We want to determine if Hannah was speeding on a highway with a speed limit of 65 miles per hour. We are given an equation that models the stopping distance d in feet of a car traveling at a speed of v miles per hour.
d=0.05v^2+1.1v
Let's find the speed of Hannah's car if it stopped after 250 feet. To do it, we need to solve the equation below.
250=0.05v^2+1.1vWe need to write it in standard form.
Now, we will solve it by using the Quadratic Formula. We first need to identify the values of a, b, and c.
0.05v^2+1.1v-250=0 ⇔ 0.05n^2+ 1.1n+( - 250)=0
We see that a= 0.05, b= 1.1, and c= - 250. Let's substitute these values into the Quadratic Formula.
The solutions for this equation are v= - 1.1 ± sqrt(51.21)0.1. Let's separate them into the positive and negative cases.
v= - 1.1 ± sqrt(51.21)0.1
v_1=- 1.1 + sqrt(51.21)/0.1
v_2=- 1.1 - sqrt(51.21)/0.1
v_1=- 1.1/0.1 + sqrt(51.21)/0.1
v_2=- 1.1/0.1 - sqrt(51.21)/0.1
v_1 ≈ 61
v_2 ≈ - 83
Using the Quadratic Formula, we found that the solutions of the given equation are v_1≈ 61 and v_2≈ - 83. We need to disregard the negative solution because the speed is a positive quantity. Thus, Hannah was traveling at about 61 miles per hour. This is less than the speed limit meaning that she was not speeding.
