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Chapter 5: Problem 16

The time \(t\), in minutes, required to drive \(d\) miles on a congested road can be calculated from $$ t=\frac{60 d}{v_{0}-a q}. $$ where \(v_{0}\) is the average speed in miles per hour when the road is uncongested, \(q\) is the traffic flow in vehicles per hour, and \(a\) is a constant that depends on various characteristics of the roadway. For central London, the estimates \({ }^{72} v_{0}=28\) miles per hour and \(a=0.008\) mile per vehicle have been used. a. According to this model, how long will it take to travel 2 miles in central London if traffic flow is 1500 vehicles per hour? b. This model can be valid only when \(a q

Short Answer

Expert verified
a. It takes 7.5 minutes. b. The largest traffic flow is 3500 vehicles per hour. c. Travel time approaches infinity as flow nears 3500 vehicles per hour.

Step by step solution

01

Plug in Given Values to Formula

To calculate the time \( t \) to travel 2 miles, we use the formula: \[ t = \frac{60d}{v_0 - aq} \] where \( d = 2 \), \( v_0 = 28 \) miles per hour, \( a = 0.008 \) miles per vehicle, and \( q = 1500 \) vehicles per hour. Plug these values into the formula: \[ t = \frac{60 \times 2}{28 - 0.008 \times 1500} \]
02

Perform the Calculation for Part a

Calculate the value of \( aq \): \[ aq = 0.008 \times 1500 = 12 \] Then compute the denominator: \[ v_0 - aq = 28 - 12 = 16 \] Now calculate \( t \): \[ t = \frac{120}{16} = 7.5 \text{ minutes} \] Thus, the time to travel 2 miles is 7.5 minutes.
03

Determine the Largest Traffic Flow (Part b)

The model is valid when \( aq < v_0 \). Set up the inequality to solve for \( q \): \[ 0.008q < 28 \] Divide both sides by 0.008: \[ q < \frac{28}{0.008} = 3500 \] Thus, the largest traffic flow that the roads can handle is 3500 vehicles per hour.
04

Analyze the Effect of Traffic Flow Near the Limit (Part c)

When the traffic flow \( q \) approaches 3500 vehicles per hour, the denominator \( v_0 - aq \) approaches zero. This implies that \( t \), which depends on the inverse of the denominator, becomes extremely large. Therefore, as \( q \) gets close to 3500, the travel time \( t \) approaches infinity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Traffic Modeling
Traffic modeling is a mathematical approach used to understand and predict traffic patterns on a road network. Traffic flow in urban areas, such as central London, can be complex because of varying conditions like congestion. This exercise uses a formula to model traffic by considering variables like the average uncongested speed (\(v_0\)), traffic flow (\(q\)), and a constant (\(a\)) that captures road characteristics.
  • Variables Explained:
  • \(v_0\) represents the speed vehicles travel when there is no congestion, measured in miles per hour.
  • Traffic Flow \(q\): The amount of vehicles passing a point on a roadway, measured in vehicles per hour.
  • Constant \(a\): A value reflecting road characteristics and driver behavior, given as miles per vehicle.
To predict travel time on a congested road, the formula uses these variables to balance factors like speed and congestion, helping in planning and management of traffic flow effectively.
Average Speed Calculation
Average speed calculation is crucial for predicting how long a trip will take under given conditions. The formula used in traffic modeling for calculating average travel time involves several components:
  • Travel Time Calculation
  • The travel time (\(t\)) formula is: \\[ t = \frac{60d}{v_0 - aq} \]Here, \(d\) is the distance in miles, \(v_0\) is the average uncongested speed, and \(aq\) represents the congestion factor.
  • Example:
  • For a 2-mile travel distance in central London with traffic flow at 1500 vehicles per hour:
  • \(aq\) is calculated as \((0.008 \times 1500) = 12\). This impacts the speed reduction due to congestion.
  • The denominator becomes \(v_0 - aq = 28 - 12 = 16\).
  • \(t = \frac{120}{16} = 7.5\) minutes.
Understanding average speed with various flow rates helps in determining travel times accurately.
Inequality Solving
Inequality solving is used to establish the conditions under which the traffic model remains valid. The model requires that the congestion factor \(aq\) remains less than the average speed \(v_0\).
  • The Inequality:
  • We set up the inequality: \(aq < v_0\).
  • The specific inequality is \(0.008q < 28\).
  • Solving the Inequality:
  • To find the limit for \(q\):
  • Divide 28 by 0.008 to get the upper bound of vehicles per hour:
  • \(q < \frac{28}{0.008} = 3500\).
This tells us that the traffic flow should not exceed 3500 vehicles per hour for the model to stay valid. If \(q\) approaches 3500, travel time (\(t\)) trends towards infinity, indicating severe congestion.

