It's possible that the way in which these terms are used varies from person to person, even among professionals in the field. However, in the usage I'm familiar with, **displacement is the change in position**, period. Definition #2 is correct and #1 is wrong. (The length and direction of a line from a fixed reference point is just called position.)

In this usage the proper form of the constant acceleration kinematic equation would be $\Delta \vec{x} = \vec{v}t + \frac{1}{2}\vec{a}t^2$, or $\vec{x} = \vec{x}_0 + \vec{v}t + \frac{1}{2}\vec{a}t^2$, where $\vec{x}$ is position and $\vec{v}$ is initial velocity. It would be valid to write $\vec{x} = \vec{v}t + \frac{1}{2}\vec{a}t^2$ *if* you always choose the origin to be at the initial position, but that seems like an unnecessary restriction.

Alternatively, you could write the equation in terms of displacement. If you use $\vec{s}$ for displacement, the equation would be $\vec{s} = \vec{v}t + \frac{1}{2}\vec{a}t^2$. That is because $\vec{s} = \Delta\vec{x}$ (displacement equals change in position). If these textbooks you're using are using this notation in which $\vec{s}$ is displacement, then it seems very strange to write $\vec{v} = \Delta\vec{s}/\Delta t$. That is unconventional and probably unclear notation, although it might not necessarily be wrong.

The Wikipedia usage is fine, though, because in that formula the displacement is $\Delta\vec{d}$, not just $\vec{d}$. In $\Delta\vec{d} = \vec{d}_f - \vec{d}_i$, the vectors $\vec{d}_i$ and $\vec{d}_f$ could be either positions or displacements.

If your position is in 3D space (which means your position vector must be defined), then there is no distinction between displacement and *change in* position.

$s=\boldsymbol{R_f-R_i}=\Delta\boldsymbol{R}$
, where $s$ is displacement and $R$ is position.

However, in $v = ds/dt$, $ds$ does not mean change in displacement but rather an infinitesimally small displacement (infinitesimally small change in position). Since $v=ds/dt$, so the instantaneous velocity is the gradient to the tangent of the curve at that particular point.

In my opinion I believe it's just an issue of wording. I don't think change in displacement means $ΔΔR$, but rather the actual displacement(change in position) that occurred during a time $ΔT$

## Best Answer

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.