Finding Acceleration from a v^2 vs. t Graph

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To find acceleration from a velocity squared versus time graph, the slope does not directly represent acceleration, as it would for a velocity versus time graph. The correct approach involves using kinematic equations, specifically V^2 = 2a(xf - xi), to calculate acceleration. The user initially obtained a slope that suggested an incorrect acceleration value, which was significantly higher than the theoretical value. After clarification, it was determined that using the kinematics equation yielded results consistent with theoretical expectations. Understanding the correct relationships between variables is crucial for accurate calculations in physics experiments.
ethinh
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Homework Statement



when graphing a velocity squared vs. time graph how do I find the acceleration. I realize the slope gives me units of acceleration but this is not correct because the theoretical acceleration is no where near this and I didnt botch this lab up.

Theoretical
a = gsin(3.03) +/- 0.085 m/s^2

when i graph my data from the photogate the linear trend line gives me a slope of 1.0481 which is the acceleration in this case but like I said that is far beyond the theoretical.

also i did this experiment by using a stopwatch and I calculated an acceleration using the equation
a=2s/t^2 and got an acceleration of 0.58 +/- 0.10 m/s^2

So again how do I calculate the correct acceleration using a v^2 vs. x graph?
 
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ethinh said:

Homework Statement



when graphing a velocity squared vs. time graph how do I find the acceleration. I realize the slope gives me units of acceleration but this is not correct because the theoretical acceleration is no where near this and I didnt botch this lab up.
The slope of the velocity-time graph is the acceleration.

But you wrote: "velocity-squared vs time" ... the slope of that graph will not give you the acceleration.
If you means "displacement-squared vs time" then that is also wrong ... as you can see by rearranging your own equations from later: a=2s/t^2 means s=(a/2)t^2 ... i.e. a plot of s vs t^2 gives a slope of a/2.

But it may be a typo - in which case there is probably something else going on that you have not taken into account with the theory.
 
Thanks Simon but I figured out that I had to use a kinematics equation V^2=2a(xf-xi) to find acceleration and it matched the theoretical value pretty well!
 
OK. So what did you plot?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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