Displacement derived from Acceleration

AI Thread Summary
To find displacement from acceleration data, one must first calculate the area under the acceleration-time curve using trapezoids, which provides the change in velocity. This involves integrating the acceleration data to derive velocity, and then integrating the velocity data to obtain displacement. The area under the curve can be computed despite differing units, as the process involves multiplying time intervals by the corresponding changes in velocity. Testing the method with constant acceleration can help validate the approach before applying it to experimental data. Ultimately, double integration of the acceleration data will yield the desired displacement.
tomtomtom1
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Hello all

I was wondering if someone could help shed some light on a physics problem that I am having difficultly with.

I am trying to find displacement of a moving object. The only data I have is acceleration (ms/^2) and time (seconds).

The acceleration of the object was measured once per 0.25 seconds. I have plotted acceleration against time.

So derive displacement I am trying to find the area under the curve using trapezoids between each time interval – this gives me velocity and by repeating the process I get displacement.

The problem I have is that one of the sides of my trapezoid is time and the other is m/s2. Can I still find the area under the curve even though the units are different?

Attached is a sample of my data.

Can anyone help?

Thanks
 

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The area under the curve is m/s^2 multiplied by s which gives you m/s, or change in velocity, not displacement.

If you were given an equation that describes acceleration you would integrate to get an equation for velocity at any particular time. Integrate the velocity equation and you get position at any particular time.
 
What i am trying to do is determine displacement by double integration.

I have plotted acceleration (m/s^2) on my Y axis and time (seconds) along my X axis.

So if i find the area under the curve between t=0 y= 2.189873122 and t= 0.25 y = 2.388952497, then should should give me velocity.

By plotting velocity (m/s) along the Y axis and time (seconds) along my X axis and repeating the above steps but for the relevant Y values i should get displacement between t=0 and t = 0.25.

That is the theroy, i do not have an equation, i just have a list of data points the data i have was produced by an accelerometer.
 
Recommend you implement your method using a spreadsheet program to minimize unintentional math errors.

You should test your method (and your spreadsheet) by using the same constant acceleration value for each interval. Your result should match the result you would get if you used a standard constant acceleration kinematic equation.

When you get that working, use your experimental acceleration values.
 
tomtomtom1 said:
What i am trying to do is determine displacement by double integration.

I have plotted acceleration (m/s^2) on my Y axis and time (seconds) along my X axis.

So if i find the area under the curve between t=0 y= 2.189873122 and t= 0.25 y = 2.388952497, then should should give me velocity.

By plotting velocity (m/s) along the Y axis and time (seconds) along my X axis and repeating the above steps but for the relevant Y values i should get displacement between t=0 and t = 0.25.

That is the theroy, i do not have an equation, i just have a list of data points the data i have was produced by an accelerometer.



Sorry I did not see your attached graph.

What the previous poster said. I just got to add its change in velocity for that time interval. Just like its displacement, not position, over those time periods.
 
tomtomtom1 said:
The problem I have is that one of the sides of my trapezoid is time and the other is m/s^2. Can I still find the area under the curve even though the units are different?
Yes, as posted above, the first step is to multiply change in time by change in velocity per unit time, so for each trapezoid, the units would be related to Δt · (Δv / Δt) = Δv, or (s) · (m/s^2) = (m/s), Then by repeating the process a second time, you're using the same time intervals and calling displacement "p" for position, then for each trapezoid, the units would be related to Δt · (Δp / Δt) = Δp or (s) · (m/s) = m (meters).
 
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