Question

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?
21 feet
21 miles
239 feet
239 miles

Explanation:

1. First, identify the values of \(v\) and \(f\) given in the problem:

• We are given that \(v = 47\) mph and \(f=0.34\).
• The formula for the car's - stopping distance \(d(v)\) is \(d(v)=\frac{2.15v^{2}}{58.4f}\).

1. Then, substitute the values of \(v\) and \(f\) into the formula:

• \(d(v)=\frac{2.15\times47^{2}}{58.4\times0.34}\).
• First, calculate \(47^{2}=47\times47 = 2209\).
• Then, \(2.15\times47^{2}=2.15\times2209 = 2.15\times(2000 + 200+9)=2.15\times2000+2.15\times200 + 2.15\times9=4300+430 + 19.35 = 4749.35\).
• Next, \(58.4\times0.34=(58 + 0.4)\times0.34=58\times0.34+0.4\times0.34=(60 - 2)\times0.34+0.136=(60\times0.34-2\times0.34)+0.136=(20.4 - 0.68)+0.136 = 19.856\).
• Now, \(d(v)=\frac{4749.35}{19.856}\approx239\) feet.

Answer:

C. 239 feet