A car is parked on a cliff overlooking the ocean on an incline that makes an
angle of 23.0° 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 3.67 m/s2 for a distance of 35.0 m
to
the edge of the cliff, which is 50.0 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
24
To solve this problem, we can break it down into two parts: (a) finding the car's position relative to the base of the cliff when it lands in the ocean, and (b) finding the length of time the car is in the air.
(a) To find the car's position relative to the base of the cliff when it lands in the ocean, we need to determine the horizontal and vertical components of the car's motion.
First, we can find the horizontal component of the car's motion using the equation:
x = v₀ₓt + (1/2)at²
where
x is the horizontal distance (35.0 m),
v₀ₓ is the initial velocity in the x-direction (which is 0 since the car starts from rest),
a is the acceleration (3.67 m/s²),
t is the time it takes for the car to reach the edge of the cliff.
Rearranging the equation to solve for t:
t = (x - (1/2)at²) / v₀ₓ
Since the car starts from rest, v₀ₓ = 0, so the equation simplifies to:
t = √(2x/a)
Substituting the given values:
t = √(2 * 35.0 m / 3.67 m/s²)
t ≈ 3.16 s
Next, we can find the vertical component of the car's motion. The car travels a vertical distance of 50.0 m downward, so the time it takes to fall can be found using the equation:
y = v₀ᵧt + (1/2)gt²
where
y is the vertical distance (-50.0 m since it is downward),
v₀ᵧ is the initial velocity in the y-direction (which is 0 since the car starts from rest),
g is the acceleration due to gravity (-9.8 m/s²),
t is the time it takes for the car to fall.
Rearranging the equation to solve for t:
t = √(2y/g)
Substituting the given values:
t = √(2 * (-50.0 m) / -9.8 m/s²)
t ≈ 3.19 s
Since the time it takes for the car to reach the edge of the cliff (3.16 s) is slightly less than the time it takes for the car to fall (3.19 s), the car will not yet reach the ocean. Instead, it will reach the cliff wall.
To find the horizontal distance from the cliff wall to the base of the cliff, we can use the equation:
x = v₀ₓt + (1/2)at²
where
x is the horizontal distance we want to find,
v₀ₓ is the initial velocity in the x-direction (which is 0 since the car starts from rest),
a is the acceleration (3.67 m/s²),
t is the time it takes for the car to fall (3.19 s).
Substituting the given values:
x = 0 * 3.19 s + (1/2) * 3.67 m/s² * (3.19 s)²
x ≈ 18.73 m
Therefore, when the car lands, its position relative to the base of the cliff is approximately 18.73 m from the base.
(b) To find the length of time the car is in the air, we already determined that the car falls for a time of approximately 3.19 s. Therefore, the length of time the car is in the air is approximately 3.19 s.