Displacement derived from Acceleration

Click For Summary

Homework Help Overview

The discussion revolves around finding the displacement of a moving object using acceleration data. The original poster is attempting to derive displacement from acceleration (measured in m/s²) and time (in seconds) by calculating the area under the curve of a plotted graph of acceleration versus time.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of using trapezoids to find the area under the curve to derive velocity and subsequently displacement. Questions are raised about the validity of calculating the area when the units of the axes differ. Some participants suggest double integration as a theoretical approach to determine displacement.

Discussion Status

Some guidance has been offered regarding the integration process and the relationship between the units involved. Participants are exploring the implications of their methods and considering testing their approaches with constant acceleration values to validate their results.

Contextual Notes

The original poster has only a list of data points produced by an accelerometer and does not have a specific equation for acceleration. There is an emphasis on ensuring the method is correctly implemented to avoid errors.

tomtomtom1
Messages
160
Reaction score
8
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
 

Attachments

Physics news on Phys.org
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).
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
12
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
25
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K