--- **Automobile Traffic 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 determined that the average waiting time \( T \) until there is a gap in traffic of at least \( t \) seconds is approximately \( T = \frac{1}{e^{Rt}} \) seconds. A pedestrian walking at a speed of 3.3 ft/s requires \( t = \frac{w}{3.3} \) seconds to cross the road. Therefore, the average time the pedestrian will have to wait before crossing is \( f(w, R) = \left( \frac{w}{3.3} \right)e^{wR/3.3} \). **Problem 1:** What is the pedestrian's average waiting time if \( w = 24 \) ft and \( R = 0.2 \) vehicles per second? *Provide answers in decimal notation to two decimal places.* We need to compute: \[ t = \frac{24}{3.3} \approx 7.27 \] \[ f(24, 0.2) = t \cdot e^{24 \times 0.2 / 3.3} = 7.27 \cdot e^{1.4545} \approx 31.15\] t = **31.15** --- **Problem 2:** Use the Linear Approximation to estimate the increase in waiting time if \( w \) is increased to 26 ft. *Provide answers in decimal notation to two decimal places.* Using linear approximation, we find the differential: \[ \Delta f \approx \frac{\partial f}{\partial w} \cdot \Delta w \] Let's approximate: \[ \Delta f = 7.54 **Incorrect** \] --- **Problem 3:** Estimate the waiting time if the width is increased to 26 ft and \( R \) decreases to 0.18 vehicles per second. *Provide answers in decimal notation to two decimal places.* New values: \[ w = 26 \] \[ R = 0.18 \] Recalculate: \[ t = \frac{26}{3.3} \approx 7.88 \] \[ f(26, 0.18)
--- **Automobile Traffic 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 determined that the average waiting time \( T \) until there is a gap in traffic of at least \( t \) seconds is approximately \( T = \frac{1}{e^{Rt}} \) seconds. A pedestrian walking at a speed of 3.3 ft/s requires \( t = \frac{w}{3.3} \) seconds to cross the road. Therefore, the average time the pedestrian will have to wait before crossing is \( f(w, R) = \left( \frac{w}{3.3} \right)e^{wR/3.3} \). **Problem 1:** What is the pedestrian's average waiting time if \( w = 24 \) ft and \( R = 0.2 \) vehicles per second? *Provide answers in decimal notation to two decimal places.* We need to compute: \[ t = \frac{24}{3.3} \approx 7.27 \] \[ f(24, 0.2) = t \cdot e^{24 \times 0.2 / 3.3} = 7.27 \cdot e^{1.4545} \approx 31.15\] t = **31.15** --- **Problem 2:** Use the Linear Approximation to estimate the increase in waiting time if \( w \) is increased to 26 ft. *Provide answers in decimal notation to two decimal places.* Using linear approximation, we find the differential: \[ \Delta f \approx \frac{\partial f}{\partial w} \cdot \Delta w \] Let's approximate: \[ \Delta f = 7.54 **Incorrect** \] --- **Problem 3:** Estimate the waiting time if the width is increased to 26 ft and \( R \) decreases to 0.18 vehicles per second. *Provide answers in decimal notation to two decimal places.* New values: \[ w = 26 \] \[ R = 0.18 \] Recalculate: \[ t = \frac{26}{3.3} \approx 7.88 \] \[ f(26, 0.18)
Calculus: Early Transcendentals
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I need help with the last three problem, they are unlined red
![---
**Automobile Traffic 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 determined that the average waiting time \( T \) until there is a gap in traffic of at least \( t \) seconds is approximately \( T = \frac{1}{e^{Rt}} \) seconds.
A pedestrian walking at a speed of 3.3 ft/s requires \( t = \frac{w}{3.3} \) seconds to cross the road. Therefore, the average time the pedestrian will have to wait before crossing is \( f(w, R) = \left( \frac{w}{3.3} \right)e^{wR/3.3} \).
**Problem 1:**
What is the pedestrian's average waiting time if \( w = 24 \) ft and \( R = 0.2 \) vehicles per second?
*Provide answers in decimal notation to two decimal places.*
We need to compute:
\[ t = \frac{24}{3.3} \approx 7.27 \]
\[ f(24, 0.2) = t \cdot e^{24 \times 0.2 / 3.3} = 7.27 \cdot e^{1.4545} \approx 31.15\]
t = **31.15**
---
**Problem 2:**
Use the Linear Approximation to estimate the increase in waiting time if \( w \) is increased to 26 ft.
*Provide answers in decimal notation to two decimal places.*
Using linear approximation, we find the differential:
\[ \Delta f \approx \frac{\partial f}{\partial w} \cdot \Delta w \]
Let's approximate:
\[ \Delta f = 7.54 **Incorrect** \]
---
**Problem 3:**
Estimate the waiting time if the width is increased to 26 ft and \( R \) decreases to 0.18 vehicles per second.
*Provide answers in decimal notation to two decimal places.*
New values:
\[ w = 26 \]
\[ R = 0.18 \]
Recalculate:
\[ t = \frac{26}{3.3} \approx 7.88 \]
\[ f(26, 0.18)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faccc7427-2c55-44a5-b68e-19581b7568be%2F344af148-bb1e-4e12-8700-6c008d2558a3%2Ferw0jv_processed.png&w=3840&q=75)
Transcribed Image Text:---
**Automobile Traffic 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 determined that the average waiting time \( T \) until there is a gap in traffic of at least \( t \) seconds is approximately \( T = \frac{1}{e^{Rt}} \) seconds.
A pedestrian walking at a speed of 3.3 ft/s requires \( t = \frac{w}{3.3} \) seconds to cross the road. Therefore, the average time the pedestrian will have to wait before crossing is \( f(w, R) = \left( \frac{w}{3.3} \right)e^{wR/3.3} \).
**Problem 1:**
What is the pedestrian's average waiting time if \( w = 24 \) ft and \( R = 0.2 \) vehicles per second?
*Provide answers in decimal notation to two decimal places.*
We need to compute:
\[ t = \frac{24}{3.3} \approx 7.27 \]
\[ f(24, 0.2) = t \cdot e^{24 \times 0.2 / 3.3} = 7.27 \cdot e^{1.4545} \approx 31.15\]
t = **31.15**
---
**Problem 2:**
Use the Linear Approximation to estimate the increase in waiting time if \( w \) is increased to 26 ft.
*Provide answers in decimal notation to two decimal places.*
Using linear approximation, we find the differential:
\[ \Delta f \approx \frac{\partial f}{\partial w} \cdot \Delta w \]
Let's approximate:
\[ \Delta f = 7.54 **Incorrect** \]
---
**Problem 3:**
Estimate the waiting time if the width is increased to 26 ft and \( R \) decreases to 0.18 vehicles per second.
*Provide answers in decimal notation to two decimal places.*
New values:
\[ w = 26 \]
\[ R = 0.18 \]
Recalculate:
\[ t = \frac{26}{3.3} \approx 7.88 \]
\[ f(26, 0.18)
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