One question is how to translate from time rates to space rates and vice versa. Consider scalar space and scalar time, and designate the spatial position, *s*, initial spatial position, *s*_{0}, temporal position, *t*, initial temporal position, *t*_{0},, velocity, v, initial velocity, *v*_{0}, acceleration *a* (assumed constant over time), legerity, u, initial legerity, *u*_{0}, and expedience, *b* (assumed constant over space). Then the linear equations of motion are as follows (see also the equations of motion at the top menu or *here*):

*s *=* s*_{0} +* vt; **v *=* v*_{0} +* at; **s *=* s*_{0} +* v*_{0}*t *+ ½*at*²; *v² *=* v*_{0}² + 2*a*(*s *–* s*_{0});

*t *=* t*_{0} +* us; **u *=* u*_{0} + *bs*; *t *=* t*_{0} +* u*_{0}*s *+ ½*bs*²; *u*² =* u*_{0}² + 2*b*(*t *–* t*_{0})

From these we have the following derivatives:

d*s*/d*t* = *v* = 1/*u*; d*v*/d*t* = *a*; d*t*/d*s* = *u* = 1/*v*; d*u*/d*s* = *b;*

d*u*/d*v* = –1/*v*² = –*u*² and d*v*/d*u* = –1/*u*² = –*v*².

If we are given the acceleration, *a*, what is the expedience, *b*? This may be determined as follows:

*b* = d*u*/d*s* = d*v*/d*t* * d*t*/d*s* * d*u*/d*v* = *a* * (1/*v*) * (–1/v²) = –*a*/*v*³ = –*a*/(*v*_{0} +* at*)³.

If *v*_{0} = 0, then *b*(*t*) = –1/(*a*²*t*³).

So the expedience is a function of scalar time.

Similarly, if we are given the expedience, *b*, what is the acceleration, *a*? This may be determined as follows:

*a* = d*v*/d*t* = d*u*/d*s* * d*s*/d*t* * d*v*/d*u* = *b* * (1/*u*) * (–1/u²) = –*b*/*u*³ = –*b*/(*u*_{0} + *bs*)³.

If *u*_{0} = 0, then *a*(*s*) = –1/(*b*²*s*³).

So the acceleration is a function of scalar space.