4) The speed of sound in air on a given day is 340 m/s. If a person shouts at a building and hears an echo 4 seconds later, how far away is the building?
4) The speed of sound in air on a given day is 340 m/s. If a person shouts at a building and hears an echo 4 seconds later, how far away is the building?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Question
#4
![### Problem Statement
The speed of sound in air on a given day is 340 m/s. If a person shouts at a building and hears an echo 4 seconds later, how far away is the building?
### Explanation
When a person shouts, the sound travels to the building and the echo is the sound reflecting back to the person. The total time taken for the sound to travel to the building and back is 4 seconds.
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Speed of sound (\( v \)) = 340 m/s
- Total time for echo (\( t \)) = 4 seconds
2. **Understand the Total Distance Traveled by Sound:**
- The sound travels to the building and back, so the total distance traveled is the sum of the distance to the building and the distance back to the person.
- Let's denote the distance to the building as \( d \). Therefore, the total distance covered by the sound \( 2d \) (to the building and back).
3. **Calculate the Distance Using the Formula:**
- The formula to calculate distance using speed and time is: \( \text{distance} = \text{speed} \times \text{time} \).
- From the given information, the total distance covered by the sound (\( 2d \)) is:
\[
2d = v \times t
\]
- Substitute the given values:
\[
2d = 340 \, \text{m/s} \times 4 \, \text{s}
\]
\[
2d = 1360 \, \text{m}
\]
4. **Solve for the Distance to the Building:**
- Divide both sides of the equation by 2 to find \( d \):
\[
d = \frac{1360 \, \text{m}}{2}
\]
\[
d = 680 \, \text{m}
\]
### Conclusion
The building is 680 meters away from the person.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F310c196c-1192-4fce-aa56-471375ee573b%2Fef2aeafc-5dc6-484c-af47-e78d17be244a%2Fv451iy7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
The speed of sound in air on a given day is 340 m/s. If a person shouts at a building and hears an echo 4 seconds later, how far away is the building?
### Explanation
When a person shouts, the sound travels to the building and the echo is the sound reflecting back to the person. The total time taken for the sound to travel to the building and back is 4 seconds.
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Speed of sound (\( v \)) = 340 m/s
- Total time for echo (\( t \)) = 4 seconds
2. **Understand the Total Distance Traveled by Sound:**
- The sound travels to the building and back, so the total distance traveled is the sum of the distance to the building and the distance back to the person.
- Let's denote the distance to the building as \( d \). Therefore, the total distance covered by the sound \( 2d \) (to the building and back).
3. **Calculate the Distance Using the Formula:**
- The formula to calculate distance using speed and time is: \( \text{distance} = \text{speed} \times \text{time} \).
- From the given information, the total distance covered by the sound (\( 2d \)) is:
\[
2d = v \times t
\]
- Substitute the given values:
\[
2d = 340 \, \text{m/s} \times 4 \, \text{s}
\]
\[
2d = 1360 \, \text{m}
\]
4. **Solve for the Distance to the Building:**
- Divide both sides of the equation by 2 to find \( d \):
\[
d = \frac{1360 \, \text{m}}{2}
\]
\[
d = 680 \, \text{m}
\]
### Conclusion
The building is 680 meters away from the person.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON