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All over the internet, I am seeing different people defining velocity as the derivative of either position, distance, or displacement and it is really confusing me. I can understand how the derivative of position is velocity because the very definition of velocity is (change in position)/(change in time) or (displacement)/(change in time). So how could (CHANGE in displacement)/(change in time) or (CHANGE in distance)/(change in time) give you velocity as well? Could someone tell me what is the correct way to define velocity.

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In one dimension, one can say "velocity is the derivative of distance" because the directions are unambiguous. In higher dimensions it is more correct to say it is the derivative of position. One can also say that it is the derivative of displacement because those two derivatives are identical.

If I say the position of an object is $p(t)$, then its displacement from any arbitrary initial point $p_0$ is $p(t) - p_0$. The derivative of that, $\frac{d}{dt}(p(t)-p_0)$ is exactly equal to $\frac{dp}{dt}$, which is the derivative of $p(t)$ as well.

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  • $\begingroup$ Ok I get that 𝑑𝑝/𝑑𝑡 equals (change in position)/(change in time) which equals velocity. so how could the derivative of displacement, namely (change in change in position)/(change in time) equal velocity was well? $\endgroup$ – user532874 May 3 at 4:09
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    $\begingroup$ @user532874 They are both equal because position and displacement differ by a constant. Two functions that differ by a constant always have the same derivative. $\endgroup$ – Cort Ammon May 3 at 4:14
  • $\begingroup$ sorry but would you be able to post a concrete scenario involving a moving object where you can show both derivatives are the same? I understand about half of what you say but I think a real scenario would be really helpful $\endgroup$ – user532874 May 3 at 6:55
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    $\begingroup$ velocity is the derivative of distance I travel along a straight road from my house to a friend and back to my house. The distance I traveled was 200 km and my displacement and change of position are both zero. I understand that I am then going to find an average but I think that it should be position or displacement which should be used when evaluating a velocity?. $\endgroup$ – Farcher May 3 at 7:55
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    $\begingroup$ Even in 1D, velocity as derivative of the distance is ambiguous. Since distance from a point increases when one is going away from the point, it would turn out that the velocity of a point moving with uniform speed along a line would have a jump (from negative to positie) when passing through the origin. Not very useful! $\endgroup$ – GiorgioP May 3 at 9:25

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