Author Topic: Acceleration Time Graphs  (Read 9499 times)

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Offline Munchor

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Acceleration Time Graphs
« on: June 08, 2012, 05:44:04 pm »
I wonder, in an acceleration (y axis) per time (x axis) graph, does the area below the line mean anything?

Like, in speed per time, the area is the space walked, right?

Thank you.

Offline aeTIos

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Re: Acceleration Time Graphs
« Reply #1 on: June 08, 2012, 05:46:24 pm »
Somewhere in my mind, the meaning of it is saved. But I can't recall exactly what it means.
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Offline Adriweb

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Re: Acceleration Time Graphs
« Reply #2 on: June 08, 2012, 07:16:51 pm »
Not sure that I understood correctly, but you mean :
acceleration plotted over time (y is accel, x is time), then the area below the curve can be interpreted as the integral, and so we know that the derivative of velocity is acceleration, so it would be the speed (velocity) you're looking for ?
« Last Edit: June 09, 2012, 08:07:21 am by adriweb »
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Offline AngelFish

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Re: Acceleration Time Graphs
« Reply #3 on: June 08, 2012, 07:29:26 pm »
A handy guide to the terminology of derivatives and integrals of the distance from an object to the origin (displacement or position):

First derivative is speed.
Second derivative is acceleration.
Third derivative is the change in acceleration (Jerk).
Fourth derivative is sometimes called jounce.

First integral is absement.
Second integral is absity.
Third integral is abseleration.
Fourth integral is abserk.
Fifth integral is absounce.

If want the integral of acceleration, it's speed (velocity is a vector, integration returns a scalar).
« Last Edit: June 08, 2012, 07:30:31 pm by Qwerty.55 »
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Offline cyanophycean314

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Re: Acceleration Time Graphs
« Reply #4 on: June 08, 2012, 09:34:30 pm »
Isn't it the change in velocity? That's the first fundamental theorem of calculus, no?

v(x1)-v(x2) = Integral(x1,x2) a dt

where v'(t) = a(t)

Offline thepenguin77

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Re: Acceleration Time Graphs
« Reply #5 on: June 09, 2012, 12:05:32 am »
Cyano's correct, it's velocity.

One easy way to look at it is that integrals (area under the curve) are basically multiplication. I mean, when you write an integral for this, it's [itegral sign]a(t)*dt. So, it's really just acceleration * time which equals velocity.

On the other hand, derivatives are division. So the derivative of velocity with respect to time is velocity/time = acceleration. Again, dv/dt shows this pretty well.
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