3. Here is a position graph of an object: At t = 1.5 s, the object's velocity is x (m) 20 A. 40 m/s В. 20 m/s C. 10 m/s 10 D. -10 m/s E. None of the above -t (s) 4 1 3
3. Here is a position graph of an object: At t = 1.5 s, the object's velocity is x (m) 20 A. 40 m/s В. 20 m/s C. 10 m/s 10 D. -10 m/s E. None of the above -t (s) 4 1 3
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
Show work/give explanation.
![**Question 3:**
Here is a position graph of an object. At \( t = 1.5 \) s, the object's velocity is:
A. 40 m/s
B. 20 m/s
C. 10 m/s
D. -10 m/s
E. None of the above
**Graph Description:**
The graph is a position-time graph depicting an object’s position \( x \) in meters (m) against time \( t \) in seconds (s).
- The graph starts at the origin (0,0).
- From \( t = 0 \) to \( t = 2 \) seconds, the graph is a straight line increasing to a peak at \( x = 20 \) meters.
- From \( t = 2 \) to \( t = 4 \) seconds, the graph slopes downward back to 0 meters at \( x = 0 \).
**Analysis:**
The graph forms a triangle with a base from \( t = 0 \) to \( t = 4 \) seconds and a peak at \( t = 2 \) seconds. The slope of the line indicates the velocity of the object. The velocity is positive as the object moves forward and negative as it returns. To find the velocity at \( t = 1.5 \) seconds, calculate the slope of the line in the interval from \( t = 0 \) to \( t = 2 \) seconds.
The slope (velocity) is calculated as:
\[ \text{Velocity} = \frac{\text{Change in position (Δx)}}{\text{Change in time (Δt)}} = \frac{20 \, \text{m} - 0 \, \text{m}}{2 \, \text{s} - 0 \, \text{s}} = 10 \, \text{m/s} \]
Thus, the object's velocity at \( t = 1.5 \) seconds is 10 m/s. The correct answer is C. 10 m/s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70932241-1ad7-48fc-b0c5-52d574ff0aa2%2Fb38508b5-a432-43f0-9277-aae261b6904a%2Fucq7nf0s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 3:**
Here is a position graph of an object. At \( t = 1.5 \) s, the object's velocity is:
A. 40 m/s
B. 20 m/s
C. 10 m/s
D. -10 m/s
E. None of the above
**Graph Description:**
The graph is a position-time graph depicting an object’s position \( x \) in meters (m) against time \( t \) in seconds (s).
- The graph starts at the origin (0,0).
- From \( t = 0 \) to \( t = 2 \) seconds, the graph is a straight line increasing to a peak at \( x = 20 \) meters.
- From \( t = 2 \) to \( t = 4 \) seconds, the graph slopes downward back to 0 meters at \( x = 0 \).
**Analysis:**
The graph forms a triangle with a base from \( t = 0 \) to \( t = 4 \) seconds and a peak at \( t = 2 \) seconds. The slope of the line indicates the velocity of the object. The velocity is positive as the object moves forward and negative as it returns. To find the velocity at \( t = 1.5 \) seconds, calculate the slope of the line in the interval from \( t = 0 \) to \( t = 2 \) seconds.
The slope (velocity) is calculated as:
\[ \text{Velocity} = \frac{\text{Change in position (Δx)}}{\text{Change in time (Δt)}} = \frac{20 \, \text{m} - 0 \, \text{m}}{2 \, \text{s} - 0 \, \text{s}} = 10 \, \text{m/s} \]
Thus, the object's velocity at \( t = 1.5 \) seconds is 10 m/s. The correct answer is C. 10 m/s.
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
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