The distance it takes a truck to stop can be modeled by the function $d(v)=\frac{2.15 v^2}{64.4 f}$.

$d =$ stopping distance in feet
$v =$ initial velocity in miles per hour
$f =$ 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 $f$, rounded to the nearest hundredth.

$f \approx$

Question

03-11-2026
Mathematics

Answered

The distance it takes a truck to stop can be modeled by the function $d(v)=\frac{2.15 v^2}{64.4 f}$.

$d =$ stopping distance in feet
$v =$ initial velocity in miles per hour
$f =$ 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 $f$, rounded to the nearest hundredth.

$f \approx$

Verified Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-03T15:02:34-05:00

Substitute the given values $v = 40$ and $d = 138$ into the formula $d = \frac{2.15 v^2}{64.4 f}$.
Simplify the expression to get $138 = \frac{3440}{64.4 f}$.
Isolate $f$ by solving the equation $8887.2 f = 3440$.
Round the resulting value to the nearest hundredth to obtain $f \approx \boxed{0.39}$.

Explanation

Problem Analysis and Data Overview

We are given the model for the stopping distance of a truck: $d(v) = \frac{2.15 v^2}{64.4 f}$, where $d$ is the stopping distance in feet, $v$ is the initial velocity in miles per hour, and $f$ is a constant related to friction. We are provided with the condition that when the truck's velocity is $v = 40 \text{ mph}$, the stopping distance is $d = 138 \text{ ft}$. Our objective is to determine the value of the constant $f$ and round it to the nearest hundredth.

Substitution and Simplification

To find the value of $f$, we substitute the known values $v = 40$ and $d = 138$ into the given formula:
\[138 = \frac{2.15 \times (40)^2}{64.4 f}\]
First, calculate the numerator: $(40)^2 = 1600$, so $2.15 \times 1600 = 3440$.
The equation becomes:
\[138 = \frac{3440}{64.4 f}\]

Calculating the Value of f

Next, isolate the variable $f$. Multiply both sides of the equation by $64.4 f$:
\[138 \times 64.4 \times f = 3440\]
Calculate the product: $138 \times 64.4 = 8887.2$.
Now the equation is:
\[8887.2 \times f = 3440\]
Solve for $f$ by dividing both sides by $8887.2$:
\[f = \frac{3440}{8887.2}\]
Performing this division gives $f \approx 0.38707$. Rounding to the nearest hundredth, we get $f \approx 0.39$.

Examples

Understanding the relationship between speed, braking, and road conditions is vital for road safety. If you are designing safety features or calculating braking distances for driving theory, you rely on formulas like this one. Knowing the friction coefficient helps traffic engineers design safer roads, and drivers can better understand why slowing down significantly reduces the distance required to come to a complete stop. This math allows for precise, life-saving predictions in real-world scenarios.