At your job with an insurance company, you have been asked to help with the inve

Question

At your job with an insurance company, you have been asked to help with the investigation of a tragic "accident." At the scene is a road that runs straight down a hill with a slope of 10 degrees below the horizontal. At the bottom of the hill, the road goes horizontally for a very short distance, then ends in a parking lot overlooking a cliff. The cliff has a vertical drop of 400 feet to the horizontal ground below where the wrecked car lies 30 feet from the base of the cliff. The only witness claims that the car was parked somewhere on the hill, he can't exactly remember where, and the car just began coasting down the road. The witness did not hear an engine and thinks that the driver was drunk and passed out knocking off his emergency brake. The witness also remembers that the car took about 3 seconds to get down the hill. The lead investigator drops a stone from the edge of the cliff and, from the sound of it hitting the ground below, determines that it takes 5.0 seconds to fall to the bottom. Based on that information, 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. You are reminded to write down all of your assumptions so the investigation team can evaluate the applicability of your calculation to this situation.

Explanation / Answer

Assumptions - there is no friction with the road or parking lot and we are ignoring drag 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(10)*3.0 = 5.105m/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 = 5.105m/s*4.686s = 23.9m So if the car had rolled without the engine running it would be 23.9m from the cliff, but since it was only 30ft (9.14m) from the cliff. The car could not have come off the hill For the car to hit 9.14m from the parking lot it would have had to have an initial velocity of 9.14/4.686 = 1.95m/s

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