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The braking distance, in feet of a car a Travling at v miles per hour is given.

D= 2.6+ V^2/22


a. What is the braking distance, in feet, if the car is going 25 mph? 55 mph?
85 mph? Show your work.



b. Suppose that the car took 450 feet to brake. Use your computations in part (a)
to make a prediction about how fast it was going when the brakes were applied.

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

Learn more here:

https://brainly.com/question/18591940

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