Notes below outline (but do not complete)
a novel solution to this problem that makes use of
Galilean chase-plane time T, and Galilean-kinematic velocity V = dx/dT.
These obey Galileo's equations for constant (proper)
acceleration *at any speed*, even the classical
expression for kinetic energy K=½mV^{2}.
A solution in terms of the more widely used (and useful)
coordinate-velocity v = dx/dt may instead be found
here.
In both cases, we also find it useful to define from Minkowski's
spacetime Pythagoras' theorem, i.e. the
metric equation: (c dτ)^{2} = (c dt)^{2} - (dx)^{2},
the proper or traveler time τ and the proper velocity w = dx/dτ.

where the speed of light *c* is 3x10^8[m/s^{2}] or 1[ly/yr],
proper acceleration α is 9.8[m/s^{2}] or
1.03[ly/yr^{2}], and
*w*_{f} & *w*_{o} denote final & initial
*proper velocities* (w = dx/dτ) in units of "distance
traveled per unit of
traveler time". Relativists will recognize proper velocity
*w* as *c* Sqrt[*γ*^{2}-1], where
*γ* = E/m*c*^{2}. Proper velocities here can be
figured by the conversion

from "Galilean velocities" *V*_{f} and *V*_{o} which
obey the classical equations for constant acceleration. The standard
equation

(where d*x* is distance traveled) in particular should do the job.
Given initial/final proper velocities, the maptime d*t* elapsed
is simply (*w*_{f}-*w*_{o})/α,
and coordinate velocity *v*=d*x*/d*t*
is *w*/*γ*. Can you show that finite proper velocities w
require that *v* is always less than *c*?

**Note:** The "Galilean velocity" *V* is the velocity familiar
from introductory physics books extended to all velocities *as
Galileo might have presumed* through the kinetic energy equation
K = (1/2)*mV*^{2}. It thus becomes the derivative of *x* with
respect to a "Galilean chase-plane time" *T* which can be used to
track events during 1D acceleration, but which at high speed follows
*neither* traveler *nor* earth based clocks. At high speeds
Galilean velocity *V* is not the Lorentz/Minkowski coordinate
velocity *v* = dx/dt.

* The Andromeda Galaxy is one of the most distant objects visible to
the naked eye. Total distance traveled is
2.2x10^{6} [ly] x 9.46x10^{15} [m/ly] = 2.08x10^{22} [meters].

Send comments, your answers to problems posed, and/or complaints, to philf@newton.umsl.edu. This page contains original material, so if you choose to echo in your work, in print, or on the web, a citation would be cool.

` (Thanks. /philf :)`