SOLUTION: A typical car's stopping distance on dry pavement "d" in feet can be approximated by the function d= 0.034s2(squared) + 0.56s

Click here to see ALL problems on Quadratic Equations

Question 303155: A typical car's stopping distance on dry pavement "d" in feet can be approximated by the function

d= 0.034s2(squared) + 0.56s - 17.11. Where "s" is the speed in miles per hour, of the car before braking.

A. How fast is the car going if it requires 100 feet for the car to stop after the brakes are applied?

Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
A typical car's stopping distance on dry pavement "d" in feet can be approximated by the function
d= 0.034s2(squared) + 0.56s - 17.11. Where "s" is the speed in miles per hour, of the car before braking.

A. How fast is the car going if it requires 100 feet for the car to stop after the brakes are applied?
.
Replace d with 100 and solve for s:
d= 0.034s^2 + 0.56s - 17.11
100= 0.034s^2 + 0.56s - 17.11
0= 0.034s^2 + 0.56s - 117.11
Solve using the quadratic formula. Doing so yields:
s = {51.0, -67.5}
Toss out the negative solution -- doesn't make sense.
so, the car was moving at 51 mph.
.
Details of quadratic formula:
.

Solved by pluggable solver: SOLVE quadratic equation with variable

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=16.24056 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 51.0287914140646, -67.4993796493587. Here's your graph:

SOLUTION: A typical car's stopping distance on dry pavement "d" in feet can be approximated by the function d= 0.034s2(squared) + 0.56s - 17.11. Where "s" is the sp