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
An object is at x = 0 at t = 0 and moves along the x axis according to the velocity-time graph in Figure P2.62. (a) What is the object’s acceleration between 0 and 4.0 s? (b) What is the object's acceleration between 4.0 s and 9.0 s? (c) What is the object's acceleration between 13.0 s and 18.0 s? (d) At what time(s) is the object moving with the lowest speed? (e) At what time is the object farthest from x = 0? (1) What is the final position x of the object at t = 18.0 s? (g) Through what total distance has the object moved between t = 0 and t = 18.0 s?
I am doing a free-fall experiment for Physics and I need to make a velocity vs. time graph using the average value of the final velocity. Do you know how I would be able to find the values for the velocity vs. time graph? The average value of the final velocity is 6.21954 m/s.    It is question #9.
The velocity vs time graph for an object is shown. How far has it gone in 4.5 seconds starting at t = 0s? The figure shows the velocity versus time graph for an object. From t = -3 to 0s, the velocity is constant at 45 m/s. From t=0 to 4.5s, velocity decreases linearly from 45 m/s to 0 m/s. 101 m 225 m 150 m 112 m

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
Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY