SOLUTION: On wet concrete, the stopping distance s, in feet, of a car traveling v miles per hour is given by s(v) = 0.055v^2 + 1.1v. At what speed could a car be traveling and still stop at

Algebra ->  Equations -> SOLUTION: On wet concrete, the stopping distance s, in feet, of a car traveling v miles per hour is given by s(v) = 0.055v^2 + 1.1v. At what speed could a car be traveling and still stop at       Log On


   

Question 1058954: On wet concrete, the stopping distance s, in feet, of a car traveling v miles per hour is given by s(v) = 0.055v^2 + 1.1v. At what speed could a car be traveling and still stop at a stop sign 44 feet away?
Answer by ikleyn(52781) About Me  (Show Source):
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On wet concrete, the stopping distance s, in feet, of a car traveling v miles per hour is given by s(v) = 0.055v^2 + 1.1v.
At what speed could a car be traveling and still stop at a stop sign 44 feet away?
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0.055v^2 + 1.1v = 44.

Simplify and solve for v.
As a first step, divide both sides by 0.055. You will get

v^2 + 20v - 800 = 0.

Factor

(v-20)*(v+40) = 0.

Take the positive root and disregard the negative.

Answer.  v = 20 mph is the unique solution.