Line XY was drawn on a plane. Using a compass and a straightedge the student wanted to construct an equilateral triangle. What w
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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
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
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
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;
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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