AcademyMotion in Space
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Position and Velocity Vectors

Level 1 - Physics topic page in Motion in Space.

Principle

A position vector locates a particle from an origin; velocity is the time derivative of that vector.

Earlier one-dimensional motion used one coordinate. In space, the same idea is applied component by component along fixed coordinate axes.

Notation

\(\vec r\)
position vector from the origin
\(\mathrm{m}\)
\(x,y,z\)
Cartesian position components
\(\mathrm{m}\)
\(\Delta\vec r\)
displacement vector
\(\mathrm{m}\)
\(\vec v\)
velocity vector
\(\mathrm{m\,s^{-1}}\)
\(v\)
speed, the magnitude of velocity
\(\mathrm{m\,s^{-1}}\)

Method

Derivation 1: Build the position vector

Choose an origin and fixed axes. A point with coordinates \((x,y,z)\) is reached by moving \(x\) along \(\hat\imath\), \(y\) along \(\hat\jmath\), and \(z\) along \(\hat k\).

Add coordinate displacements
\[\vec r=x\hat{\imath}+y\hat{\jmath}+z\hat{k}\]
Compare two positions
\[\Delta\vec r=\vec r_f-\vec r_i\]
Displacement depends on two positions, not on the path between them.
Write displacement in components
\[\Delta\vec r=\Delta x\hat{\imath}+\Delta y\hat{\jmath}+\Delta z\hat{k}\]

Derivation 2: Differentiate the position vector

The unit vectors are fixed for Cartesian axes. That means the derivative acts on the scalar components.

Let each component depend on time
\[\vec r(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k}\]
Differentiate component by component
\[\vec v(t)=\frac{dx}{dt}\hat{\imath}+\frac{dy}{dt}\hat{\jmath}+\frac{dz}{dt}\hat{k}\]
Name the velocity components
\[\vec v=v_x\hat{\imath}+v_y\hat{\jmath}+v_z\hat{k}\]
Here \(v_x=dx/dt\), \(v_y=dy/dt\), and \(v_z=dz/dt\).

Derivation 3: Interpret speed and tangent direction

Over a short time interval, displacement points along a small chord of the path. In the limit as the interval shrinks, that chord becomes tangent to the path.

Average velocity
\[\vec v_{\mathrm{avg}}=\frac{\Delta\vec r}{\Delta t}\]
Instantaneous velocity
\[\vec v=\lim_{\Delta t\to0}\frac{\Delta\vec r}{\Delta t}=\frac{d\vec r}{dt}\]
Speed is magnitude
\[v=|\vec v|=\sqrt{v_x^2+v_y^2+v_z^2}\]

The sketch shows the geometric interpretation: position points from the origin, while velocity is tangent to the path at the particle.

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The position vector points from the origin; the velocity vector is tangent to the path.

Rules

These are the compact results from the derivations above.

Position vector
\[\vec r=x\hat{\imath}+y\hat{\jmath}+z\hat{k}\]
Displacement vector
\[\Delta\vec r=\vec r_f-\vec r_i\]
Velocity vector
\[\vec v=\frac{d\vec r}{dt}\]
Speed
\[v=|\vec v|=\sqrt{v_x^2+v_y^2+v_z^2}\]

Examples

Question
For
\[\vec r(t)=2t\hat{\imath}+t^2\hat{\jmath}\]
find
\[\vec v\]
and speed at
\[t=3\,\mathrm{s}\]
Answer
Differentiate the position vector:
\[\vec v=2\hat{\imath}+2t\hat{\jmath}\]
At
\[t=3\]
\[\vec v=2\hat{\imath}+6\hat{\jmath}\,\mathrm{m\,s^{-1}}\]
The speed is
\[v=\sqrt{2^2+6^2}=\sqrt{40}\,\mathrm{m\,s^{-1}}\]

Checks

  • Position depends on the chosen origin.
  • Displacement compares two position vectors.
  • Velocity is tangent to the path.
  • Speed is a magnitude, so it is never negative.

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Calculus Motion

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Acceleration Vectors