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

the distance it takes a truck to stop can be modeled by the function $d(v)=\frac{2.15v^{2}}{64.4f}$. $d =$ stopping distance in feet. $v =$ initial velocity in miles per hour. $f =$ a constant related to friction. when the trucks initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft. determine the value of $f$, rounded to the nearest hundredth.

the distance it takes a truck to stop can be modeled by the function $d(v)=\frac{2.15v^{2}}{64.4f}$. $d =$ stopping distance in feet. $v =$ initial velocity in miles per hour. $f =$ a constant related to friction. when the trucks initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft. determine the value of $f$, rounded to the nearest hundredth.

Answer

Explanation:

Step1: Substitute given values into formula

Given $d(v)=\frac{2.15v^{2}}{64.4f}$, $v = 40$ and $d=138$. Substitute these values: $138=\frac{2.15\times40^{2}}{64.4f}$.

Step2: Simplify the numerator

First, calculate $2.15\times40^{2}=2.15\times1600 = 3440$. So the equation becomes $138=\frac{3440}{64.4f}$.

Step3: Cross - multiply

Cross - multiplying gives $138\times64.4f=3440$. Then $8887.2f = 3440$.

Step4: Solve for f

Divide both sides by 8887.2: $f=\frac{3440}{8887.2}\approx0.39$.

Answer:

$0.39$