A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 3.82 m/s2 for a distance of 30.0 m to the edge of the cliff, which is 20.0 m above the ocean.
(a) What is the car's position relative to the base of the cliff when the car lands in the ocean.
(b) What is the length of time the car is in the air.
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To solve this problem, we need to break it down into two parts: the motion along the incline and the motion in the air.
(a) First, let's find the time it takes for the car to roll down the incline and reach the edge of the cliff. We can use the kinematic equation:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity (which is 0 in this case), a is the acceleration (3.82 m/s^2), and t is the time.
Plugging in the values, we have:
30.0 m = (1/2)(3.82 m/s^2)t^2
Rearranging the equation, we get:
t^2 = (2 * 30.0 m) / 3.82 m/s^2
t^2 = 15.71 s^2
Taking the square root of both sides, we find:
t = 3.96 s
So it takes approximately 3.96 seconds for the car to reach the edge of the cliff.
Next, we calculate the horizontal distance traveled by the car during this time. The horizontal distance can be found using the formula:
d = v * t
where d is the distance, v is the horizontal velocity, and t is the time. The horizontal velocity can be found using the formula:
v = a * t
Plugging in the values, we get:
v = 3.82 m/s^2 * 3.96 s
v = 15.12 m/s
Now we can find the horizontal distance traveled by the car:
d = 15.12 m/s * 3.96 s
d = 59.83 m
Therefore, when the car lands in the ocean, its position relative to the base of the cliff is 59.83 meters.
(b) To find the time the car is in the air, we can use the second kinematic equation:
v = u + at
where v is the final velocity (which is 0 as the car is in the air), u is the initial velocity (which is the horizontal velocity we calculated earlier), a is the acceleration (9.8 m/s^2, due to gravity), and t is the time.
Plugging in the values, we have:
0 = 15.12 m/s + (-9.8 m/s^2)t
Simplifying the equation, we get:
9.8 m/s^2t = 15.12 m/s
t = 15.12 m/s / 9.8 m/s^2
t = 1.54 s
Therefore, the car is in the air for approximately 1.54 seconds.