The stopping distance d of a car in feet is related to the speed in mph by the equation d(s)=.02s^2+1.1s

Calculate the speed when your stopping distance is:
1.) 51ft
2.) 315.12ft

You must solve two quadratic equations.

0.02 s ^ 2 + 1.1 s = 51

and

0.02 s ^ 2 + 1.1 s = 315.12

If you don't know how solve equations in google type:

quadratic equation online

When you see list of results click on:

Free Online Quadratic Equation Solver:Solve by Quadratic Formula

When page be open in rectangle type:

0.02 s ^ 2 + 1.1 s = 51

and click option: solve it!

You wil see solution step-by-step

Then type:

0.02 s ^ 2 + 1.1 s = 315.12

and click option: solve it!

Remark:

Speed Can't be negative.

You must use only positive solutions.

To find the speed when the stopping distance is given, we need to solve the equation d(s) = .02s^2 + 1.1s for the speed 's'. We can do this by rearranging the equation and solving for 's'.

1) Stopping distance = 51ft:
Setting d(s) = 51, we get:
51 = .02s^2 + 1.1s

Rearranging the equation:
.02s^2 + 1.1s - 51 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = .02, b = 1.1, and c = -51. We can solve this equation using the quadratic formula:

s = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values, we get:
s = (-(1.1) ± √((1.1)^2 - 4(.02)(-51))) / (2(.02))

Calculating the values inside the square root:
s = (-1.1 ± √(1.21 + 4.08)) / 0.04
s = (-1.1 ± √5.29) / 0.04

Calculating the values inside the square root:
s = (-1.1 ± 2.3) / 0.04
s ≈ (-3.4 / 0.04) or (1.2 / 0.04)
s ≈ -85 or 30

Since speed cannot be negative, the approximate speed when the stopping distance is 51ft is 30 mph.

2) Stopping distance = 315.12ft:
Setting d(s) = 315.12, we get:
315.12 = .02s^2 + 1.1s

Rearranging the equation:
.02s^2 + 1.1s - 315.12 = 0

Again, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = .02, b = 1.1, and c = -315.12. We can solve this equation using the quadratic formula:

s = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values, we get:
s = (-(1.1) ± √((1.1)^2 - 4(.02)(-315.12))) / (2(.02))

Calculating the values inside the square root:
s = (-1.1 ± √(1.21 + 25.2048)) / 0.04
s = (-1.1 ± √26.4148) / 0.04

Calculating the values inside the square root:
s = (-1.1 ± 5.139) / 0.04
s ≈ (-6.239 / 0.04) or (4.039 / 0.04)
s ≈ -160 or 100

Again, speed cannot be negative, so the approximate speed when the stopping distance is 315.12ft is 100 mph.