A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 21.0° 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 3.37 m/s2 for a distance of 60.0 m to the edge of the cliff, which is 30.0 m above the ocean.
(a) Find the car's position relative to the base of the cliff when the car lands in the ocean.
(b) Find the length of time the car is in the air.
60 = (1/2)(3.37) t^2
solve fot t, the rolling time
v = a t = 3.37 t
that v is the speed at the cliff edge
u = v cos 21 = horizontal speed until bottom
Vi = -v sin 21 = initial speed up (negative of course)
h = Hi + Vi t + (1/2) a t^2
now t is the time airborne
0 = 30 - v (sin 21) t - 4.9 t^2
solve for time in the air
d = u t = v ( cos 21) t
What would be the answers to A and B?
To find the car's position relative to the base of the cliff when it lands in the ocean, we need to first break down the motion of the car into horizontal and vertical components.
(a) Horizontal Motion:
Since the car is rolling down the incline with a constant acceleration, we can use the equation of motion:
d = v₀t + (1/2)at²,
where d is the horizontal displacement, v₀ is the initial velocity (which is zero in this case), t is the time, and a is the acceleration.
In this case, the horizontal displacement is given as 60.0 m. The acceleration (a) is given as 3.37 m/s².
Using the equation of motion, we can rearrange it to solve for time:
t = √(2d/a).
Substituting the given values, we get:
t = √(2 * 60.0 / 3.37) ≈ 5.329 seconds.
Now that we have the time it takes for the car to reach the edge of the cliff, we can determine the horizontal position relative to the base of the cliff.
The horizontal distance (x) covered by the car is given by:
x = v₀t + (1/2)at²,
where v₀ is the initial velocity (zero in this case), t is the time, and a is the acceleration (3.37 m/s²).
Substituting the values, we have:
x = 0 * 5.329 + (1/2) * 3.37 * (5.329)²,
x ≈ 0 + 0.5 * 3.37 * 28.403,
x ≈ 48.019 m.
Therefore, the car's horizontal position relative to the base of the cliff when it lands in the ocean is approximately 48.019 meters.
(b) Vertical Motion:
To find the length of time the car is in the air, we need to calculate the time it takes for the car to fall from the cliff to the ocean, taking into account the vertical distance and the acceleration due to gravity.
The vertical distance is given as 30.0 m. The acceleration due to gravity (g) is approximately 9.8 m/s².
To obtain the time for vertical motion, we can use the equation of motion:
d = v₀t + (1/2)gt²,
where d is the vertical displacement, v₀ is the initial velocity (zero in this case), t is the time, and g is the acceleration due to gravity.
Simplifying the equation, we have:
d = (1/2)gt².
Substituting the given values, we get:
30.0 = (1/2) * 9.8 * t²,
60 = 9.8t²,
t² = 60 / 9.8,
t ≈ √6.122.
Therefore, the length of time the car is in the air is approximately √6.122 seconds.