The braking distance is the distance the car travels before coming to a stop after the brakes are applied
a. The braking distances are as follows;
- The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
- The braking distance at 55 mph, is approximately <u>298.35 ft.</u>
- The braking distance at 85 mph, is approximately <u>708.92 ft.</u>
b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s
Reason:
The given function for the braking distance is D = 2.6 + v²/22
a. The braking distance if the car is going 25 mph is therefore;
25 mph = 36.66339 ft./s
![D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.](https://tex.z-dn.net/?f=D%20%3D%202.6%20%2B%20%5Cdfrac%7B36.66339%5E2%7D%7B22%7D%20%3D%2063.7%20%5C%20ft.)
At 25 mph, the braking distance is approximately <u>63.7 ft.</u>
At 55 mph, the braking distance is given as follows;
55 mph = 80.65945 ft.s
![D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.](https://tex.z-dn.net/?f=D%20%3D%202.6%20%2B%20%5Cdfrac%7B80.65945%5E2%7D%7B22%7D%20%5Capprox%20298.35%20%5C%20ft.)
At 55 mph, the braking distance is approximately <u>298.35 ft.</u>
At 85 mph, the braking distance is given as follows;
85 mph = 124.6555 ft.s
![D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.](https://tex.z-dn.net/?f=D%20%3D%202.6%20%2B%20%5Cdfrac%7B124.6555%5E2%7D%7B22%7D%20%5Capprox%20708.92%20%5C%20ft.)
At 85 mph, the braking distance is approximately <u>708.92 ft.</u>
b. The speed of the car when the braking distance is 450 feet is given as follows;
![450 = 2.6 + \dfrac{v^2}{22}](https://tex.z-dn.net/?f=450%20%3D%202.6%20%2B%20%5Cdfrac%7Bv%5E2%7D%7B22%7D)
v² = (450 - 2.6) × 22 = 9842.8
v = √(9842.2) ≈ 98.211 ft./s
The car was moving at v ≈ <u>98.211 ft./s</u>
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brainly.com/question/18591940