In the lesson, you investigated the relationship between velocity and braking distance. You will now investigate the relationship between the coefficient of friction and the braking distance.
Recall that the formula for the braking distance of a car is $\displaystyle {d}=\frac{{{V}^{{2}}}}{{{2}{g{{\left({f}+{G}\right)}}}}}$ 

1. Which of the letters in the formula represents a constant?
   • G, road grade as a %
   • d, distance in ft
   • f, the friction coefficient
   • g, the force of gravity in ft/sec/sec
   • V, velocity in ft/sec
   
   
   
2. To investigate the relationship between the coefficient of friction and the braking distance, you need to hold the other variables fixed. Let G = 0.05. Which of the following is a correct interpretation of the value G = 0.05?
   • The grade of a road is 5%, which is a vertical increase of 5 feet for every 100 feet of horizontal increase.
   • The grade of a road is 0.05%, which is a vertical increase of 0.05 feet for every 100 feet of horizontal increase.
   • The grade of a road is 5%, which is a vertical increase of 5 feet for every 1 foot of horizontal increase.
   • The grade of a road is 0.05%, which is a vertical increase of 0.05 feet for every 1 foot of horizontal increase.
   
   
   
3. Let V = 55 mph and use the value of G in Part (b). Which of the following expressions represents the simplified form of the formula using these values?
   NOTE: g = acceleration due to gravity (32.2 ft/sec/sec)
   • $\displaystyle {d}=\frac{{{6507.11}}}{{{64.4}{f}+{0.05}}}$ ft
   • $\displaystyle {d}=\frac{{{6507.11}}}{{{64.4}{f}}}+{2020.84}$ ft
   • $\displaystyle {d}=\frac{{{6507.11}}}{{{64.4}{f}+{3.22}}}$ ft
4. Create a Google spreadsheet for the following table:

   surface (assuming good tires)	friction	distance
   Dry asphalt/concrete	1
   Wet asphalt	0.7
   Dry earth road	0.65
   Gravel	0.6
   Wet earth road	0.55
   Hard packed snow	0.15

   
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