With work Please! You have a job with an insurance company helping with the inve

Question

With work Please!

You have a job with an insurance company helping with the investigation of a car accident. At the scene you find a road running straight down a hill which is sloped =5° above the horizontal. At the bottom of the hill, the road goes horizontally for a very short distance to the edge of a cliff. The cliff has a height h= 420.0 ft vertical drop to the horizontal ground below. At a distance d= 25.0 ft from the base of the cliff is the wrecked car. A witness claims that the car parked somewhere on the hill, began coasting down the road on its own, taking about t1= 3.0 s before going over the edge of the cliff. The witness thinks that the driver was drunk and passed out, releasing the brake. Your boss drops a stone from the edge of the cliff and from the sound of its landing, determines that it takes t= 5.0 to hit the bottom. You are told to calculate the car's average acceleration coming down the hill using the statement of the witness and the other facts in the case. Your boss seems to suspect foul play.

Explanation / Answer

the acceleration down the hill is g*sin()

If the car traveled down the ramp for 3.0s then it speed at the bottom would be

v = a*t = g*sin()*t = 9.80*sin(5)*3.0 = 2.562m/s

The height of the cliff is determined by the stone falling. Let t1 be the time to fall and t2 be the time for the sound to return.

For t1 we have y = 1/2*g*t1^2 for t2 we have y = v*t2 where v is the speed of sound (use 343m/s)

So we have 1/2*g*t1^2 = 343*t2 and that t1 + t2 = 5.0s

So 1/2*9.8*t1^2 - 343*(5.0 - t1) = 0 or 4.9*t1^2 + 343*t1 - 1715 = 0

Therefore t1 = (-343 +-sqrt(343^2 - 4*4.9*(-1715)))/(2*4.9)) = 4.686s

This means the cliff is 1/2*g*t1^2 high

But it will take the car 4.686s to fall to the ground

and it would be v*t1 meters from the base

Therefore x = 2.562m/s*4.686s = 12m

So if the car had rolled without the engine running it would be 12m from the cliff, but since it was only 25ft (7.62m) from the cliff. The car could not have come off the hill

For the car to hit 7.62m from the parking lot it would have had to have an initial velocity of
Average acceleration = 7.62/4.686 = 1.626m/s

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