How do I get the position at time $t$, if acceleration depends on position?

Question

In my visit to more realistic particle motion animation, $F=\frac{kq_1q_2}{d^2}$, $F=ma$, so: $$a=\frac{kq_1q_2}{md^2}$$ My velocity, integrating the above (from a site because I forgot how to do it), is $v=-kq_1q_2md+v_0$. Curiously, the integral of that, which should give displacement (!), is $d=d_0+(v_0d)-(kq_1q_2/m)(ln|d|)$. Does that work? If so, what do I do to get $v$, $d$ or $a$ at a certain time t? (I thought of getting $W=F\cdot d$ and trying to connect work to time thru another equation, maybe power?, etc., but now $F$ varies and also depends on $d$ (duh...) and I kinda stopped knowing what to do. Most (practically all?) stuff I saw deals with either varying acceleration in terms of time, or circular motion. Thanks in advance :)

ADDENDUM: This looks suspiciously close to gravity related questions, but in my case I have ions or charged particles in mind, where both masses matter, as well as the movements of both bodies. What brought up the issue was the question "were ions hard balls, when they 'collide' is the ricochet a fact, or is the attraction so great that it is already generating a -v higher than v of the conservation of momentum ricochet resulting from the collision? If there is a 'bounce', What is it like? etc.". (My initial model has only an Li+ and a F- 6000 picometers apart in a 1 cubic meter 'universe', in case you wonder what I am doing.)

Answer 1

$$a = \frac{dv}{dt}$$

$$a = \frac{dv}{dt} \frac{dx}{dx}$$

$$a = \frac{dv}{dx} \frac{dx}{dt}$$

$$a = v\frac{dv}{dx}$$

$$\int_{x_0}^x a.dx = \int_{v_0}^{v} v. dv$$

Here, $x$ is not the final position; it is a variable. Note that we are solving the differential equation for a general case.

$$2\int_{x_0}^x a .dx = v^2 - v^2_0$$

$$v^2 = v^2_0 + 2\int_{x_0}^x a .dx$$

$$v = \sqrt{v^2_0 + 2\int_{x_0}^x a .dx}$$

$$\frac{dx}{dt} = \sqrt{v^2_0 + 2\int_{x_0}^x a .dx}$$

$$\int_{0}^t dt = \int_{x_i}^{x_f} \frac{1}{\sqrt{v^2_0 + 2(\int_{x_0}^x a .dx)}} dx$$

Here, $x_f$ is the final position. If you need a general equation, you can replace $x_f$ with $x$ but be careful not to confuse the upper limit of the outside integral with the upper limit of the inner integral. They are two different things and the upper limit of the inner integral must be a variable for the sake of the outside integral.