OBJECTIVE To use rational functions and proportionality to analyze the problem of road capacity.

The best way to keep traffic moving is to avoid accidents, and the best way to avoid accidents is for drivers to maintain a safe following distance. To give drivers enough time to react to unforeseen events, the safe following distance is greater at higher speeds. You may be familiar with the car-length rule (at least one car length for every $10$ miles per hour) or the $2$-second rule (the car ahead must pass some object at least $2$ seconds before you do). In this exploration we investigate how algebra can help us find how many cars a road can carry at different speeds—assuming, of course, that all drivers maintain a safe following distance.

I. Carrying Capacity and Safe Following Distance

We’ll investigate how many cars a road can safely carry at different speeds. Let’s assume that each car is $20$ feet long and that the safe following distance $F(s)$ is one car length for every $10$ miles per hour of the speed $s$.

The following diagram is helpful in visualizing traffic.

1. Let’s find a formula for the safe following distance at any speed.
   1. Complete the table for the safe following distance.
      

      Speed $s$	$10$	$20$	$30$	$40$	$50$	$60$
      Following distance $F$	$20$

   2. So at a speed of $s$ miles per hour, the safe following distance is
      $F(s)$ $=$ __________.
   3. At a speed of $s$ miles per hour the road distance $D(s)$ that each car uses is

      $D(s) = \text{length of car} + \text{safe following distance}$
      
                $=$ _______________ $+$ _______________

2. Since we are measuring car lengths in feet, let’s convert speeds from the familiar miles per hour to feet per minute.
   1. Use the fact that $1$ mi/h = $88$ ft/min to complete the table.
      

      Speed (mi/h)	$10$	$20$	$30$	$40$	$50$	$60$
      Speed (ft/min)	$880$

   2. From the pattern in the table we see that $s$ mi/h is the same as
      $V(s)$ $=$ __________ ft/min.
3. If traffic is moving at $s$ miles per hour, then the number $N(s)$ of cars that pass a given point every minute is given by $$N(s) = \frac{\text{distance cars travel in one minute}}{\text{distance each car uses}}$$
   $$= \frac{V(s)}{D(s)} = \frac{\color{red}{\fbox{          }}}{\color{red}{\fbox{                         }}}$$ We’ll call the number $N(s)$ the carrying capacity of the road at speed $s$.
4. Graph the rational function you found in Question 3.
   1. What happens to the carrying capacity of the road as the speed increases?
   2. At what speed is the carrying capacity greatest?

Note that the stopping distance, the actual distance required to stop a car, is greater than the braking distance because of the reaction time needed before the brakes are applied.

II. Using Braking Distance to Find Maximum Carrying Capacity

For a moving car the braking distance is the distance required to bring the car to a complete stop after the brakes are applied. We can infer from physical principles that the braking distance $T(s)$ is proportional to the square of the speed $s$.

Using the braking distance as the safe following distance, let’s try to find the speed at which the maximum carrying capacity occurs. The following diagram is helpful.

1. Let’s find a formula for the braking distance at speed $s$.
   1. Express the above proportionality statement as an equation:
      

      $T(s)$ $=$ __________

   2. Suppose that the braking distance at $20$ mi/h is $20$ ft. Use this fact to find the proportionality constant. So the braking distance is given by the formula
      

      $T(s)$ $=$ __________

   3. Let’s assume that each car maintains the proper braking distance from the next car. In this case, at speed $s$ the road distance $D(s)$ each car uses is

      $D(s) = \text{length of car} + \text{braking distance}$
      
                $=$ _______________ $+$ _______________

2. If traffic is moving at $s$ miles per hour, then in this case the carrying capacity $N(s)$ of the road is $$N(s) = \frac{\text{distance cars travel in $1$ minute}}{\text{distance each car uses}}$$ $$ = \frac{V(s)}{D(s)}$$ $$= \frac{\color{red}{\fbox{          }}}{\color{red}{\fbox{                         }}}$$
3. Graph the rational function you found in Question 2.
   1. What happens to the carrying capacity as the speed increases?
   2. What is maximum carrying capacity? At what speed (in mi/h) does it occur?