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Physics: physical representation of the math

Posted on 2011-09-21
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r(t) is the position function of a particle moving in 2 or 3 space.

**I don't know how to do integral symbol, so I'll just denote that symbol as integ(f(x))**

integ( ||dr/dt|| ) from t0 to t1

I know dr/dt = v(t) which is the velocity function, and the norm of v(t) is the magnitude of the vector, aka, the numerical speed of the particle at time t, but when we integrate the speed, wouldn't we just end up with the position function again?

That is what my intuition is telling me, but my intuition has failed me before, so I would rather be shown why my thinking is wrong, if it even is. Thanks.
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Question by:Zenoture
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by:Zenoture
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Wait, i think it just hit me as I posted the question.

If I were to assume some random value for || v(t) || to be 10 m/s, upon integration I would end up with 10t, where t is in seconds, and 10 is still in meters per second. So by simple plug and chug, where t = 1...

10 m/s * 1 s = 10m therefore proving that the above statement is the displacement of the particle after t seconds.


Right?
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abbright earned 167 total points
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I believe this is true if the velocity is a straight one, that is only in one direction. If it isn't then the integral denotes the length of the way the particle has gone from t0 to t1, even if it was a circle and the particle ends up at the same point at t1 compared to when it started at t0.
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by:TommySzalapski
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It is important to note that in this entire discussion we are all integrating with respect to time.
The integral of dr/dt is of course (dr/dt)dt and so is just dr.
dr is the change in position so you are correct.

The indefinite integral of the velocity is the position function (of course were you to compute it, you would get that + C which would be the initial position).

The definite integral of the velocity is the displacement (distance from start to end).

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by:TommySzalapski
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Of course, if the velocity is not constant, then the integral would be the total distance traveled, not the final displacement (as abbright mentioned).
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