. Simplify the left-hand side:
[tex]\[ 8887.2 f = 3440 \][/tex]

6. Now, divide both sides by [tex]\( 8887.2 \)[/tex] to isolate [tex]\( f \)[/tex]:
[tex]\[ f = \frac{3440}{8887.2} \][/tex]

7. Perform the division:
[tex]\[ f \approx 0.3870735439733549 \][/tex]

Finally, round the value of [tex]\( f \)[/tex] to the nearest hundredth:
[tex]\[ f \approx 0.39 \][/tex]

Thus, the value of [tex]\( f \)[/tex], rounded to the nearest hundredth, is:

[tex]\[ f \approx 0.39 \][/tex]