# Displacement derived from Acceleration

1. Jul 7, 2013

### tomtomtom1

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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2. Jul 7, 2013

### pgardn

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.

3. Jul 7, 2013

### tomtomtom1

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.

4. Jul 7, 2013

### lewando

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.

5. Jul 7, 2013

### pgardn

Sorry I did not see your attached graph.

What the previous poster said. I just gotta add its change in velocity for that time interval. Just like its displacement, not position, over those time periods.

6. Jul 7, 2013

### rcgldr

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).