Constant velocity, constant acceleration?

[jfnn]

Homework Statement

Hi, I have a quick question about an experiment I am doing.

I did an experiment where I rolled a ball over a flat surface and dropped a ball from a height, and I was asked if the ball rolling over the flat surface (1-d) motion is subject to constant velocity? I was also asked if the ball dropped from a height is subject to constant acceleration? Here is my picture attached too. Both of the graphs are position vs. time. The parabolic graph is the graph where I have to know if the ball is subject to constant acceleration. The linear graph is the one that asks if it is constant velocity.

Homework Equations

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I know if an object is at constant velocity, the position vs. time graph will be linear. This graph, the best fit line is linear; however, the velocity between individual points is not the exact same as the velocity of the best fit line (got this from slope), thus I said the velocity deviates and is not always constant due to forces, such as wind in my home or friction.

For the acceleratinon position vs time graph, I see that it is parabolic, which means that it should be constant acceleration. However, when I compare the quadratic equation for the line of best fit to the kinematic quadratic equation for position, i get different values of the acceleration. For example, the acceleration is not -9.8? It is -8.6...

Please help. Thank you.

The Attempt at a Solution

Above. [/B]

 

[mfb]

Measurements are never exact, so small deviations can be expected. 10% is probably more than you would expect.

How did you measure time and position for the falling ball? How did you get the -8.6 (I guess m/s²)?

 

[jfnn]

Measurements are never exact, so small deviations can be expected. 10% is probably more than you would expect.

How did you measure time and position for the falling ball? How did you get the -8.6 (I guess m/s²)?

Okay so the small deviations exist. I then have come to the conclusion that velocity is conserved for the first graph and acceleration is for the second; however, deviation exist because of error and limitation. ˇhe error could be the breeze in my house, the human error from deciding the centre of the ball when using the software, and etc.. Did i forget any error btw?

Then I am asked to figure out the value for acceleration based on the parabola graph. I know these values from the program: A = -4.231, B = -1.542 and C = 6.33*10^-3

where the following equation must be satisfied: At^2 + Bt + c = y

I know that for kinematics, that y = 1/2ayt^2 + v0y*t + y0 therefore At^2 must equal 1/2at^2 so after plugging in A, and getting rid of time, I get the value for ay=g= -8.46 m/s^2?

Is that logic right?

Thank you,

 

[jfnn]

Measurements are never exact, so small deviations can be expected. 10% is probably more than you would expect.

How did you measure time and position for the falling ball? How did you get the -8.6 (I guess m/s²)?

Oh, I used a program software called Tracker where I uploaded the video and it did it for me.

 

[jfnn]

Oh, I used a program software called Tracker where I uploaded the video and it did it for me.

Your help is greatly appreciated. Thank you!

 

[mfb]

Okay so the small deviations exist. I then have come to the conclusion that velocity is conserved for the first graph and acceleration is for the second; however, deviation exist because of error and limitation.

I think that is a good conclusion.

ˇhe error could be the breeze in my house, the human error from deciding the centre of the ball when using the software, and etc.. Did i forget any error btw?

Your position measurement won't be perfectly linear, especially towards the edges of the camera field of view.
Drag in general is something you didn't consider so far.

 

[jfnn]

I think that is a good conclusion.Your position measurement won't be perfectly linear, especially towards the edges of the camera field of view.
Drag in general is something you didn't consider so far.

Oh yes. I determined that friction acts constantly on the original ball across the horizontal displacement thus it doesn't impact the velocity. drag would impact the acceleration by increasing the velocity in the downwards direction until the velocity is constant with the drag at a terminal velocity?

Did you know if I found the value of g correct or not by the way?

Thanks again,

 

[jfnn]

Oh yes. I determined that friction acts constantly on the original ball across the horizontal displacement thus it doesn't impact the velocity. drag would impact the acceleration by increasing the velocity in the downwards direction until the velocity is constant with the drag at a terminal velocity?

Did you know if I found the value of g correct or not by the way?

Thanks again,

Oops drag would decrease the velocity by acting against the lotion, making the acceleration potentially change some.. sorry for confusion.

 

[mfb]

Oops drag would decrease the velocity by acting against the lotion, making the acceleration potentially change some.. sorry for confusion.

Right.

Concerning the value of g: How sure are you that the measured distances are correct? This includes the camera setup. If your camera was closer than expected you underestimate the position differences.

 

[jfnn]

Right.

Concerning the value of g: How sure are you that the measured distances are correct? This includes the camera setup. If your camera was closer than expected you underestimate the position differences.

The measured differences were all taken via the computer software. I guess the error would come from the camera maybe being too close? I didn't think it was too close but it could have potentially been as nothing else would cause this difference right?

 

[jfnn]

The measured differences were all taken via the computer software. I guess the error would come from the camera maybe being too close? I didn't think it was too close but it could have potentially been as nothing else would cause this difference right?[/QUOTE

 

[mfb]

I don't know, you had the setup, I didn't.