Question

A velocity-time graph for an object moving along the x axis is shown in the figure. Every division along the vertical axis corresponds to 1.50 m/s and each division along the horizontal axis corresponds to 1.50 s. (a) Plot a graph of the acceleration versus time. (b) Determine the average acceleration of the object in the time interval t = 7.50 s to t = 22.5 s. m/s" (c) Determine the average acceleration of the object in the time interval t = 0 to t = 30.0 s. m/s" 

          A velocity-time graph for an object moving along the x axis is shown in the figure. Every division along the vertical axis corresponds to 1.50 m/s and each division along the horizontal axis corresponds to 1.50 s.
(a) Plot a graph of the acceleration versus time.
(b) Determine the average acceleration of the object in the time interval t = 7.50 s to t = 22.5 s.
m/s"
(c) Determine the average acceleration of the object in the time interval t = 0 to t = 30.0 s.
m/s"
Show more…
A velocity-time graph for an object moving along the x axis is shown in the figure. Every division along the vertical axis corresponds to 1.50 m/s and each division along the horizontal axis corresponds to 1.50 s. (a) Plot a graph of the acceleration versus time. (b) Determine the average acceleration of the object in the time interval t = 7.50 s to t = 22.5 s. m/s" (c) Determine the average acceleration of the object in the time interval t = 0 to t = 30.0 s. m/s"

Added by Ronald A.

Close

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
A velocity-time graph for an object moving along the x axis is shown in the figure. Every division along the vertical axis corresponds to 1.50 m/s and each division along the horizontal axis corresponds to 1.50 s. (a) Plot a graph of the acceleration versus time. (b) Determine the average acceleration of the object in the time interval t = 7.50 s to t = 22.5 s. (c) Determine the average acceleration of the object in the time interval t = 0 to t = 30.0 s.
Close icon
Play audio
Feedback
Powered by NumerAI
David Collins Jennifer Stoner
Kathleen Carty verified

Supratim Pal and 95 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question
*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs
*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.
*

Recommended Videos

-
a-velocity-time-graph-for-an-object-moving-along-the-x-axis-is-shown-in-figure-p213-a-plot-a-graph-o-91112

A velocity-time graph for an object moving along the $x$ axis is shown in Figure P2.13. (a) Plot a graph of the acceleration versus time. Determine the average acceleration of the object (b) in the time interval $t=5.00 \mathrm{s}$ to $t=15.0 \mathrm{s}$ and (c) in the time interval $t=0$ to $t=20.0 \mathrm{s}$

Adi S.
a-velocity-time-graph-for-an-object-moving-along-the-x-axis-is-shown-in-figure-mathrmp-213-a-plot-a-

A velocity-time graph for an object moving along the $x$ axis is shown in Figure $\mathrm{P} 2.13 .$ (a) Plot a graph of the acceleration versus time. Determine the average acceleration of the object $(b)$ in the time interval $t=5.00 \mathrm{s}$ to $t=15.0 \mathrm{s}$ and (c) in the time interval $t=0$ to $t=20.0 \mathrm{s}$

Physics for Scientists and Engineers with Modern Physics

an-object-moves-along-the-axis-according-to-the-equation-yt-seconds_-determine-the-following_-average-velocity-of-the-object-during-the-time-interva-ms-where-b-270-m-c-220-ms-and-yt-will-be-54673

An object moves along the y axis according to the equation y(t) = b + ct³, where b = 2.70 m, c = 2.20 m/s³, and y(t) will be in meters when t is entered in seconds. Determine the following. (a) average velocity of the object during the time interval t = 1.00 s to t = 3.00 s m/s (b) instantaneous velocity at the time t = 2.00 s m/s (c) acceleration of the object at the time t = 2.00 s m/s² (d) In this case, the average velocity over the time interval t = 1.00 s to t = 3.00 s is not equal to the instantaneous velocity at the midpoint of the time interval (t = 2.00 s) because of which of the following? The acceleration is not a constant. The acceleration is negative. The acceleration is positive. The average velocity is never equal to the instantaneous velocity.

Rachel B.
*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,129 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,377 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,903 solutions
*

Transcript

-
00:01 Hello and welcome to this video solution of numerate.
00:04 So here it's given a velocity time graph of an object moving along the x -axis as is shown in this figure, right? and every division along the vertical axis corresponds to 1 .5 and each direction along the horizontal axis correspond to 1 .5 seconds to right? so based on this i have taken, so if you see there are 8 divisions in the downward directions, so based on 8 times of 1 .5 gives you a negative of 12 meters per second and similarly for eight divisions here so this maximum velocity is plus 12 right i've taken some time plots here specifically on the portions where the velocity is changing right and rest time it is stationary mean sorry it's it's constant right the horizontal means it's constant and when it's moving upwards in this slope there that means the body's accelerating right now you have to plot a acceleration time graph based on this one right so this is pretty simple you can easily draw that right so this is acceleration time graph right now initially this was at rest so there will be no acceleration because the velocity is constant till let's say seven seconds right and beyond which started accelerating the body because the velocity is increasing with time so what we can do is uh you can take this 7 .5 from here right it started accelerating which will have a certain constant acceleration right till 22 .5 22 .5 right this is 7 .5 and 22 .5 right okay so the value of acceleration let's find out so acceleration will be equal to change in velocity over the change in time right now change in velocity will be it's pretty simple so negative 12 to 12 so 12 minus of minus 12 over the change in time is 22 .5 minus 7 .5 and this is equal to 24 over 1 .6 meters per second square is a value of acceleration...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade

Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever