## Displacement

The displacement of a body in a time interval is equivalent to the change of its position in that interval. Since the position of a body is a vector quantity, the displacement of a body is also a vector.

The displacement vector, or simply the displacement, of a body between the positions Pi and Pf is defined as the vector difference of the position vectors of the defined points Pi and Pf. Its expression, in Cartesian coordinates, is given by:

$∆\stackrel{\to }{r}={\stackrel{\to }{r}}_{f}-{\stackrel{\to }{r}}_{i}=\left({x}_{f}-{x}_{i}\right)\stackrel{\to }{i}+\left({y}_{f}-{y}_{i}\right)\stackrel{\to }{j}+\left({z}_{f}-{z}_{i}\right)\stackrel{\to }{k}$

Where:

• $∆\stackrel{\to }{r}$ : Displacement vector or displacement
• ${\stackrel{\to }{r}}_{i}$, ${\stackrel{\to }{r}}_{f}$ : Position vectors of the points where the body is located at the beginning (Pi) and the end (Pf) of the movement
• xi, xf,yi, yf, zi, zf: coordinates x, y and z of the points Pi and Pf

The unit of measurement of displacement is the meter [m] and in three dimensions its magnitude is given by the following expression:

$\left|∆\stackrel{\to }{r}\right|=\sqrt{{\left({x}_{f}-{x}_{i}\right)}^{2}+{\left({y}_{f}-{y}_{i}\right)}^{2}+{\left({z}_{f}-{z}_{i}\right)}^{2}}$

It is important to realize that a body may move between two instants of time and yet its displacement may be 0. This always happens when the initial and final positions of the body, in the time range studied, are the same. In the following picture, you can see the displacement vector and the various concepts presented in a three-dimensional space. In the case where we are analyzing the problem in fewer dimensions, we can eliminate the unnecessary coordinates, simplifying the above expressions. The displacement vector expression would become:

• In two dimensions $∆\stackrel{\to }{r}={\stackrel{\to }{r}}_{f}-{\stackrel{\to }{r}}_{i}=\left({x}_{f}-{x}_{i}\right)\stackrel{\to }{i}+\left({y}_{f}-{y}_{i}\right)\stackrel{\to }{j}+\overline{)\left({z}_{f}-{z}_{i}\right)\stackrel{\to }{k}}=\left({x}_{f}-{x}_{i}\right)\stackrel{\to }{i}+\left({y}_{f}-{y}_{i}\right)\stackrel{\to }{j}$, and its magnitude would be given by the expression $\left|∆\stackrel{\to }{r}\right|=\sqrt{{\left({x}_{f}-{x}_{i}\right)}^{2}+{\left({y}_{f}-{y}_{i}\right)}^{2}+\overline{){\left({z}_{f}-{z}_{i}\right)}^{2}}}=\sqrt{{\left({x}_{f}-{x}_{i}\right)}^{2}+{\left({y}_{f}-{y}_{i}\right)}^{2}}$, since (zf-zi)=0
• In one dimension $∆\stackrel{\to }{r}={\stackrel{\to }{r}}_{f}-{\stackrel{\to }{r}}_{i}=\left({x}_{f}-{x}_{i}\right)\stackrel{\to }{i}+\overline{)\left({y}_{f}-{y}_{i}\right)\stackrel{\to }{j}}+\overline{)\left({z}_{f}-{z}_{i}\right)\stackrel{\to }{k}}=\left({x}_{f}-{x}_{i}\right)\stackrel{\to }{i}$, and its magnitude would be given by the expression $\left|∆\stackrel{\to }{r}\right|=\sqrt{{\left({x}_{f}-{x}_{i}\right)}^{2}+\overline{){\left({y}_{f}-{y}_{i}\right)}^{2}}+\overline{){\left({z}_{f}-{z}_{i}\right)}^{2}}}={x}_{f}-{x}_{i}$, since (yf-yi)=0 and (zf-zi)=0

Experiment and Learn

Displacement Vector

The graph shows the trajectory followed by a body over the time considered.

With the mouse, drag the starting (Pi) and final (Pf) position of the body to any position you want. Then press the play button.

Observe how the body move from the initial position to the final position and the displacement vector is obtained.

### Now...

We have not found any exercise for this section. If you think that it is due to an error, we will appreciate your feedback.

## Formulas worksheet

Here is a full list of formulas for the section Displacement. By understanding each equation, you will be able to solve any problem that you may encounter at this level.

Click on the icon   to export them to any compatible external program.

#### Displacement vector in three dimensions, Cartesian coordinates

$∆\stackrel{\to }{r}={\stackrel{\to }{r}}_{f}-{\stackrel{\to }{r}}_{i}=\left({x}_{f}-{x}_{i}\right)\stackrel{\to }{i}+\left({y}_{f}-{y}_{i}\right)\stackrel{\to }{j}+\left({z}_{f}-{z}_{i}\right)\stackrel{\to }{k}$

#### Magnitude of the displacement vector in two-dimensional Cartesian

$\left|∆\stackrel{\to }{r}\right|=\sqrt{{\left({x}_{f}-{x}_{i}\right)}^{2}+{\left({y}_{f}-{y}_{i}\right)}^{2}}$

#### Displacement vector in two dimensions, Cartesian coordinates

$∆\stackrel{\to }{r}={\stackrel{\to }{r}}_{f}-{\stackrel{\to }{r}}_{i}=\left({x}_{f}-{x}_{i}\right)\stackrel{\to }{i}+\left({y}_{f}-{y}_{i}\right)\stackrel{\to }{j}$

#### Magnitude of the displacement vector in three-dimensional Cartesian

$\left|∆\stackrel{\to }{r}\right|=\sqrt{{\left({x}_{f}-{x}_{i}\right)}^{2}+{\left({y}_{f}-{y}_{i}\right)}^{2}+{\left({z}_{f}-{z}_{i}\right)}^{2}}$