A car is parked on a cliff overlooking the ocean on an incline that makes an ang

Question

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 17.0° 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 2.75 m/s2 for a distance of 65.0 m to the edge of the cliff, which is 35.0 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

Explanation / Answer

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:

vf2=v02+2ad

where vf is final speed
v0=initial speed = 0
a = accel = 2.75 m/s2
d=distance = 65m

so the speed on leaving the cliff is:

vf2=0+2(2.75)(65)
vf2 = 357.5
vf=18.91 m/s

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

v(horizontal) = 18.91 cos 17 =18.0837m/s
v(vertical) = -18.91 sin 17 = -5.528 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 gt2

y(t)=height at any time t,

y=initial position
v0y=initial y speed

so we have:

y(t)=35-5.528t-4.9t2

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

0=35-5.528t-4.9t2

this is a quadratic equation with solution

t= 2.167s

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 2.167 s is:

x=18.08m/s x 2.167s =39.187m

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