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Sorry for the long question..

Thanks!!

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- Thread starter dumbboy340
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- #1

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- 1

Sorry for the long question..

Thanks!!

- #2

HallsofIvy

Science Advisor

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As I said, the derivative is the "instantaneous rate of change" so we can define "instantaneous" speed as the derivative of the distance with respect to time and define "instantaneous" acceleration as the derivative of the speed with respect to time.

For example, suppose we are looking at an object moving along the parabola [itex]y= x^2[/itex] such that at any time "t", x= 4t and [itex]y= 16t^2[tex]. At time t= 1, we are at (4, 16) and at time t= 2 we are at (8, 64), Our

A slightly more "geometric" way of putting this is that, as long as something is moving with constant speed, v, its position from some give "starting point" is d= vt+ d0. If you were to graph that, with t on the horizontal axis, d on the vertical, then the graph is a straight line with

- #3

Svein

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You just introduce a change - first you calculated for x= 2 and the for x= 3. Thus, the

Sorry for the long question..

Thanks!!

As an aside, in physics we try to stay away from derivatives as much as possible. They are OK in theory, but if your "function" consists of a set of measurements, those measurements will contain noise and measurement errors. A derivative of something with noise in it will tend to show the noise, not the real "function".

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