$$\frac{mv^2}{2}=\frac{mc^2}{2}(1-e^{-2\frac{\Delta Uq}{mc^2}})$$

The kinetic energy is derived under the assumption that Coulomb's force is proportional to $$\ 1-v^2/c^2$$

For simplicity, I took the case when there is no initial velocity.

Accelerated particle is emitted from the origin

**Derivation:**

From Newton's Second Law: $$m\frac{dv}{dt}=qE(1-v^2/c^2)$$

Here $$E$$ - the electric field intensity

We integrate:

$$\frac{dv}{(1-v^2/c^2)}=\frac{qE}{mc}dt$$

$$\frac{v}{c}=th(\frac{qE}{mc}t)$$ [1]

Once again, we integrate

$$\frac{dx}{dt}=c\ th(\frac{qE}{mc}t)$$

$$x=\frac{mc^2}{qE}\ ln\{ch(\frac{qE}{mc}t)\}$$

Next we do recalculate the electric field intensity into a potential difference

$$\Delta U=xE$$

$$\Delta Uq=mc^2\ ln\{ch(\frac{qE}{mc}t)\}$$

$$e^{\frac{\Delta Uq}{mc^2}}=ch(\frac{qE}{mc}t)$$

We use the equation [1], to eliminate the time

$$\frac{mv^2}{2}=\frac{mc^2}{2}th^2(\frac{qE}{mc}t)=\frac{mc^2}{2}\frac{e^{2\frac{\Delta Uq}{mc^2}}-1}{e^{2\frac{\Delta Uq}{mc^2}}}=\frac{mc^2}{2}(1-e^{-2\frac{\Delta Uq}{mc^2}})$$