# How to integrate acceleration to find velocity

I am really confused on how to convert a formula I have for acceleration to find the velocity at that point. The equation is:

$$a=\omega^2\cdot52.24\cdot\cos(\theta)$$

where $\omega$ is the angular velocity in radians per second and $\theta$ is the angle it is at (starting at 0)

I want to find the velocity at this point but have no idea how. I know I have to integrate, but what do I integrate exactly? And how would I write it out?

I have entered $\int_0^1\omega(t)^2\cdot52.24\cdot cos(0)\,dt\,$ into Wolfram Alpha but it just spits out the same as the answer.

The equation you have is probably giving you the acceleration in circular motion. The acceleration is towards the center and the velocity is tangential. In that case you should not be integrating the acceleration to get the velocity. You should use the fact that in uniform circular motion $a=\omega^2r$, so $r=52.24$. Then the velocity is just $v=\omega r$
• Thank you. I have just edited my question a little. It actually includes a cos function at the end, but I removed it because the angle was zero however I now realise I can't just take it out of the equation. This makes my velocity zero when $\theta$ is zero, but also makes my displacement negative for some reason, if I am doing trig integration rules properly. The displacement when $\theta$ is zero should be zero. Do you know what I have done wrong there? – user88720 Apr 26 '17 at 16:07
• So the acceleration is angular? Then you have a differential equation where $a=\frac {d\omega}{dt}$ and $\omega=\frac {d\theta}{dt}$ That doesn't look like a reasonable one to solve analytically to me, so you would have to use a numeric solver. If $\omega(0)=0$ there is no acceleration and the particle just stays there. – Ross Millikan Apr 26 '17 at 16:13