Please show all work! Just answer part D!
Chapter1: Equations, Inequalities, And Mathematical Modeling
Section1.1: Graphs Of Equations
Problem 9ECP
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Please show all work! Just answer part D!
Transcribed Image Text:(c) Estimate the waiting time if the width is increased to 27 ft and \( R \) decreases to 0.18.
(d) What is the rate of increase \( \delta \) in waiting time per 1-ft increase in width when \( w = 30 \) ft and \( R = 0.3 \) vehicle per second?
![### Traffic and Pedestrian Waiting Time Analysis
Automobile traffic passes a point \(P\) on a road of width \(w\) feet with an average rate of \(R\) vehicles per second. Although the arrival of automobiles is irregular, traffic engineers have found that the average waiting time \(T\) until there is a gap in traffic of at least \(t\) seconds is approximately \( T = t e^{Rt} \) seconds.
A pedestrian walking at a speed of 3.5 ft/s requires \( t = w / 3.5 \) s to cross the road. Therefore, the average time the pedestrian will have to wait before crossing is \( f(w, R) = \left( \frac{w}{3.5} \right) e^{wR / 3.5} \) seconds.
#### Problems
**(a) What is the pedestrian’s average waiting time if \( w = 25 \) ft and \( R = 0.2 \) vehicle per second?**
**(b) Use the Linear Approximation to estimate the increase in waiting time if \( w \) is increased to 27 ft.**
### Solutions
**(a) Calculating the pedestrian’s average waiting time:**
Given:
- \( w = 25 \) feet
- \( R = 0.2 \) vehicle per second
We use the formula:
\[ f(w, R) = \left( \frac{w}{3.5} \right) e^{wR / 3.5} \]
Substituting \( w = 25 \) and \( R = 0.2 \):
\[ f(25, 0.2) = \left( \frac{25}{3.5} \right) e^{(25 \times 0.2) / 3.5} \]
\[ f(25, 0.2) = \left( 7.142857 \right) e^{5/3.5} \]
\[ f(25, 0.2) = 7.142857 e^{1.42857} \]
\[ f(25, 0.2) \approx 7.142857 \times 4.177 \]
\[ f(25, 0.2) \approx 29.82 \, \text{seconds} \]
**(b) Est](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa40558bc-afda-4e2b-be1a-c4211c27f73c%2F0e61b5d2-a9e8-43a6-ac5a-69caa8621b5c%2Ftqu4m9_processed.png&w=3840&q=75)
Transcribed Image Text:### Traffic and Pedestrian Waiting Time Analysis
Automobile traffic passes a point \(P\) on a road of width \(w\) feet with an average rate of \(R\) vehicles per second. Although the arrival of automobiles is irregular, traffic engineers have found that the average waiting time \(T\) until there is a gap in traffic of at least \(t\) seconds is approximately \( T = t e^{Rt} \) seconds.
A pedestrian walking at a speed of 3.5 ft/s requires \( t = w / 3.5 \) s to cross the road. Therefore, the average time the pedestrian will have to wait before crossing is \( f(w, R) = \left( \frac{w}{3.5} \right) e^{wR / 3.5} \) seconds.
#### Problems
**(a) What is the pedestrian’s average waiting time if \( w = 25 \) ft and \( R = 0.2 \) vehicle per second?**
**(b) Use the Linear Approximation to estimate the increase in waiting time if \( w \) is increased to 27 ft.**
### Solutions
**(a) Calculating the pedestrian’s average waiting time:**
Given:
- \( w = 25 \) feet
- \( R = 0.2 \) vehicle per second
We use the formula:
\[ f(w, R) = \left( \frac{w}{3.5} \right) e^{wR / 3.5} \]
Substituting \( w = 25 \) and \( R = 0.2 \):
\[ f(25, 0.2) = \left( \frac{25}{3.5} \right) e^{(25 \times 0.2) / 3.5} \]
\[ f(25, 0.2) = \left( 7.142857 \right) e^{5/3.5} \]
\[ f(25, 0.2) = 7.142857 e^{1.42857} \]
\[ f(25, 0.2) \approx 7.142857 \times 4.177 \]
\[ f(25, 0.2) \approx 29.82 \, \text{seconds} \]
**(b) Est
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