I would like to calculate the $R(t)$ of an object that accelerates at a constant magnitude, but at a changing direction (making the acceleration always tangent to the trajectory. Something like a rocket engine, or aerodynamic drag(with neglect for complex factors that make acceleration non-constant)).
I would pick a predetermined $R(t)$, say $R(t)=ti-\frac{9.81t^2}{2}j$. Then, I would like to apply my acceleration, which probably cannot be done with $W(R(t)) = R(t)-\frac{dR}{ds}$, because the derivative changes too.
I thought of making an equation that looks like this: $$ R(t) = R(t) + C\frac{dR}{ds} $$ where $C$ denotes the magnitute of the constant acceleration. I am not sure whether this would give me what I am looking for though.
I also heard that I might need differential equations for this problem, but I don't know how to construct one.
Your $R(t)$ is really a point with three components. You have a set of coupled differential equations. You should also define $V(t)=\frac d{dt}R(t)$ as the velocity and $A(t)=\frac d{dt}V(t)$ as the acceleration, each with three components. You compute $A(t)$ by dividing the force on the object by its mass.
The naive way to approach this is to take a small time step $\Delta t$. At the start you have the conditions $V(t), R(t)$ and you update the conditions by $$V(t+\Delta t)=V(t)+A(t)\Delta t\\R(t+\Delta t)=R(t)+V(t)\Delta t$$ This is called the Euler method of integrating differential equations. It is conceptually simple-the change in position is just the velocity times the time step. The problem is also easy to see-the velocity is changing during the time step and we are ignoring that fact in updating the position. This leads to accumulating error in the solution. Reducing the time step can help with this, but you run into numerical accuracy and maybe compute time issues if you make it too small. There are more complex algorithms that are more accurate. You can consult any numerical analysis textbook for a discussion. I like chapter 16 of Numerical Recipes as a starting point. It is free online. Others will have their favorites.