Question

a cars stopping distance in feet is modeled by the equation ( d(v)=\frac{2.15 v^{2}}{58.4 f} ) 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?
21 feet
21 miles
239 feet
239 miles

Explanation:

Step1: Substitute the values into the formula

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

Step2: Calculate \(v^{2}\)

\(v^{2}=47^{2}=2209\)

Step3: Calculate the numerator and denominator

Numerator: \(2.15\times2209 = 2.15\times(2000 + 200+9)=2.15\times2000+2.15\times200 + 2.15\times9=4300+430 + 19.35=4749.35\)
Denominator: \(58.4\times0.34=(60 - 1.6)\times0.34=60\times0.34-1.6\times0.34 = 20.4-0.544 = 19.856\)

Step4: Calculate the value of \(d(47)\)

\(d(47)=\frac{4749.35}{19.856}\approx239\)

Answer:

C. 239 feet