Fasten your belts

According to Newton (and that does not change in relativity):

$\dot{p} = F = ma_{0}.$

But, of course, the momentum for Einstein is a little bit different than it is for Newton: $p = \gamma m v$ , where

$\gamma = \frac{1}{\sqrt{1 - \left ( \frac{v}{c} \right )^{2}}}.$

Then it is easy to see that $\dot{p} = \gamma^{3} m \dot{v} = ma_{0}$ from which we find

$\frac{dv}{\left ( 1 - \left ( \frac{v}{c} \right )^{2} \right )^{3/2}} = a_{0} dt.$

Integrating and isolating the velocity we'll find:

$v(t) = \frac{a_{0}t}{\sqrt{1 + \left ( \frac{a_{0}t}{c} \right )^{2}}}.$

As we know the total length of the trip, it would be nice if we could find a relation between time and position. It is easy to find, since $dx = v(t) dt$ . Therefore

$L = \int_{0}^{L} dx = 2 \int_{0}^{t_f/2} \frac{a_{0}t}{\sqrt{1 + \left ( \frac{a_{0}t}{c} \right )^{2}}} \, dt = \frac{2c^{2}}{a_{0}} \left ( \left ( \left ( \frac{a_{0}t_f}{2c} \right )^{2} + 1 \right )^{1/2} - 1 \right ).$

The integration on $t$ is bounded in the interval $\left [ 0, t_f/2 \right ]$ because the force changes direction exactly at $x = L/2$ . But, as this is a symmetrical configuration, the total time integral is just twice the value we get (therefore the "2" before the integral). Isolating for $t_f$ :

$t_f = \frac{2c}{a_{0}} \left ( \left ( \frac{a_{0}L}{2c^{2}} + 1 \right )^{2} - 1 \right )^{1/2}.$

So $A = 2$ , $B = 2$ , $C = 1$ and $D = - 1$ . The sum is $4$ .