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A car’s stopping distance in feet is modeled by the equation d(v)= 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?

A21 feet

B21 miles

C239 feet

D239 miles

Verified Answer :

JeanaShupp

Answer: C is the correct answer.The approximate stopping distance of the car =239 feet.


Step-by-step explanation:

Given: A car’s stopping distance in feet is modeled by the equation

\(d(v)=2.15\frac{v^2}{58.4f}\) 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 v= 47 mph and f = 0.34 then

The distance of the car \(d(47)=2.15\frac{{47}^2}{58.4\times0.34}=2.15\frac{2209}{19.856}=2.15\times111.25=239.18\)

≈239 feet           [ round to nearest ones]

Therefore ,the approximate stopping distance of the car =239 feet.

Answer:

239 pts (Letter C)

Step-by-step explanation:

Got it correct

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