Propagating Uncertainties in acceleration when using a video software

• Iwanttolearnphysics
In summary: You did not follow my instructions and proceeded to make up your own. Include a lab report that states the data generated by LoggerPro cannot be this precise.

Iwanttolearnphysics

Homework Statement
How do you calculate the uncertainty in acceleration given that the values for initial velocity, final velocity and time were from a video analysis software? The values given by the software are the following: v = 0.9015303150, u = 0.134909200, t = 0.7123523333.
Relevant Equations
a = v-u/t
Here's why I'm confused:
• I recorded a video of a ball moving across the floor and uploaded it to the video analysis software I used.
• I had to track the moving ball while it moved and I did this while the video was playing in slow motion through the video analysis software.
• put a dot in the center of the ball every time it moved, so that the software could calculate the final and initial velocities of the ball.
• I'm confused because the values it gave me are very small! with 10 significant digits, and if I calculate the uncertainties, they would be really small.
• I don't think I should use the uncertainty ±0.0000000001 because if it was just the software tracking the ball without my input (I had to put blue dots on the ball every time it moved), then this would be believable.
• But I had to put dots on the ball and I don't think it was possible for me to know for sure if I was putting the dots in the exact center (or the exact same place as before). There was no ruler I could use. The software literally just said to put dots on the ball so it could track its movement.
• Should I include human reaction time? If I do, the uncertainties will be really huge.
Here's my calculations:
Acceleration:
• a = v-u/t
• a = 0.9015303150-0.134909200/0.7123523333
• a = 1.0761825 m/s^2
Uncertainties:
• in v = (0.0000000001/0.9015303150)x100 = 1.10922504×10^−8%
• in u = (0.0000000001/0.134909200)x100 = 7.41239293×10^−8%
• in t = (0.0000000001/0.7123523333)x100 = 1.40379971×10^−8%
Summation of uncertainties: 9.92541768×10^−8%
which is just ± 0.0000001.

So my final answer would be: 1.076183 m/s ± 0.0000001
This is such a small uncertainty and I don't think this is right. I'm helping my cousin with his homework.

Here's what we did but with a ball:

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To measure the error in placing the dots, you could deliberately place a dot just enough offset that you can see it isn't right. Looking at what that does to the recorded position data then gives you the bound on the error.

Using this tool the way shown, I would worry whether the scaling had all worked out. You could estimate what the average speed should be, given the total distance and the time taken, and compare that with the LoggerPro number.

Iwanttolearnphysics said:
Homework Statement:: How do you calculate the uncertainty in acceleration given that the values for initial velocity, final velocity and time were from a video analysis software? The values given by the software are the following: v = 0.9015303150, u = 0.134909200, t = 0.7123523333.
Relevant Equations:: a = v-u/t

if it was just the software tracking the ball without my input (I had to put blue dots on the ball every time it moved), then this would be believable.
I disagree, such an uncertainty would definitely not be believable because it would be way more precise than you could expect from any reasonable instrument.

haruspex said:
To measure the error in placing the dots, you could deliberately place a dot just enough offset that you can see it isn't right. Looking at what that does to the recorded position data then gives you the bound on the error.

Using this tool the way shown, I would worry whether the scaling had all worked out. You could estimate what the average speed should be, given the total distance and the time taken, and compare that with the LoggerPro number.
Would it be reasonable to just not include the error here because it's almost non quantifiable? I could maybe redo the video and put a meter stick on the floor if it's not possible to ignore the uncertainties here. But I don't believe that's part of the things my cousin's class needs to do. Thank you.

Orodruin said:
I disagree, such an uncertainty would definitely not be believable because it would be way more precise than you could expect from any reasonable instrument.
Thank you. Would it be possible to say in the lab report that the data generated by LoggerPro cannot be this precise? I can't imagine why the company who made it would give such small uncertainties for no reason at all.

Iwanttolearnphysics said:
Thank you. Would it be possible to say in the lab report that the data generated by LoggerPro cannot be this precise? I can't imagine why the company who made it would give such small uncertainties for no reason at all.
They are not giving any uncertainties at all. Applying reasonable uncertainties is your job. The software just takes input and computes things based on that input. You need to estimate the uncertainty in the input you are giving, such as the uncertainty in placing the markers etc. Many uncertainties (such as exposure time for frames, frame time difference, etc) will most likely be negligible in comparison to this.

Iwanttolearnphysics said:
because it's almost non quantifiable
In post #2 I gave you a perfectly reasonable way to quantify it.

Orodruin