Velocity of particle due to acceleration. Special relativity I understand how to do transformations of four velocities, acceleration and so on under Lorentz boosts. However after all I have learned in special relativity I still don't know how just an ordinary particle's velocity changes due to an external force. 
That is if I'm given a particle with some velocity and we apply a force in the same direction of the velocity or in any direction for a zero initial velocity. How would I find the velocity at time t later in my frame.
For Newtonian mechanics it is obviously 
$v = v_0 + at$.
Can someone point me in the right direction or to some literature on the topic. 
 A: You just need to apply $\vec{F}=\frac{d}{dt}(\gamma m\vec{v})$. For example suppose the second case you say, that I apply a force in some direction, let's say the x-direction, to a particle with zero initial velocity. You would get
$$F_x=m\frac{d}{dt}(\gamma v_x)$$
$$0=\frac{d}{dt}(\gamma mv_y)$$
$$0=\frac{d}{dt}(\gamma mv_z)$$
From the second equation you get $m\gamma v_y=constant=0$, because at $t=0$, $\gamma=1$ and $v_y=0$. From this, since $\gamma\ge1$ always, you get $v_y=0$ for all $t$. The same can be done for the $z$ component, and you get $v_z=0$ for all $t$.
For the x component, just as in Newtonian mechanics, things can get as complicated as you want if the force has some dependence on time or position, you will be faced to solve a very ugly integral. Let's say that the force is constant, then
$$\frac{F_x}{m}=\frac{d}{dt}(\gamma v_x)$$
and
$$\gamma v_x=\frac{F_x}{m}t+[\gamma v_x]|_{t=0}=\frac{F_x}{m}t$$
You can now substitute $\gamma$
$$\frac{v_x}{\displaystyle\sqrt{1-\frac{v_x^2}{c^2}}}=\frac{F_x}{m}t$$
and solve for $v_x$.
