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I have a question in my Precalculus class. Here is an example-
A deer was struck by a car in the road. The driver left skid marks of 185 feet.
The police know that the length of skid marks (when brakes are applied) varies directly
to the square of the speed of the car. They know that under normal conditions
a car going 30 miles/hour leaves skid marks of 40 feet.
How fast was the car going before applying brakes before the deer was hit?
Round answer to two decimal places.
The equation I was told to use is L=Ks^2
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First step is to determine the value of coefficient K in your equation.
So, you write this equation, based on the second scenario
40 = ( <<<---=== in this case, L = 40 ft, s = 30 mph, the speed)
From this equation,
K = = 0.0444... = 0.0444 (rounded).
Thus you know your function in all details: it is
L = 0.0444*s^2, where s is the speed of the car, in miles per hour.
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| At this point, first step is complete, |
| and I start the second step. |
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Now in your base case with the unhappy deer, you write the corresponding equation
185 = , where v is the unknown speed of the car,
From the equation, you get
= = 4166.67,
v = = 64.55 miles per hour (rounded). ANSWER
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Solved.
Is everything clear to you in my solution ?