Acceleration Position vs. Time Graph

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SUMMARY

The discussion focuses on calculating the x-component of acceleration between two points, M and S, using the formula a = Δv/Δt. The user calculated the change in position as -8 m and the change in time as 3 s, leading to an initial acceleration estimate of -0.889 m/s². However, after further analysis involving the velocities at points M and S, the correct acceleration was determined to be approximately -6.33 m/s². The conversation highlights the challenges of obtaining exact answers from graphical data and emphasizes the importance of calculating instantaneous velocities.

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  • Understanding of kinematics, specifically acceleration and velocity calculations.
  • Familiarity with interpreting position vs. time graphs.
  • Knowledge of the formula a = Δv/Δt for acceleration.
  • Ability to calculate instantaneous velocity from given data points.
NEXT STEPS
  • Study the concept of instantaneous velocity and its calculation from position vs. time graphs.
  • Learn how to accurately read and interpret graphical data in physics problems.
  • Explore kinematic equations and their applications in different scenarios.
  • Review examples of uniform motion and how it affects acceleration calculations.
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Students studying physics, particularly those focusing on kinematics, educators teaching motion concepts, and anyone interested in mastering acceleration calculations from graphical data.

Anony-mouse
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Homework Statement


http://img370.imageshack.us/img370/9488/96191446sb5.jpg

Evaluate Ax, the x-component of the acceleration between M and S.

Homework Equations


a=x/t^2

The Attempt at a Solution


M= 48 m at 8.5 s
S= 40 m at 11.5 s

so change of x=-8 m, and change of t = 3s

so a= -8 / (3^2), so -8/9 m/s^2 (or -0.889 m/s^2). Where did I go wrong?
 
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Hint: figure out the velocities at M and at S, then the change in velocity between M and S, and then...
 
Would the velocity at M be the velocity between R and M, and the velocity at S be the velocity between S and N, or do I have to calculate the instantaneous velocity at each point?
 
Anony-mouse said:
Would the velocity at M be the velocity between R and M, and the velocity at S be the velocity between S and N, ...
In this specific case, yes.

... or do I have to calculate the instantaneous velocity at each point?
In general, yes.
 
Anony-mouse said:
Would the velocity at M be the velocity between R and M, and the velocity at S be the velocity between S and N, or do I have to calculate the instantaneous velocity at each point?

They are the same thing since the velocity is uniform on those parts of the graph.
 
So I ended up calculating 20/3 for the M velocity, and -20/1.5 for the S velocity, which gave me a change of velocity of -20. If the change in time was 3 seconds, then the x-component of the acceleration would be -20/3 m/s^2. But that's not the right answer given.

Is the x component just the -20? I can't figure out what I did wrong.
 
Your values look correct, within any reasonable accuracy that can be read from this graph. You should be only slightly off from the "correct" answer. Just how far from the given answer is your value of -6.7 m/s^2?
 
I gave this answer: -6.67 m/s^2, so I guess it was a mistake.
 
I'm not convinced you made a mistake. You got -6.67 m/s^2. Earlier you said this is not "the right answer given". So what is the right answer given, and how far off are you from it?
 
  • #10
I don't really know, because it is automated and answers aren't revealed till monday, but I can find out within a couple of days. I would like to know what the answer is myself. All I know is that the computer said my answer was wrong and I can't change it.
 
  • #11
It's a little odd that they would require an "exact" answer, given the approximations inherent in reading a graph. You really have correctly solved the problem, but have probably read the numbers on the graph a little differently than the problem author.

By the way, I get -12 m/s for the velocity at S.
 
  • #12
I found out that the correct answer was -6.33 m/s^2... seems kind of unfair to have an exact answer like that without having exact numbers.
 
  • #13
Yes, I agree.
 

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