What is the average velocity if the distance vector is * s* and the duration vector is

*? Let the vectors be represented with rectilinear components:*

**t*** s* = (

*s*) =

_{1}, s_{2}, s_{3}**e**

_{1}*s*

_{1}+**e**

_{2}*s*

_{2}+**e**

_{3}*s*,

_{3}* t* = (

*t*) =

_{1}, t_{2}, t_{3}**e**

_{1}*t*

_{1}+**e**

_{2}*t*

_{2}+**e**

_{3}*t*.

_{3}where **e _{1}**,

**e**, and

_{2}**e**are orthogonal unit vectors. Then is the velocity vector the distance vector divided by the magnitude of the time vector? That would be equal to the ratio of the magnitudes times the unit direction vector of the distance:

_{3}* s* / |

*| = (|*

**t****| / |**

*s**|) (*

**t***/ |*

**s***|).*

**s**Or should the direction of the duration vector be taken into account as well? If the velocity were the resultant of three orthogonal movements, we would have:

* v* = (

*v*) = (

_{1}, v_{2}, v_{3}*s*) =

_{1 }/ t_{1}, s_{2 }/ t_{2}, s_{3 }/ t_{3}**e**

_{1}*s*

_{1 }/ t_{1}+**e**

_{2}*s*

_{2 }/ t_{2}+**e**

_{3}*s*.

_{3 }/ t_{3}Then consider that every vector is the sum of its component vectors and we have the answer. So vector division is a division of parallel components. If *t _{1} = t_{2} = t_{3}*, then the velocity vector equals the spatial vector divided by a scalar.

Note that if a vehicle moves East a distance *s _{a}* in time

*t*, then moves North a distance of

_{a}*s*in a time

_{b}*t*, the resulting position could have distance and duration that are not parallel. The distance angle would be atan(

_{b}*s*) and the duration angle would be atan(

_{b}/s_{a}*t*). The angle of the velocity could be different from either. This is because they are in different ‘spaces’ because they have different units. The space of distance is commonly taken as the real space even though they all are real.

_{b}/t_{a}