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+1.1v
We 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.
