A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 25.0° 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 3.97 m/s2 for a distance of 50.0 m to the edge of the cliff, which is 40.0 m above the ocean.
(a) Find the car's position relative to the base of the cliff when the car lands in the ocean.
m
(b) Find the length of time the car is in the air.
s
To find the car's position relative to the base of the cliff when it lands in the ocean, we need to break down the problem into two parts: horizontal and vertical motion.
(a) Horizontal Motion:
The horizontal motion of the car is constant, as there are no forces acting upon it in the horizontal direction. The distance covered horizontally is given as 50.0 m. Therefore, the car's horizontal position relative to the base of the cliff when it lands in the ocean is 50.0 m.
(b) Vertical Motion:
To determine the length of time the car is in the air, we need to find the time it takes for the car to fall from the cliff to the ocean.
Let's use the following kinematic equation for vertical motion:
d = v_i * t + (1/2) * a * t^2
Where:
d = vertical distance (40.0 m)
v_i = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time in seconds
Since the car rolls down the incline, the initial vertical velocity is 0 m/s. The acceleration will be the gravitational acceleration, but with a negative sign due to the downward direction.
Using the equation, we can rearrange it to solve for time:
t = sqrt((2 * d) / (-a))
Substituting the values given:
t = sqrt((2 * 40.0 m) / (-(-9.8 m/s^2)))
Simplifying the equation:
t = sqrt(80.0 m / 9.8 m/s^2)
t = sqrt(8.16 s^2)
t ≈ 2.86 s
Therefore, the length of time the car is in the air is approximately 2.86 seconds.