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.
( square ) 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.)
( square ) mph
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Explanation:

Part (a)

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

We have the stopping distance formula \( d = 0.05r^2 + 0.75r \). Substitute \( r = 60 \) into this formula:
\( d = 0.05\times(60)^2 + 0.75\times60 \)

Step2: Calculate each term

First, calculate \( 0.05\times(60)^2 \): \( 0.05\times3600 = 180 \)
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 into the formula \( 135 = 0.05r^2 + 0.75r \)
Rearrange it to the standard quadratic form: \( 0.05r^2 + 0.75r - 135 = 0 \)
Multiply through by 20 to eliminate the decimal: \( r^2 + 15r - 2700 = 0 \)

Step2: Use the quadratic formula

The quadratic formula for \( ax^2 + bx + c = 0 \) is \( r=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \)
Here, \( a = 1 \), \( b = 15 \), \( c = - 2700 \)
Calculate the discriminant \( \Delta=b^2 - 4ac=(15)^2-4\times1\times(-2700)=225 + 10800 = 11025 \)

Step3: Solve for \( r \)

\( r=\frac{-15\pm\sqrt{11025}}{2\times1}=\frac{-15\pm105}{2} \)
We have two solutions:
\( r_1=\frac{-15 + 105}{2}=\frac{90}{2}=45 \)
\( r_2=\frac{-15 - 105}{2}=\frac{-120}{2}=-60 \) (speed can't be negative, so we discard this)

Answer:

225

Part (b)