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The following formula \({ }^{71}\) can be used to approximate the average transit speed \(S\) of a public transportation vehicle: $$ S=\frac{C D}{C T+D+C^{2}\left(\frac{1}{2 a}+\frac{1}{2 d}\right)} $$ Here \(C\) is the cruising speed in miles per hour, \(D\) is the average distance between stations in miles, \(T\) is the stop time at stations in hours, \(a\) is the rate of acceleration in miles per hour per hour, and \(d\) is the rate of deceleration in miles per hour per hour. For a certain subway the average distance between stations is 3 miles, the stop time is 3 minutes \((0.05\) hour), and rate of acceleration and deceleration are both \(3.5\) miles per hour per second (12,600 miles per hour per hour). a. Using the information provided, express \(S\) as a rational function of \(C\). b. As a rule of thumb, we might use the cruising speed as an estimate of transit speed. Discuss the merits of such a rule of thumb. How does increasing cruising speed affect such an estimate? (Suggestion: Compare the graph of the given rational function with that of the graph of \(S=C\).) c. What cruising speed will yield a transit speed of 30 miles per hour?

Step by step solution

01

Substitute Given Values

Insert the provided values for \(D\), \(T\), \(a\), and \(d\) into the transit speed equation. We have: \[ D = 3, \ T = 0.05, \ a = d = 12,600. \] Replacing these in the formula, we obtain:\[ S = \frac{C \times 3}{C \times 0.05 + 3 + C^2 \left( \frac{1}{2 \times 12,600} + \frac{1}{2 \times 12,600} \right)}.\] Where the terms for acceleration and deceleration simplify to:\[ \frac{1}{12,600}.\]

02

Simplify the Expression

Simplify the equation for \(S\):\[ \frac{1}{2a} + \frac{1}{2d} = \frac{1}{12,600} + \frac{1}{12,600} = \frac{2}{12,600} = \frac{1}{6,300}. \] Thus, the equation simplifies to:\[ S = \frac{3C}{0.05C + 3 + \frac{C^2}{6,300}}. \] This is the rational function of \(C\).

03

Analyze Transit Speed Estimation

To analyze the rule of thumb, compare the formula \(S = C\) with the function derived in Step 2. "Using cruising speed as transit speed assumes all downtime and acceleration/deceleration effects are negligible. However, this is unrealistic as stop times and acceleration phases reduce average speed. Thus, increasing \(C\) without reducing these factors might result in noticeable discrepancies between \(C\) and \(S\)."

04

Solve for Desired Transit Speed

To find \(C\) for \(S = 30\), set the equation from Step 2 equal to 30:\[ 30 = \frac{3C}{0.05C + 3 + \frac{C^2}{6,300}}. \]Cross multiply to eliminate the fraction:\[ 30(0.05C + 3 + \frac{C^2}{6,300}) = 3C. \]Simplify and solve this equation for \(C\), typically using numerical methods since it is quadratic in nature.

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Key Concepts

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

Transit Speed

Transit speed in public transport systems measures how effectively a vehicle moves passengers between destinations. It reflects the practical speed of the vehicle accounting for various factors that influence its operation. Transit speed is not merely the cruising or top speed of a vehicle, but it considers:

• The average distance between stops: Shorter distances between stops typically result in a lower average speed since the vehicle spends more time accelerating and decelerating.
• The stop time: Longer stop times at each station lead to slower transit speeds.
• The rates of acceleration and deceleration: These determine how quickly a vehicle can resume cruising speed after a stop.

The formula provided for calculating transit speed (\[S = \frac{C D}{C T + D + C^2 \left(\frac{1}{2a} + \frac{1}{2d}\right)}\]) encapsulates these elements, showcasing how they interplay to affect transit efficiency. A practical understanding of transit speed helps optimize routes and schedules for better passenger experiences.

Quadratic Equations

Quadratic equations are crucial in solving for specific conditions within rational function-based problems like transit speed calculations.
They take the general form:\[ax^2 + bx + c = 0.\]In the context of the transit speed problem, you encounter a quadratic equation when solving for cruising speed, given a desired transit speed.\[30 = \frac{3C}{0.05C + 3 + \frac{C^2}{6,300}}.\]Setting this equation involves steps to remove fractions, typically resulting in a quadratic form after simplifying. Solving quadratic equations:

• Begin by using algebraic manipulation to rearrange the equation into standard form.
• Utilize methods like factoring, completing the square, or the quadratic formula\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]to find potential values for C.

Quadratic equations present solutions that are generally numerical, especially if the polynomials do not factor neatly. Understanding these processes is vital for resolving functions that model more complex real-world phenomena.

Estimation in Mathematics

Estimation is a valuable mathematical tool, particularly when simplifying complex equations or making quick decisions based on available data. In transit speed calculations, estimations help anticipate scenarios without detailed computation.
For instance, using cruising speed (\(C\)) as an estimate for transit speed (\(S\)) is a rough calculation, assuming minimizing factors like stop time and deceleration.

• This approach is beneficial for quick assessments such as determining potential journey times in real-time scenarios.
• While it simplifies calculations, it may introduce significant errors if applied to detailed planning or optimization.

Comparing the rational function graphed against the line \(S = C\) reveals how much factors like stop times actually affect speeds. Estimations are an indispensable part of advanced mathematical reasoning and problem-solving, offering a means to approach problems when precise calculations are not possible or practical.