Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $\vec{v} = \Delta\vec{s}/\Delta\vec{t}$. 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.

share|cite|improve this answer
I am slightly confused, probably my ignorance. But displacement is $\vec{s}$, not $\Delta \vec{s}$, right? If displacement is $\vec{s}$, I interpret the constant acceleration equation ($\Delta \vec{s}=\vec{u}t+\frac{1}{2}\ \vec{a}t^2$) in your answer as the change in the change in position? Wrong? – toksing May 1 '12 at 2:15
Here I'm using $\vec{s}$ to mean position with respect to a fixed origin. If you are using $\vec{s}$ to mean displacement (that is, a change in position), then things are different; in particular, $\vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2$ is a perfectly fine way to write the equation. (I wouldn't do it that way, though.) – David Z May 1 '12 at 2:18
Actually, I edited the answer to be a little clearer. – David Z May 1 '12 at 2:23

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.

share|cite|improve this answer

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

share|cite|improve this answer
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – Asher May 28 at 16:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.