Question 1204951: A car with good tire tread can stop in less distance than a car with poor tread. The formula for the stopping distance d, in feet, of a car with good tread on dry cement is approximated by
d = 0.04v2 + 0.5v,
where v is the speed of the car. If the driver must be able to stop within 74 ft, what is the maximum safe speed, to the nearest mile per hour, of the car?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A car with good tire tread can stop in less distance than a car with poor tread. The formula for the stopping distance d, in feet, of a car with good tread on dry cement is approximated by
d = 0.04v2 + 0.5v,
where v is the speed of the car. If the driver must be able to stop within 74 ft, what is the maximum safe speed, to the nearest mile per hour, of the car?
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d = 0.04v^2 + 0.5v = 74
v^2 + 12.5v - 1850 = 0
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=7556.25 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 37.2133466267843, -49.7133466267843.
Here's your graph:
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Ignore the negative result
v = 37 mi/hr
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