algebra 2 question read description

Question

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 75 ft, what is the maximum safe speed, to the nearest mile per hour, of the car?

Answer 1

To begin with, I set d=75:

75 = 0.04v² + 0.5v

Subtract 75 from both sides to get a quadratic equation in standard form: 0 = 0.04v² + 0.5v - 75.

The quadratic formula is a good tool to solve equations like this. If you're not familiar with it, here's a website explaining what quadratics are and how to solve them: https://www.mathsisfun.com/algebra/quadratic-equation.html

In this equation, a=.04, b=.05, and c=-75.

Solving the equation results in two possibilities: v= -50, and v= 37.5. The maximum velocity will not be negative, so v=37.5 mph is your answer.

I hope this helps, and have fun mathing!

Liz Z.

Answer 2

75=0.04v^2+0.5v

75=v^2/25 + v/2

75 = (37.5)^2/25 + 37.5/2

75 = 56.25 + 18.75 =75

max speed = 37.5 mph