- #1
Instananeous accerlerations on a velocity vs time graph
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Homework Help Overview
The discussion revolves around finding instantaneous accelerations at specific time points on a velocity vs. time graph. The subject area includes kinematics and the interpretation of graphical data related to motion.
Discussion Character
- Exploratory, Conceptual clarification, Problem interpretation
Approaches and Questions Raised
- Participants discuss the concept of acceleration as the slope of the velocity vs. time graph and explore how to determine instantaneous acceleration at given time points. Some express uncertainty about the calculations and the method to apply.
Discussion Status
The discussion is ongoing, with participants sharing insights about interpreting the graph and the relationship between velocity and acceleration. There is a mix of attempts to clarify concepts and some confusion regarding the calculation process. Guidance has been offered regarding the interpretation of the graph and the application of the slope concept.
Contextual Notes
There are indications of imposed homework rules, as some participants mention that calculations should not be provided directly. This has led to some confusion and questions about the appropriateness of sharing specific calculations.
- #2
Doc Al
Mentor
- 45,584
- 2,460
Hint: Another way to express this is to say that acceleration is the slope of the velocity vs time graph.mandy9008 said:
- #3
Axiom17
- 70
- 0
As Doc Al said, a good way to go about understanding this question is to simply think of the slope of the velocty vs. time graph as representing the acceleration. So the greater the gradient of the slope, i.e. the steaper the line, then the greater the acceleration at that time.
OK so you've got the graph plotted of what's going on, that's good. It's much easier to visualise what's going on and know if you're right when able to look at a graph.
This is stating, rather more confusingly than it could be, that as has been said already that the acceleration is equal to the change in x, i.e. the change in the velocity, with respect to the change in y, i.e. the change in time. Hopefully that's clearer now if it wasn't already.
OK so you've got the first 2 correct, not sure if that was through applying the formula, a bit of guesswork, or just intuition. They're simple once you understand what's going on here.
For the acceleration at [itex]10.0s[/itex], just apply the formula, using the value for the velocity at this time that can be read off the graph.
Hope that helps now
OK so you've got the graph plotted of what's going on, that's good. It's much easier to visualise what's going on and know if you're right when able to look at a graph.
This is stating, rather more confusingly than it could be, that as has been said already that the acceleration is equal to the change in x, i.e. the change in the velocity, with respect to the change in y, i.e. the change in time. Hopefully that's clearer now if it wasn't already.
OK so you've got the first 2 correct, not sure if that was through applying the formula, a bit of guesswork, or just intuition. They're simple once you understand what's going on here.
For the acceleration at [itex]10.0s[/itex], just apply the formula, using the value for the velocity at this time that can be read off the graph.
Hope that helps now
- #4
Axiom17
- 70
- 0
Apologies just realized that method won't work very well 
I guess I'm not too with it at the moment! 
Because the gradient of the line doesn't change around the time [itex]10.0s[/itex] what you can do is pick a point just before or after when [itex]v \neq 0[/itex] and thus you can get a non-zero value for the velocity and an accompanying time, then can find the acceleration using that simple formula.
I guess I'm not too with it at the moment! Because the gradient of the line doesn't change around the time [itex]10.0s[/itex] what you can do is pick a point just before or after when [itex]v \neq 0[/itex] and thus you can get a non-zero value for the velocity and an accompanying time, then can find the acceleration using that simple formula.
- #5
inky
- 99
- 1
mandy9008 said:
I wonder I don't know who delete my calculation .
- #6
Axiom17
- 70
- 0
inky said:
.. pardon?!
- #7
- 12,179
- 186
Your calculation was deleted because you are not supposed to do the calculation. The student who asked the question is supposed to do the calculation.inky said:
Did you not see the message I sent you?
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