Answer:
(a) The velocity of the car before the brakes were applied is 77.46 ft/s
(b) The time required for the car to stop is 7.8 s
Explanation:
Given;
acceleration of the car, a = 10 ft/s²
distance traveled by the car, d = 300 ft
(a) the velocity of the car before the brakes were applied is given;
v² = u² + 2ad
v² = 0 + 2(10 x 300)
v² = 6000
v = √6000
v = 77.46 ft/s
(b) the time required for the car to stop
d = ut + ¹/₂at²
d = 0 + ¹/₂at²
d = ¹/₂at²
t² = 2d / a
t = √ ( 2d / a)
t = √ ( 2 x 300 / 10)
t = 7.8 s
Therefore, the time required for the car to stop is 7.8 s
The speed of the car immediately before the brakes were applied was 77.46 ft/s and it took 7.75 s for the car to stop.
Using the formula:
v² = u² + 2as
Where S is the distance = 300 ft, u is the initial velocity, a is the acceleration = -10 ft/s², v is the final velocity = 0 ft/s (stops)
0² = u² + 2(-10)(300)
0 = u² - 6000
u² = 6000
u = 77.46 ft/s
b)
v = u + at
0 = 77.46 - 10t
t = 7.75 s
The speed of the car immediately before the brakes were applied was 77.46 ft/s and it took 7.75 s for the car to stop.
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