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 4.00 m/s2 for a distance of 45.1 m to the edge of the cliff, which is 19.4 m above the ocean. Find (a) the cars position relative to the base of the cliff when the car lands in the ocean
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 distance the car travels during its descent.
Step 1: Find the vertical distance traveled by the car
Using the given information, we know that the car travels a horizontal distance of 45.1 m and the cliff is 19.4 m above the ocean. Since the car starts from rest and accelerates at a constant rate of 4.00 m/s², we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (we assume it to be 0 since the car lands in the ocean)
vi = initial velocity (0 m/s since the car starts from rest)
a = acceleration (4.00 m/s²)
d = unknown distance (we'll solve for this)
Rearranging the equation to solve for d:
d = (vf^2 - vi^2) / (2a)
Substituting the given values:
d = (0 - 0) / (2 * 4.00)
d = 0 / 8.00
d = 0 m
Therefore, the car has traveled vertically 0 m.
Step 2: Find the horizontal distance traveled by the car
Since the car's position relative to the base of the cliff when it lands in the ocean is the same as the horizontal distance traveled by the car, the car's position is simply the horizontal distance traveled.
Therefore, the car's position relative to the base of the cliff when it lands in the ocean is 45.1 m.
To find the position of the car relative to the base of the cliff when it lands in the ocean, we can divide the problem into two parts: the horizontal distance traveled by the car and the vertical distance fallen by the car.
1. Horizontal Distance:
The car travels a distance of 45.1 m on the incline. The angle of the incline is given as 24.0° below the horizontal. Since the car is rolling down the incline, the horizontal distance is equal to the distance traveled along the incline. We can use the equation for distance traveled with constant acceleration:
𝑑 = 𝑢𝑡 + 0.5𝑎𝑡^2
Here, 𝑑 is the distance, 𝑢 is the initial velocity, 𝑡 is the time, and 𝑎 is the acceleration.
Initially, the car is at rest, so 𝑢 = 0 m/s.
The acceleration is given as 4.00 m/s^2.
The time can be found using the equation of motion: 𝑣 = 𝑢 + 𝑎𝑡, where 𝑣 is the final velocity.
The final velocity can be found using the equation: 𝑣^2 = 𝑢^2 + 2𝑎𝑑 (since initial velocity 𝑢 = 0).
Rearranging the above equations, we can find the time it takes for the car to travel the horizontal distance.
2. Vertical Distance:
The car falls vertically from the edge of the cliff, which is 19.4 m above the ocean. To find the vertical distance fallen by the car, we can use the equation for distance traveled:
𝑑 = 𝑢𝑡 + 0.5𝑎𝑡^2
Here, 𝑑 is the vertical distance, 𝑢 is the initial velocity (which is 0 m/s as the car starts from rest), 𝑡 is the time, and 𝑎 is the acceleration due to gravity (-9.8 m/s^2 since it is acting downward).
Now, let's calculate the horizontal and vertical distances:
1. Horizontal Distance:
Using the given values and equations mentioned above, we can determine the time it takes for the car to travel the horizontal distance of 45.1 m.
2. Vertical Distance:
Using the equation and given values, we know that the car falls a vertical distance of 19.4 m.
Finally, the car's position relative to the base of the cliff when it lands in the ocean is the horizontal distance traveled (found in step 1) measured horizontally from the base of the cliff, and the vertical distance fallen (found in step 2) measured downward from the cliff's edge.