A car’s stopping distance in feet is modeled by the equation d times v is equal to the fraction with numerator 2 point 1 5 v squared and denominator 58 point 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 is equal to 0 point 3 4 , what is the approximate stopping distance of the car?
Substitute v = 47 and f = 0.34 into the model: d = (2.15 × 47²) ÷ (58.4 × 0.34). This gives d = 4749.35 ÷ 19.856 ≈ 239.2 feet, so the approximate stopping distance is 239 feet. The answer is choice A.
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