These traveler-frame results follow using a routine protocol for converting from values for the Galilean kinematic, via w=V*Sqrt[1+0.25*(V/c)2] and Δτ=[ArcSinh(wf/c)-ArcSinh(wo/c)]*c/α where c=3x10^8 m/s is the speed of light. In this particular problem, the traveler values are quite different from the Newtonian values.
Relativistic Inertial-Frame (Map Kinematic) Results:
These inertial-frame results follow by routine conversion from the traveler-frame values, using v=w/SQRT[1+(w/c)^2], and dt=[wf-wo]/α. In this particular problem, the map kinematic values are quite different from the Galilean values.
Final Discussion: You were presented with two possible solutions -- choose the one for which Vf is positive. The other solution would yield the velocity of the spacecraft prior to the start of the race had it decelerated (in the forward direction) to the finish line. The GK chaseplane time elapsed is ΔT = 8.788x10^7 seconds, and the final GK velocity is Vf = 8.618x10^8 meters per second. Unfortunately, this method of parameterizing the world-line of the dragster with time corresponds to no single observer's physical clocks, since time itself proceeds at different rates for observers traveling at significantly different speeds.
To consider the clocks of the driver, note that the traveler frame (or proper) time elapsed is 2.247 traveler years! Moreover, her final velocity is 5.028 roddenberries -- meaning that she is now traveling more than 5 lightyears of distance (as measured in the rest frame of our solar system) per year of her own time. Speaking of the spectator or solar-system point of view, however, the time elapsed on earth during this race is 4.874 inertial years. Thus her friends back home are now almost 5 years older at the end of the race, even though fewer than 27 months have elapsed for her. Moreover, from their point of view, she has a coordinate velocity of only 0.981 c, since to inertial observers the interconnected fabric of spacetime does indeed limit travelers to lightspeed or less.
Picturing It: An approximate graphical solution using distance-time relationships between the variables (and letting α = 1[ly/y^2] exactly) is shown below. Actually, the solution is exact if the acceleration is 0.97 g = 1[ly/y^2], instead of 1 g exactly. A blowup near the origin of this plot can be used to graphically solve acceleration problems at lower everyday speeds.
Making Quadruplets: This graphical strategy also works if we combine constant acceleration segments. For example, one of the famous "twin paradox" problems might involve a 4 ly acceleration leg like that discussed in this problem, followed by a second 4 ly leg for decelerating to a stop, a 3rd 4 ly leg for accelerating in the return direction, and a 4th deceleration leg bringing an inertial twin's traveling sister back home. After such an adventure, our traveling twin will have aged 4 times 2.247 or about 9 years, while the twin brother who stayed home will have aged nearly 20 years. Some cool things about this approach to calculating and visualizing the adventure are: (i) only one inertial frame is needed, so confusions built into our language about things like length contraction and simultaneity aren't involved ; (ii) the Galilean solution of this problem remains part of the overall solution, and in fact can serve as the means to actually do the calculation, followed by conversion to velocities and elapsed times associated with the other kinematics, and (iii) all variables can be shown on a single graph, allowing visualization and estimation without total dependence on equations that some find abstract at best.
The resulting twin-adventure graph is also (I think) pleasant to look at! To keep the details of the plot from obscuring its symmetry, to keep the onus on the reader to fill in the blanks, and of course to conserve bandwidth on a very busy web, we provide a reduced version of this twin plot for your enjoyment below. Can you guess what the curves represent without downloading the full sized image?