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Most popular questions from this chapter

Suppose a population is growing according to the logistic formula \(N=\frac{500}{1+3 e^{-0.41 t}}\), where \(t\) is measured in years. a. Suppose that today there are 300 individuals in the population. Find a new logistic formula for the population using the same \(K\) and \(r\) values as the formula above but with initial value 300 . b. How long does it take the population to grow from 300 to 400 using the formula in part a?

A queue, or line, of traffic can form when a feeder road meets a main road with a high volume of traffic. In this exercise we assume that gaps in traffic on the main road (allowing cars to enter from the feeder road) appear randomly and that cars on the feeder road arrive at the intersection randomly also. We let \(s\) be the average rate at which a gap in traffic appears and \(a\) the average arrival rate, both per minute. Assume that \(s\) is greater than \(a\). a. The average length \(L\) of a queue is given by the rational function $$ L=\frac{a^{2}}{s(s-a)} . $$ Explain what happens to average queue length if arrival rate and gap rate are nearly the same. b. The average waiting time \(w\) (the time in minutes spent in the queue) is given by $$ w=\frac{a}{s(s-a)} . $$ If the gap rate is 3 per minute, what arrival rate will result in a driver's average waiting time of 2 minutes?

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let \(A\) be the amount of litter present, in grams per square meter, as a function of time \(t\) in years. If the litter falls at a constant rate of \(L\) grams per square meter per year, and if it decays at a constant proportional rate of \(k\) per year, then the limiting value of \(A\) is \(R=L / k\). For this exercise and the next, we suppose that at time \(t=0\), the forest floor is clear of litter. a. If \(D\) is the difference between the limiting value and \(A\), so that \(D=R-A\), then \(D\) is an exponential function of time. Find the initial value of \(D\) in terms of \(R\). b. The yearly decay factor for \(D\) is \(e^{-k}\). Find a formula for \(D\) in terms of \(R\) and \(k\). Reminder: \(\left(a^{b}\right)^{c}=a^{b c}\). c. Explain why \(A=R-R e^{-k t}\).

The accompanying table shows the relationship between the length \(L\), in centimeters, and the weight \(W\), in grams, of the North Sea plaice (a type of flatfish). \({ }^{32}\) a. Find a formula that models \(W\) as a power function of \(L\). (Round the power to one decimal place.) b. Explain in practical terms what \(W(50)\) means, and then calculate that value. c. If one plaice were twice as long as another, how much heavier than the other should it be? $$ \begin{array}{|c|c|} \hline L & W \\ \hline 28.5 & 213 \\ \hline 30.5 & 259 \\ \hline 32.5 & 308 \\ \hline 34.5 & 363 \\ \hline 36.5 & 419 \\ \hline 38.5 & 500 \\ \hline 40.5 & 574 \\ \hline 42.5 & 674 \\ \hline 44.5 & 808 \\ \hline 46.5 & 909 \\ \hline 48.5 & 1124 \\ \hline \end{array} $$

When seeds of a plant are sown at high density in a plot, the seedlings must compete with each other. As time passes, individual plants grow in size, but the density of the plants that survive decreases. \({ }^{33}\) This is the process of selfthinning. In one experiment, horseweed seeds were sown on October 21 , and the plot was sampled on successive dates. The results are summarized in Table \(5.8\), which gives for each date the density \(p\), in number per square meter, of surviving plants and the average dry weight \(w\), in grams, per plant. a. Explain how the table illustrates the phenomenon of self-thinning. b. Find a formula that models \(w\) as a power function of \(p\). c. If the density decreases by a factor of \(\frac{1}{2}\), what happens to the weight? d. The total plant yield y per unit area is defined to be the product of the average weight per plant and the density of the plants: \(y=w \times p\). As time goes on, the average weight per plant increases while the density decreases, so it's unclear whether the total yield will increase or decrease. Use the power function you found in part b to determine whether the total yield increases or decreases with time. Check your answer using the table. $$ \begin{array}{|l|r|c|} \hline \text { Date } & \text { Density } p & \text { Weight } w \\ \hline \text { November 7 } & 140,400 & 1.6 \times 10^{-4} \\ \hline \text { December } 16 & 36,250 & 7.7 \times 10^{-4} \\ \hline \text { January 30 } & 22,500 & 0.0012 \\ \hline \text { April 2 } & 9100 & 0.0049 \\ \hline \text { May 13 } & 4510 & 0.018 \\ \hline \text { June } 25 & 2060 & 0.085 \\ \hline \end{array} $$
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