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bbsgirl10
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The graph is just an example to show the type of motion it travels. I just want to know whether the acceleration and initial velocity will be same for both position/time graph and Velocity time graphharuspex said:It doesn’t feel like you posted the entire question, but maybe all you left out was "Find".
How would you go about find the initial velocity and acceleration from the position/time graph?
But to answer your question, no, the two graphs represent different motions.
And I answered that (from the data in these graphs) they will not be.bbsgirl10 said:The graph is just an example to show the type of motion it travels. I just want to know whether the acceleration and initial velocity will be same for both position/time graph and Velocity time graph
may i know why? because i get around the same value for acceleration and initial velocity based on experiment I got a = 0.200 (for postion/time graph) and a =0.203 (for velocity/time graph). I also get Initial velocity = 0.0711 (for postion/time graph) and Initial velocity = 0.0811 (for velocity/time graph)haruspex said:And I answered that they will not be.
I think I misunderstood your post 3. See my edited post #4. (I don't see why you posted those two graphs with the question - it's just confusing.)bbsgirl10 said:may i know why? because i get around the same value for acceleration and initial velocity based on experiment I got a = 0.200 (for postion/time graph) and a =0.203 (for velocity/time graph). I also get Initial velocity = 0.0711 (for postion/time graph) and Initial velocity = 0.0811 (for velocity/time graph)
Welcome!bbsgirl10 said:may i know why? because i get around the same value for acceleration and initial velocity based on experiment I got a = 0.200 (for postion/time graph) and a =0.203 (for velocity/time graph). I also get Initial velocity = 0.0711 (for postion/time graph) and Initial velocity = 0.0811 (for velocity/time graph)
bbsgirl10 said:
If you read through the thread you will find the posted graphs are irrelevant to the question being asked. Ignore them.cormsby said:Assuming uniform acceleration for the top graph, then solving x = ½at2 for a , using the point (20,8000) gives a=40 (units unknown).
Eyeballing the bottom graph for P(10,65) and Q(25,140) then calculating slope gives a=5m/s2 , so not the same experimental data as in the top graph.
The points on the provided position versus time graph are very nicely patterned. The five y coordinates are at 0 hash marks, 0.5 hash marks, 2.0 hash marks, 4.5 hash marks and 8.0 hash marks. Double each of them and that is 0, 1, 4, 9, 16. It does not get much more obvious than that.haruspex said:Since you are assuming acceleration is constant you could pick three data points and solve the quadratic. That will give the acceleration too.
Yes, but, as has been discovered in the course of this thread, the posted graphs are irrelevant and should be ignored. We do not have the actual data that gave rise to the question.jbriggs444 said:The points on the provided position versus time graph are very nicely patterned.
I did in fact read the thread, including these posts asking whether acceleration will be same for both graphs:haruspex said:If you read through the thread you will find the posted graphs are irrelevant to the question being asked. Ignore them.
But that is a misreading, which I also made. @bbsgirl is asking whether in general, given a velocity time graph and a distance time graph for the same motion (which those graphs are not) the accelerations deduced from them would be the same. If you look at the numbers quoted in those posts they clearly do not correspond to either posted graph, but do closely correspond to each other.cormsby said:I did in fact read the thread, including these posts asking whether acceleration will be same for both graphs:
https://www.physicsforums.com/threads/velocity-and-acceleration-graphs.1045526/post-6800175
https://www.physicsforums.com/threads/velocity-and-acceleration-graphs.1045526/post-6800187
I did not see the phrase "in general", and I think you did not, either.haruspex said:@bbsgirl is asking whether in general
My original response to @bbsgirl10:cormsby said:I did not see the phrase "in general", and I think you did not, either.
Perhaps we should let bbsgirl explain herself, rather than speculate.
to which she replied:haruspex said:the two graphs represent different motions.
and added that:bbsgirl10 said:The graph is just an example to show the type of motion it travels.
From those data it is clear she is not using the posted graphs in her calculations. She never did post the data she was using and, unfortunately, appears to have abandoned the thread. So you can go on believing the graphs are relevant to the question but no other responder does.bbsgirl10 said:i get around the same value for acceleration and initial velocity based on experiment I got a = 0.200 (for postion/time graph) and a =0.203 (for velocity/time graph). I also get Initial velocity = 0.0711 (for postion/time graph) and Initial velocity = 0.0811 (for velocity/time graph)
I am not a "believer".haruspex said:you can go on believing
Velocity is the rate of change of an object's position with respect to time, while acceleration is the rate of change of an object's velocity with respect to time.
On a velocity vs. time graph, the slope represents acceleration. The steeper the slope, the greater the acceleration. On an acceleration vs. time graph, the area under the curve represents the change in velocity.
If the slope of the line on a velocity vs. time graph is not constant, then the object is accelerating. This can be seen as a curved line on the graph.
A horizontal line on a velocity vs. time graph represents a constant velocity. This means that the object is not accelerating and is moving at a constant speed.
The direction of an object's motion is represented by the sign of the slope on a velocity vs. time graph. A positive slope indicates motion in the positive direction, while a negative slope indicates motion in the negative direction.