College Physics, Volume 1
College Physics, Volume 1
2nd Edition
ISBN: 9781133710271
Author: Giordano
Publisher: Cengage
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 2, Problem 53P

(a)

To determine

The average velocity over the time interval t=0.0s to t=10.0s.

(a)

Expert Solution
Check Mark

Answer to Problem 53P

The average velocity over the time interval t=0.0s to t=10.0s, is 3.0m/s_.

Explanation of Solution

Write the expression for average velocity.

  vavg=ΔxΔt        (I)

Conclusion:

Substitute, 30m0m for Δx, and 10.0s0.0s for Δt in equation (I) to find the average velocity from t=0.0s to t=10.0s.

  vavg=30m0m10.0s0s=3.0m/s

Therefore, the average velocity over the time interval t=0.0s to t=10.0s, is 3.0m/s_.

(b)

To determine

The average velocity over the time interval t=0.0s to t=5.0s.

(b)

Expert Solution
Check Mark

Answer to Problem 53P

The average velocity over the time interval t=0.0s to t=5.0s is 4.5m/s_.

Explanation of Solution

Use equation (I) to find the answer.

Conclusion:

Substitute, 22.5m0m for Δx, and 5.0s0.0s for Δt in equation (I) to find the average velocity from t=0.0s to t=5.0s.

  vavg=22.5m0m5.0s0s=4.5m/s

Therefore, the average velocity over the time interval t=0.0s to t=5.0s is 4.5m/s_.

(c)

To determine

The average velocity between t=5.0s and t=10.0s.

(c)

Expert Solution
Check Mark

Answer to Problem 53P

The average velocity between t=5.0s and t=10.0s is 1.5m/s_.

Explanation of Solution

Use equation (I) to find the answer.

Conclusion:

Substitute, 30m22.5m for Δx, and 10.0s5.0s for Δt in equation (I) to find the average velocity from t=5.0s to t=10.0s.

  vavg=30m22.5m10.0s5.0s=1.5m/s

Therefore, the average velocity between t=5.0s and t=10.0s is 1.5m/s_.

(d)

To determine

The relation of answers in part (a), (b), and (c).

(d)

Expert Solution
Check Mark

Answer to Problem 53P

The relation is average of answers in part (b), and (c) is answer in part (a).

Explanation of Solution

The total average velocity over the time interval t=0.0s to t=10.0s must be equal to the average of the individual average velocities for the time intervals t=0.0s to t=5.0s and t=5.0s to t=10.0s.

Write the expression for the average velocity.

  vavg,a=vavg,b+vavg,c2        (II)

Conclusion:

Substitute, 4.5m/s for vavg,b, 1.5m/s for vavg,c in equation (II).

  vavg,a=4.5m/s+1.5m/s2=3.0m/s

Therefore, the relation is average of answers in part (b), and (c) is answer in part (a).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
For each part make sure to include sign to represent direction, with up being positive and down being negative. A ball is thrown vertically upward with a speed of 30.5 m/s. A) How high does it rise? y= B) How long does it take to reach its highest point? t= C) How long does it take the ball return to its starting point after it reaches its highest point? t= D) What is its velocity when it returns to the level from which it started? v=
Four point charges of equal magnitude Q = 55 nC are placed on the corners of a rectangle of sides D1 = 27 cm and D2 = 11cm. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative. Use a coordinate system where the positive y-direction is up and the positive x-direction is to the right. A. Which of the following represents a free-body diagram for the charge on the lower left hand corner of the rectangle? B. Calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the rectangle.Numeric   : A numeric value is expected and not an expression.Fx = __________________________________________NC. Calculate the vertical component of the net force, in newtons, on the charge which lies at the lower left corner of the rectangle.Numeric   : A numeric value is expected and not an expression.Fy = __________________________________________ND. Calculate the magnitude of the…
Point charges q1=50.0μC and q2=-35μC are placed d1=1.0m apart, as shown. A. A third charge, q3=25μC, is positioned somewhere along the line that passes through the first two charges, and the net force on q3 is zero. Which statement best describes the position of this third charge?1)  Charge q3 is to the right of charge q2. 2)  Charge q3 is between charges q1 and q2. 3)  Charge q3 is to the left of charge q1. B. What is the distance, in meters, between charges q1 and q3? (Your response to the previous step may be used to simplify your solution.)Give numeric value.d2 = __________________________________________mC. Select option that correctly describes the change in the net force on charge q3 if the magnitude of its charge is increased.1)  The magnitude of the net force on charge q3 would still be zero. 2)  The effect depends upon the numeric value of charge q3. 3)  The net force on charge q3 would be towards q2. 4)  The net force on charge q3 would be towards q1. D. Select option that…

Chapter 2 Solutions

College Physics, Volume 1

Ch. 2 - Prob. 6QCh. 2 - Prob. 7QCh. 2 - Prob. 8QCh. 2 - Prob. 9QCh. 2 - Prob. 10QCh. 2 - Prob. 11QCh. 2 - Prob. 12QCh. 2 - Prob. 13QCh. 2 - Prob. 14QCh. 2 - Prob. 15QCh. 2 - Prob. 16QCh. 2 - Prob. 17QCh. 2 - Prob. 18QCh. 2 - Prob. 19QCh. 2 - Three blocks rest on a table as shown in Figure...Ch. 2 - Two football players start running at opposite...Ch. 2 - Prob. 22QCh. 2 - In SI units, velocity is measured in units of...Ch. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Consider a marble falling through a very thick...Ch. 2 - Prob. 10PCh. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Figure P2.13 shows three motion diagrams, where...Ch. 2 - Prob. 14PCh. 2 - Figure P2.15 shows several hypothetical...Ch. 2 - Prob. 16PCh. 2 - Figure P2.17 shows several hypothetical...Ch. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - For the object described by Figure P2.24, estimate...Ch. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60P
Knowledge Booster
Background pattern image
Physics
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.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
An Introduction to Physical Science
Physics
ISBN:9781305079137
Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY