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 (b) the length of time the car is in the air
To calculate the length of time the car is in the air, we need to use the kinematic equation for displacement.
First, let's find the time it takes for the car to reach the edge of the cliff. We'll use the equation:
s = ut + (1/2)at^2
Where:
- s is the displacement (distance covered by the car) = 45.1 m
- u is the initial velocity of the car = 0 m/s (as it starts from rest)
- a is the constant acceleration = 4.00 m/s^2
- t is the time taken
Rearranging the equation, we get:
t^2 = (2s)/a
Plugging in the values, we have:
t^2 = (2 * 45.1) / 4.00
t^2 = 90.2 / 4.00
t^2 = 22.55
Taking the square root of both sides, we find:
t ≈ 4.75 seconds
Now, since we know the car is in the air when it reaches the edge of the cliff, we can calculate the time it spends in the air.
Since the car is not moving horizontally when it leaves the cliff, the time it takes to reach the ocean can be calculated using the vertical motion equation:
h = ut + (1/2)gt^2
Where:
- h is the vertical displacement (height of the cliff) = 19.4 m
- u is the vertical initial velocity = 0 m/s (as the car is not moving vertically when it leaves the cliff)
- g is the acceleration due to gravity = 9.81 m/s^2 (taking vertical direction positive upwards)
- t is the time taken
Rearranging the equation, we get:
t^2 = (2h)/g
Plugging in the values, we have:
t^2 = (2 * 19.4) / 9.81
t^2 = 38.8 / 9.81
t^2 = 3.95
Taking the square root of both sides, we find:
t ≈ 1.99 seconds
Therefore, the length of time the car is in the air is approximately 1.99 seconds.