Apply acceleration that is always opposite to the velocity

Question

I have 2 functions for the $x$ and $y$ components of the velocity of an object ($z$ should always be $0$ in this case)

$$V_x(t)=v_{xi}+\int_0^ta_x(t')dt'$$

$$V_y(t)=v_{yi}+\int_0^t(-g+a_y(t'))dt'$$

and a function for the magnitude of $a(t)$:

$$a(t)=\frac{T}{m-t\frac{T}{I\cdot g_0}}$$

$a(t)$ is always in the opposite direction of $V(t)$.

How could I replace $a_x(t)$ and $a_y(t)$ in the functions $V_x(t)$ and $V_y(t)$ so that $a$ has the magnitude defined by its function at time t but in the opposite direction of $V$? In other words, how could I get the component functions of $a(t)$ so that it is the opposite direction of $V(t)$ at all t?

I tried setting up the functions like:

$$V_x(t)=v_{xi}+\int_0^ta(t')\cos(\theta)dt'$$

$$V_y(t)=v_{yi}+\int_0^t(-g+a(t')\sin(\theta))dt'$$

but then $\theta$ would have to be equal to something like $\tan^{-1}(\frac{V_y(t)}{V_x(t)})$ which would require me to already know $V_x(t)$ and $V_x(t)$ at that t.

I also tried applying the magnitude of $a(t)$ to the total magnitude of $V(t)$ rather than trying to split it into components to get $V(t)=\sqrt{v_{xi}^2+(v_{yi}-gt)^2}-\int_0^t{a(t')dt'}$ but I need the velocity in component form since my next step is go integrate the y component for the y position and I will also need to solve for t later and not even mathematica can seem to solve for t in that equation.

For context I'm using these equations to determine how long a suicide burn would take and when to start burning for a rocket so that $V(t)=0$ at the same $t$ that the y position is 0. $a(t)$ is the acceleration from the engine.

Answer 1

You can show without much difficulty that the trajectory will be linear, so the angle is a constant, that is, is a function of the initial conditions. Imagine you rotate you axes so the initial velocity is along x. Then by symmetry, the particle will remain on the x-axis. I can prove it if your intuition is still unconvinced.

Proof: (I ignored g because you said that a is the acceleration, which means that whatever forces are acting, including g, are included in the calculation of a)

$dv_x/dt=-a(t)v_x/\sqrt{v_x^2+v_y^2}$

$dv_y/dt=-a(t)v_y/\sqrt{v_x^2+v_y^2}$

Dividing both we get: $v_x'/v_x=v_y'/v_y$

And after integration we get:

$v_y/v_x=v_{0y}/v_{0x}$

That is, the velocity stays at the same angle than the initial velocity.