# Apply acceleration that is always opposite to the velocity

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.

• Your second equation is ambiguous. You need to use brackets to clarify.
– Gert
Sep 4 '20 at 18:59
• 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. Sep 4 '20 at 19:47
• 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. Sep 4 '20 at 20:37
• You should not include g in your second equation because the acceleration is a, not a+g Sep 5 '20 at 3:12

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.

• I don't think the trajectory will be linear, since gravity is involved. Sep 4 '20 at 22:08
• 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. Sep 4 '20 at 22:11
• I does not matter what creates the force, the acceleration has the same direction than the velocity. I will post the proof later Sep 4 '20 at 22:28
• 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) Sep 5 '20 at 3:16
• @BioPhysicist sorry, I missed the division sign, thanks Sep 5 '20 at 3:17