Question

when a driver decides to stop a car, it takes time first for the driver to react and put a foot on the brake, and then it takes additional time for the car to slow down. the total distance traveled during this period of time is called the stopping distance of the car. for some cars, the stopping distance ( d ), in feet, is given by the equation ( d = 0.05r^2 + 0.75r ), where ( r ) is the speed of the car in miles per hour.
(a) find the distance needed to stop a car traveling at 60 miles per hour.
______ ft
(b) if skid marks at an accident site are 135 feet long, how fast was the car traveling? (round your answer to the nearest integer.)
______ mph
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Explanation:

Part (a)

Step1: Substitute \( r = 60 \) into the formula

We are given the stopping distance formula \( d = 0.05r^2 + 0.75r \). Substitute \( r = 60 \) into this formula.

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Step2: Calculate each term

First, calculate \( 0.05\times(60)^2 \). \( 60^2 = 3600 \), so \( 0.05\times3600 = 180 \). Then calculate \( 0.75\times60 = 45 \).

Step3: Sum the terms

Add the two results together: \( 180 + 45 = 225 \).

Step1: Set up the equation

We know that \( d = 135 \), so substitute \( d = 135 \) into the formula \( d = 0.05r^2 + 0.75r \) to get the quadratic equation:

$$ 0.05r^2 + 0.75r - 135 = 0 $$

Step2: Multiply through by 20 to simplify

Multiply each term by 20 to eliminate the decimal: \( r^2 + 15r - 2700 = 0 \)

Step3: Solve the quadratic equation

We can use the quadratic formula \( r=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) for the quadratic equation \( ax^2+bx + c = 0 \). Here, \( a = 1 \), \( b = 15 \), and \( c=-2700 \).
First, calculate the discriminant \( \Delta=b^2 - 4ac=(15)^2-4\times1\times(-2700)=225 + 10800 = 11025 \)
Then, \( \sqrt{\Delta}=\sqrt{11025}=105 \)
Now, find the roots:

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We have two solutions: \( r=\frac{-15 + 105}{2}=\frac{90}{2}=45 \) and \( r=\frac{-15 - 105}{2}=\frac{-120}{2}=-60 \)
Since speed cannot be negative, we discard \( r=-60 \).

Answer:

\( 225 \)

Part (b)