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Sagot :
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 63.7 ft.
- The braking distance at 55 mph, is approximately 298.35 ft.
- The braking distance at 85 mph, is approximately 708.92 ft.
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
[tex]D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.[/tex]
At 25 mph, the braking distance is approximately 63.7 ft.
At 55 mph, the braking distance is given as follows;
55 mph = 80.65945 ft.s
[tex]D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.[/tex]
At 55 mph, the braking distance is approximately 298.35 ft.
At 85 mph, the braking distance is given as follows;
85 mph = 124.6555 ft.s
[tex]D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.[/tex]
At 85 mph, the braking distance is approximately 708.92 ft.
b. The speed of the car when the braking distance is 450 feet is given as follows;
[tex]450 = 2.6 + \dfrac{v^2}{22}[/tex]
v² = (450 - 2.6) × 22 = 9842.8
v = √(9842.2) ≈ 98.211 ft./s
The car was moving at v ≈ 98.211 ft./s
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