Use this scenario:
An electron gains speed in the Stanford Linear Accelerator (SLA) across 3000 meters, reaching a final velocity of 0.95c due to a constant force pushing on the electron. Given the mass of the electron and the constant force pushing on the electron, what is the function for velocity in terms of time?
How do we derive this?
There are two different interpretations of "constant force" possible
In terms of trajectories, the first says that ${d\over dt} {dx\over d\tau} = f$ where f is constant, t is time, x is position and $\tau$ is proper time. The time in this situation is the total electron's momentum divided by the rate of momentum increase, f. Then
$$p(t) = {dx\over d\tau} = {mv\over \sqrt{1-v^2}} = f t $$
solving for v,
$$ v = {at\over \sqrt{1+(at)^2}}$$ where $a={f\over m}$, which can be integrated to give the constant momentum-rate trajecotry x(t) explicitly.
$$ x(t) = {1\over a} \sqrt{1+(at)^2}$$
so that
$$ x^2 - t^2 = {1\over a^2}$$
This is the hyperbola in spacetime, so it has constant curvature, and so it is also the solution to the second interpretation: $||{d^2x\over d\tau^2}|| =f$, where f is constant. The two interpretations, although it is not immediately obvious, are the same.
To understand the coincidence needs a little bit of hyperbolic geometry (it works for circles in Euclidean space just as well). Given a trajectory with rapidity (relativistic analog of angle) $\gamma(t)$, the rate of change of the momentum with respect to time is the time rate of change of $\sinh(\gamma)$, which is $\cosh(\gamma){d\over dt}{\gamma}$, where the derivative is with respect to time, but this is also $\dot{\gamma}$, where the dot is derivative with respect to proper time.
This allows you to solve the stated problem very simply. The momentum of the electron at the end of the motion, when $v=.95$, is just over 3 times the mass of the electron (times c). This divided by the force is the time required to accelerate the electron to that speed.
