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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.

Cort Ammon
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  • Ok I get that / 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? –  May 03 '19 at 04:09
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    @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. – Cort Ammon May 03 '19 at 04:14
  • 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 –  May 03 '19 at 06:55
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    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?. – Farcher May 03 '19 at 07:55
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    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! – GiorgioP-DoomsdayClockIsAt-90 May 03 '19 at 09:25
  • @user532874 You need to be clear about the difference between the derivative of position, which is velocity measured at a specific time, and the change in position/change in time, which is the average velocity over the time period. They are not the same. – gandalf61 May 03 '19 at 09:56
  • @user532874 I'm wondering if the issue may be definitions. What is the definition of "displacement" you are using? The usual definition, which I am using is "the difference between the position of an object and a reference point," and the reference point is typically an initial position or a "rest" position. For example, consider a coordinate system where position (0, 0) is a point on the floor, (1,0) refers to a point on the floor 1 foot to the right of that origin, and (0,1) refers to a point 1 foot above the origin., I have a bucket that starts on a chair, one foot up and 2 feet from the – Cort Ammon May 03 '19 at 15:35
  • ... origin. That corresponds to a point (2, 1). If I then move it to a table one foot to the left of the chair, where the table is 3 feet high, its new position is (1, 3). Using these defintiions, its position is (1, 3) and its displacement from its initial position is (-1, 2) which was calculated as (1 - 2, 3 - 1) – Cort Ammon May 03 '19 at 15:36
  • If you use that definition of displacement, the answer I wrote shows that the derivative of position with respect to time, $\frac{dp}{dt}$ and the derivative of displacement with respect to time, $\frac{d}{dt}p(t)-p_0$ are numerically identical for all values of $p_0$ because, by the usual definitions of displacement, $p_0$ does not change with time. – Cort Ammon May 03 '19 at 15:38
  • If you are using another meaning of displacement (such as the amount of water displaced by a ship while it is at rest in the water), then you will indeed find that they mean different things. But the definition of displacement I used here is very typical. It's useful in many cases where you don't actually care about the position of an object in the coordinate system, but you do want to track its velocity. If I have an object on a spring bouncing, its displacement from its resting position will always be the same function of time, regardless of where in the room I put it. – Cort Ammon May 03 '19 at 15:41
  • I know you talked about your bucket example and I understand that and the difference between displacement and position in that example. I think the root of my misunderstanding is when you say "/ and the derivative of displacement with respect to time, (()−0)/dt are numerically identical for all values of 0". I don't get how to express this in terms of numbers, like to actually prove this statement by using a real example and plugging in numbers to prove the equation (()−0)/dt = /. Could you show this to me? –  May 04 '19 at 16:36
  • @user532874 Sure. I stand on a tall cliff, holding a ball such that its position is 50m above the ground. I then let it go. Its position can be calculated as $p(t)=-4.9t^2 + 50$. Its velocity, as the derivative of position, is $\frac{dp}{dt} = -9.8t$. Now if we think about displacement, it starts at its initial position, so its displacement at t=0 is 0. Its displacement as a function of time is $d(t)=-4.8t^2$. Its velocity, as the derivative of displacement, is $\frac{dd}{dt}=-9.8t$. Both approaches yield the same results. – Cort Ammon May 05 '19 at 03:43