# How to convert acceleration to velocity when the acceleration itself is changing?

I am trying to simulate a rocket and I don't really understand how to convert acceleration to velocity when it is changing constantly. This is my acceleration chart where $$x_1$$ is time in seconds and $$a$$ is the acceleration (measured in newtons): If you need any background context on how I arrived at these values, I took the force of gravity times the mass, added the force of thrust to that and divided all of that by mass.

Now I want to convert this data to velocity. How would I do that for each of the specific accelerations at their specific times? I do know this is possible because BPS space made a flight simulation similar to this.

Also, if possible, it would also be helpful to know how to find the position from the velocity. I know it would be quite similar to finding the velocity from the acceleration, but just in case if you know it, it would be very helpful.

• What are the units of the axis of your diagram? Dec 7, 2022 at 17:25
• Have you learned integral calculus? Dec 7, 2022 at 17:37
• You probably need a written continuous expression for acceleration. Based on your description, you might say: $a=\frac{1}{m}(mg+F)=\frac{1}{m}(mg+\dot{m}V_e+(p_e-p_0)A_e)$. I'm taking your word for acceleration since I'm not knowledgeable in aeronautics. I'd start by rearranging your plot into a continuous graph and integrating over time to obtain velocity. $v=\int a dt$. The same could be done to obtain position: $x=\int v dt$. Dec 7, 2022 at 17:40
• The reason I am not using a continuous graph is that usually rocket suppliers measure thrust in time increments. The time increment here is 0.25 seconds. Also, by the way, the scale I am working at is the model rocket scale. I wouldn't say I am extremely knowledgeable in aeronautics but I just used the F=ma equation. Dec 7, 2022 at 17:49
• How would I be able to change this into a continuous graph? Dec 7, 2022 at 18:01

Now I want to convert this data to velocity. How would I do that for each of the specific accelerations at their specific times?

You cannot "convert" a single acceleration data point to a velocity. You have to integrate the acceleration to find the change in velocity. (You can not find the absolute velocity unless you know the velocity at some point in time; you will otherwise only be able to find the change in velocity.)

In one-dimension $$a = \frac{dv}{dt}\;,$$ where $$a$$ is acceleration, $$v$$ is velocity, and $$t$$ is time. (N.b., although you used the symbol "$$x$$" for time, I am using the more-conventional "$$t$$" for time.)

Therefore, in one dimension: $$\Delta v = v_2 - v_1 = \int_{t_1}^{t_2}a(t) dt$$

The integral can be discretized like: $$\int_{t_1}^{t_2} a(t)dt \to \sum_{n} a_n {\delta t}_n\;,$$ where $${\delta t}_n$$ is your time step and $$a_n$$ is the acceleration now indexed at discrete time points.

With the discretized version, we have: $$\Delta v = v_2 - v_1 \approx \sum_{n=n_1}^{n_2-1} a_n {\delta t}_n =\delta t\sum_{n=n_1}^{n_2-1} a_n\;,$$ where the last equal sign holds for evenly-spaced data points, and where the sum over $$n$$ runs over the points corresponding to the first time through the last time, inclusive of the first time, but not the last time (including the first but not the last is is a choice, a different choice could also be made). This is illustrated with examples below.

Update:

For example, denote the velocity at $$t=0$$ by $$v_0$$. Then the velocity at $$t=0.25$$ seconds can be estimated as: $$v(t=0.25) = v_0 + a_0 \delta t$$

$$= v_0 + (-9.81 \times 0.25)$$

For example, the velocity at $$t=1$$ second can be estimated as: $$v(t=1) = v_0 + a_0 \delta t + a_{0.25}\delta t + a_{0.5}\delta t + a_{0.75}\delta t$$ $$= v_0 + ((-9.81 - 3.93 + 7.84 + 4.31)\times 0.25)$$

Change in velocity is the area under the curve of acceleration vs. time.

In this case, it is easier to calculate the $$\Delta v$$ of the engine by calculating the red area below, and then subtracting the $$\Delta v$$ due to gravity, which is the blue area below. Hint: Use the trapezoidal rule to integrate the curve on the left.