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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.

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  • $\begingroup$ Your second equation is ambiguous. You need to use brackets to clarify. $\endgroup$
    – Gert
    Sep 4 '20 at 18:59
  • $\begingroup$ Why does this problem need to be two dimensional? It would simplify things to eliminate the x direction. Then you might more easily be able to see if the problem is well posed. $\endgroup$
    – Andrew
    Sep 4 '20 at 19:47
  • $\begingroup$ Its 2 dimensional since there is an initial x velocity that the acceleration must counteract. To put it into context, this is supposed to calculate the time a suicide burn would take on a rocket. If I ignored the x component and at the start of the burn and the initial x velocity was say 500m/s it would hit the ground at 500m/s which isn't ideal. $\endgroup$ Sep 4 '20 at 20:37
  • $\begingroup$ You should not include g in your second equation because the acceleration is a, not a+g $\endgroup$ Sep 5 '20 at 3:12
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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.

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  • $\begingroup$ I don't think the trajectory will be linear, since gravity is involved. $\endgroup$
    – invjac
    Sep 4 '20 at 22:08
  • $\begingroup$ Could you prove it to me then? I'm a little confused on how it's linear and if it is linear that means I got the velocity equations very wrong since without $a(t)$ it should be a perfect parabola since removing the acceleration its just a projectile falling in gravity. Also if I integrate the velocity equations (ignoring the acceleration since that my problem) I get $P_x(t)=P_{ix}+V_{ix}*t$ and $P_y(t)=P_{iy}+V_{iy}*t-0.5*g*t^2$ which looks very parabolic to me. $\endgroup$ Sep 4 '20 at 22:11
  • $\begingroup$ I does not matter what creates the force, the acceleration has the same direction than the velocity. I will post the proof later $\endgroup$ Sep 4 '20 at 22:28
  • $\begingroup$ I'm not sure how $dv/dt$ can have units of acceleration times velocity squared. Are you missing division signs in your equations? (And also negative signs if you want acceleration in the opposite direction as velocity) $\endgroup$ Sep 5 '20 at 3:16
  • $\begingroup$ @BioPhysicist sorry, I missed the division sign, thanks $\endgroup$ Sep 5 '20 at 3:17

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