a cars stopping distance in feet is modeled by the equation $d(v)=\frac{2.15v^{2}}{58.4f}$ where $v$ is the…

a cars stopping distance in feet is modeled by the equation $d(v)=\frac{2.15v^{2}}{58.4f}$ where $v$ is the initial velocity of the car in miles per hour and $f$ is a constant related to friction. if the initial velocity of the car is 47 mph and $f = 0.34$, what is the approximate stopping distance of the car?\no 21 feet\no 21 miles\no 239 feet\no 239 miles

a cars stopping distance in feet is modeled by the equation $d(v)=\frac{2.15v^{2}}{58.4f}$ where $v$ is the initial velocity of the car in miles per hour and $f$ is a constant related to friction. if the initial velocity of the car is 47 mph and $f = 0.34$, what is the approximate stopping distance of the car?\no 21 feet\no 21 miles\no 239 feet\no 239 miles

Answer

Explanation:

Step1: Substitute values into formula

Given $d(v)=\frac{2.15v^{2}}{58.4f}$, substitute $v = 47$ and $f=0.34$. So we have $d(47)=\frac{2.15\times47^{2}}{58.4\times0.34}$.

Step2: Calculate numerator

First, calculate $47^{2}=47\times47 = 2209$. Then $2.15\times47^{2}=2.15\times2209 = 4749.35$.

Step3: Calculate denominator

Calculate $58.4\times0.34 = 19.856$.

Step4: Calculate stopping - distance

$d(47)=\frac{4749.35}{19.856}\approx239$. Since the formula gives the distance in feet, the unit of the result is feet.

Answer:

239 feet