Equations of motion for an object with normal and angular acceleration I am currently coding something with a moving object and can't figure out the physics with my school knowledge only. I hope this question is not below the standards of this forum.
I have a moving object at ($x_0, y_0, \theta_0, v_0, \omega_0$), where $x$ and $y$ are the 2D-coordinates, $\theta$ the angle (to the x axis), $v$ the forward velocity and $\omega$ the angular velocity.
I want to accelerate the object with a forward acceleration of $a$ and an angular acceleration of $\alpha$ and I want to calculate the new position after a time $\Delta t$.
If $\alpha = 0$ (just as an intermediate step), the new state should be
$$
x_1 = x_0 + \Delta t (v_0 \cos \theta_0) + 0.5 a (\Delta t^2) \\
y_1 = y_0 + \Delta t (v_0 \sin \theta_0) + 0.5 a (\Delta t^2) \\ 
\theta_1 = \theta_0 + \omega_0 \Delta t  \\
v_1 = v_0 + a \Delta t \\
\omega_1 = \omega_0
$$
I am not sure if this is correct, but this is all I have. And I can't figure out how to add the angular acceleration to this equations, and I guess $\omega$ should be in the equations for $x$ and $y$ too.
 A: You need to keep track of 3 coordinates (2 positions and 1 rotation) and their derivatives. What you are missing is current velocity vector (2 components) which is not necessarily along the orientation direction unless constrainted to do so.
Your best best is a simplectic integrator such as
$$\begin{aligned}
  \begin{pmatrix} vx_1 \\ vy_1 \\ \omega_1 \end{pmatrix} & =
  \begin{pmatrix} vx_0 \\ vy_0 \\ \omega_0 \end{pmatrix} +
  \begin{pmatrix} a \cos\theta_0 \\ a\sin\theta_0 \\ \alpha \end{pmatrix} \Delta t \\
  \begin{pmatrix} x_1 \\ y_1 \\ \theta_1 \end{pmatrix} & =
  \begin{pmatrix} x_0 \\ y_0 \\ \theta_0 \end{pmatrix} + \begin{pmatrix} vx_1 \\ vy_1 \\ \omega_1 \end{pmatrix} \Delta t \\
   t_1 & = t_0 + \Delta t
\end{aligned}$$
and take small steps to make it appear smooth. The above will not artificially add energy to the system like an Euler integrator does. The difference is to take the velocity step first and then use the new velocity in the position step.
If the acceleration is constant for the entire step, the exact equations are
$$\begin{pmatrix} x_1 \\ y_1 \\ \theta_1 \end{pmatrix} =
  \begin{pmatrix} x_0 \\ y_0 \\ \theta_0 \end{pmatrix}+ \frac{1}{2} \begin{pmatrix} vx_0 \\ vy_0 \\ \omega_0 \end{pmatrix} \Delta t  + \frac{1}{2} \begin{pmatrix} vx_1 \\ vy_1 \\ \omega_1 \end{pmatrix} \Delta t \\$$
the problem with this is that typically the constant acceleration is a bad assumption.
