Questions: While driving a car, you see a child suddenly crossing the street. Your brain registers the emergency and sends a signal to your foot to hit the brake. The car travels a reaction distance D, in feet, during the time it takes you to react, where D is a function of the car's speed r, in miles per hour. That reaction distance is a linear function given below. Complete parts (a) through (d). D(r) = (33r + 16) / 32 a) Find D(5), D(20), D(50), and D(65). D(5) = 5.7 D(20) = 21.1 D(50) = 52.1 D(65) = 67.5 (Round to the nearest tenth as needed.) b) Choose the correct graph of D(r). A. c. B. D. c) Interpret the meaning of the slope in the context of this problem. For every 1-mph increase in speed, the braking distance increases by .

Find $D(5)$, $D(20)$, $D(50)$, and $D(65)$.

Substitute $r=5$ into the equation

$D(5) = \frac{33(5) + 16}{32} = \frac{165 + 16}{32} = \frac{181}{32} = 5.65625 \approx 5.7$

Substitute $r=20$ into the equation

$D(20) = \frac{33(20) + 16}{32} = \frac{660 + 16}{32} = \frac{676}{32} = 21.125 \approx 21.1$

Substitute $r=50$ into the equation

$D(50) = \frac{33(50) + 16}{32} = \frac{1650 + 16}{32} = \frac{1666}{32} = 52.0625 \approx 52.1$

Substitute $r=65$ into the equation

$D(65) = \frac{33(65) + 16}{32} = \frac{2145 + 16}{32} = \frac{2161}{32} = 67.53125 \approx 67.5$

$D(5) \approx 5.7$, $D(20) \approx 21.1$, $D(50) \approx 52.1$, $D(65) \approx 67.5$

Choose the correct graph of $D(r)$.

Analyze the equation

The equation $D(r) = \frac{33r+16}{32}$ is a linear equation with a positive slope and a positive y-intercept.

Check the graphs

Graph C is the correct graph since it's a linear equation with a positive slope and a positive y-intercept.

Graph C.

Interpret the meaning of the slope in the context of this problem.

Find the slope.

The equation is $D(r) = \frac{33r}{32} + \frac{16}{32}$, so the slope is $\frac{33}{32}$, which is approximately 1.03.

Interpret the slope.

The slope represents the change in braking distance ($D$) for every 1 mph increase in speed ($r$). In this case, for every 1 mph increase in speed, the braking distance increases by approximately $\frac{33}{32}$ feet.

For every 1-mph increase in speed, the braking distance increases by $\frac{33}{32}$ feet.

$D(5) \approx 5.7$, $D(20) \approx 21.1$, $D(50) \approx 52.1$, $D(65) \approx 67.5$ Graph C. For every 1-mph increase in speed, the braking distance increases by $\frac{33}{32}$ feet.