# What is displacement? Position relative to a reference point or change of position

What is the "official" or most useful definition of displacement in the context of kinematics? There are two common ones:

1. Displacement is the length and direction of a line from a fixed reference point. (Basically position).
2. Displacement is the change in position.

Textbooks using the first definition frequently define velocity as $v=\Delta s/\Delta t$ (where $s$ is displacement) but then for acceleration give equations such as $s=ut+1/2at^2$ (if they were consistent, they would have to use $\Delta s$). What throws me off is that are the better textbooks.

A similar confusion comes through in Wikipedia: A displacement vector is the straight path between the initial and the final position. But velocity is defined as $v=\Delta d/\Delta t$.

Background: I am writing on a learning tool for students but different textbooks require them to learn contradicting definitions and I would appreciate your help on which is the right definition to learn.

• Displacement is difference (relative vector) between end and start points of a path. Oct 11, 2022 at 14:23

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 we think as a vector difference then we could also take the negative of the difference. Does initial and final imply that we should consider time? Nov 17, 2020 at 17:21
• This answer is absolutely misleading. Displacement is change in position from a fixed reference point. Jul 30, 2022 at 18:21
• @Brethlosze I could just as well say that your comment is absolutely misleading. Do you have a reference or anything to support your assertion? Jul 31, 2022 at 5:35

The official definition is that displacement is the change in position from an original position. Note that "displacement" has other meanings such as the amount of water displaced by an object in a fluid such as the water displaced by a boat.

The distance covered by an object in particular direction is called Displacement
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