Question

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees below horizontal. The negligent driver leaves the car in neutral and the emergency brakes are defective. The car rollsf from rest down the incline with a constant acceleration of 3.82m/s*2 for a distance of 30.0 m to the edge of the cliff which is 45.0 m above ocean .

A)Find the cars positive relative to the base of the cliff when the car lands in the ocean in meters

B)Find the length of time the car is in the air in seconds

Answer 1

the information allows us to figure out the horizontal and vertical speed of the car when it goes off the edge of the cliff

we find that speed from:

vf^2=v0^2+2ad where vf is final speed
v0=initial speed = 0
a = accel = 3.82 m/s/s
d=distance = 35m

so the speed on leaving the cliff is:

vf^2=0+2(3.82)(35)
vf^2 = 267.4
vf=16.35 m/s

now, we need to find the components of the car's velocity as it leaves the cliff, they are:

v(horizontal) = 16.35 cos 24 = 14.9m/s
v(vertical) = -16.35 sin 24 = -6.65 m/s

we need to find the time the car is in the air, for this we use the equation of motion:

y(t)=y0+v0y t - 1/2 gt^2

y(t)=height at any time t, y==initial position
v0y=initial y speed

so we have:

y(t)=20-6.65t-4.9t^2

we want to find how long it takes for the car to reach y=0:

0=20-6.65t-4.9t^2

this is a quadratic equation with solution

t= 1.45s

since the horizontal speed will not change once the car leaves the clilff (since there are no horizontal forces acting), we have that the horizontal distance traveled in 1.45 s is:

x=14.9m/s x 1.45s =21.6m

Answer 2

we find that speed from:

vf^2=v0^2+2ad where vf is final speed
v0=initial speed = 0
a = accel = 3.82 m/s/s
d=distance = 30m

so the speed on leaving the cliff is:

vf^2=0+2(3.82)(30)
vf^2 = 229.2
vf=15.13 m/s

now, we need to find the components of the car's velocity as it leaves the cliff, they are:

v(horizontal) = 15.13 cos 24 = 13.83 m/s
v(vertical) = -15.13 sin 24 = -6.15 m/s

we need to find the time the car is in the air, for this we use the equation of motion:

y(t)=y0+v0y t - 1/2 gt^2

y(t)=height at any time t, y==initial position
v0y=initial y speed

so we have:

y(t)=45-6.15t-4.9t^2

we want to find how long it takes for the car to reach y=0:

0=45-6.15t-4.9t^2

this is a quadratic equation with solution

t= 2.59s

since the horizontal speed will not change once the car leaves the clilff (since there are no horizontal forces acting), we have that the horizontal distance traveled in 1.45 s is:

x=13.83m/s x 2.59s =35.8197m

Answer 3

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees 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 3.82m/s^2 for a distance of 30.0m. to the edge of the cliff. The cliff is 45.0m. above the ocean.

The equation below is very useful for determining the final velocity, when the initial velocity, acceleration, and distance are given!

Final velocity^2

Answer 4

First step is to determine the speed of the car at the moment it leaves the cliff.

The equations of motion while the car is rolling are:

d = 0.5*3.82*t^2
and
v = 3.82*t

B)
solving first for the t, which is the rolling time (length of time the car is in the air )
Given d = 30.0m

30 = 0.5*3.82*t^2

t = 3.96 seconds

so

v = 15.14 m/s
the velocity has direction, which is 24 degrees below the horizontal.

Once airborne, the car has the vertical equation of motion for position of
y(t) = 45 - 15.14*sin(24)*t - 0.5*9.81*t^2

y(t) = 45