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I was doing kinematics when a silly question came to my mind. It is as follows:

When we differentiate $x(t)$ (position as a function of time), we get $v(t)$ (instantaneous velocity). Doing the reverse,i.e taking the antiderivative we get back the position function.

Now say I've a function $\alpha(t)$ which describes the instantaneous displacement of the body at an instance of time. What do we get on differentiating this? Do we still get the instantaneous velocity function or something else? Does is it even make sense?

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    $\begingroup$ How is the instantaneous displacement different from the displacement? $\endgroup$
    – mmesser314
    Commented Jun 2, 2023 at 4:58
  • $\begingroup$ Displacement can be over a time interval. For eg: the displacement between 2s and 3s. Instantaneous displacement is the displacement at a particular time(when the time-interval approaches 0). $\endgroup$
    – PandaScientist
    Commented Jun 2, 2023 at 5:10
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    $\begingroup$ What mmesser314 is trying to tell you, is that whatever you originally called position, is equal to "instantaneous displacement from origin". The thing you call displacement between 2s and 3s is $\vec{\Delta x}$ as opposed to $\vec x(t)$ $\endgroup$
    – naturallyInconsistent
    Commented Jun 2, 2023 at 5:30
  • $\begingroup$ Thank you. I get it now $\endgroup$
    – PandaScientist
    Commented Jun 2, 2023 at 5:34

1 Answer 1

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Displacement is defined as difference between position vectors : $$ \vec s = \vec r - \vec r~' \tag 1$$

Differentiating that we get :

$$ \frac {d\vec s}{dt} = \frac {d(\vec r - \vec r~')}{dt} = \frac {d\vec r}{dt} - \frac {d\vec r~'}{dt} = \vec v - \vec v~' \tag 2 $$

So, basically what you get is relative instantaneous velocity to some reference point.

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