Question

free response
police use a formula to estimate the speed a car was traveling before an accident by measuring its skid marks. function s is used to model the speed the car was traveling in mph where d is the distance the car skidded in feet and f is the coefficient of friction which depends on the road surface and road conditions.
$s(d)=\sqrt{30df}$
a. a country road has a coefficient of friction of 0.9 when it is dry and 0.4 when it is wet. what values of distance skidded would you expect in both conditions for a car that was travelling 110 mph when the brakes were applied?
b. analyze the rate of change from a measured 70 - foot skid mark to a 160 - foot skid mark on the dry country road.
c. the distance of a skid mark is inversely proportional to the product of 30 and the coefficient of friction. using the 60 - foot skid mark on a dry country road, find the constant of proportionality. explain what it means in this context.

Explanation:

Step1: Solve part a

• Dry condition:
• Given \(S(d)=\sqrt{30df}\), we want to solve for \(d\) when \(S = 110\) and \(f = 0.9\).
• Square both sides of the equation \(S=\sqrt{30df}\) to get \(S^{2}=30df\).
• Then \(d=\frac{S^{2}}{30f}\). Substitute \(S = 110\) and \(f = 0.9\): \(d=\frac{110^{2}}{30\times0.9}=\frac{12100}{27}\approx448.15\)
• Wet condition:
• Substitute \(S = 110\) and \(f = 0.4\) into \(d=\frac{S^{2}}{30f}\).
• \(d=\frac{110^{2}}{30\times0.4}=\frac{12100}{12}\approx1008.33\)

Step2: Solve part b

• Find \(S(70)\):
• Substitute \(d = 70\) and \(f = 0.9\) into \(S(d)=\sqrt{30df}\).
• \(S(70)=\sqrt{30\times70\times0.9}=\sqrt{1890}\approx43.47\)
• Find \(S(160)\):
• Substitute \(d = 160\) and \(f = 0.9\) into \(S(d)=\sqrt{30df}\).
• \(S(160)=\sqrt{30\times160\times0.9}=\sqrt{4320}\approx65.73\)
• Calculate the average rate of change:
• Use the formula \(\frac{S(d_{2})-S(d_{1})}{d_{2}-d_{1}}\), where \(d_{1}=70\), \(d_{2}=160\), \(S(d_{1})\approx43.47\), \(S(d_{2})\approx65.73\)
• \(\frac{65.73 - 43.47}{160 - 70}=\frac{22.26}{90}\approx0.25\)

Step3: Solve part c

• Find the constant of proportionality:
• Since \(d=\frac{k}{30f}\) (inverse proportion), when \(d = 60\) and \(f = 0.9\)
• Cross - multiply to get \(k = 30df\). Substitute \(d = 60\) and \(f = 0.9\)
• \(k=30\times0.9\times60=1620\)
• Interpret the constant:
• In the context of the skid - mark formula \(S=\sqrt{30df}\), for a given road (fixed \(f\)), the relationship \(d=\frac{k}{30f}\) shows that as the coefficient of friction (\(f\)) increases (road is in better condition, e.g., dry), the skid - mark distance (\(d\)) for a given speed (related to \(k\)) decreases. Here, \(k\) represents a fixed value that relates the three variables \(d\), \(f\), and the underlying speed - related quantity in the formula.

Answer:

a. When dry: \(d=\frac{S^{2}}{30f}\), \(S = 110\) mph, \(f=0.9\). Then \(d=\frac{110^{2}}{30\times0.9}=\frac{12100}{27}\approx448.15\) feet. When wet: \(f = 0.4\), \(d=\frac{110^{2}}{30\times0.4}=\frac{12100}{12}\approx1008.33\) feet.

b. The average rate of change formula is \(\frac{S(d_{2})-S(d_{1})}{d_{2}-d_{1}}\). For \(d_{1} = 70\), \(S(70)=\sqrt{30\times70\times0.9}=\sqrt{1890}\approx 43.47\) mph. For \(d_{2}=160\), \(S(160)=\sqrt{30\times160\times0.9}=\sqrt{4320}\approx65.73\) mph. The average rate of change is \(\frac{65.73 - 43.47}{160 - 70}=\frac{22.26}{90}\approx0.25\) mph per foot.

c. If \(d=\frac{k}{30f}\) (inverse - proportion), when \(d = 60\) and \(f=0.9\), then \(k=30\times0.9\times60 = 1620\). In this context, the constant of proportionality \(k = 1620\) means that for the dry country - road (\(f = 0.9\)), the product of \(30\), the coefficient of friction (\(f\)), and the skid - mark distance (\(d\)) is always \(1620\).