### Velocity, acceleration, trajectory

$x=k\omega t\cos\omega t\\y=k\omega t\sin\omega t\\v_x=k\omega(\cos\omega t -\omega t \sin\omega t)\\v_y=k\omega(\sin\omega t + \omega t \cos\omega t)\\a_x=k\omega^2(-2\sin\omega t - \omega t \cos\omega t) \\a_y=k\omega^2(2\cos\omega t - \omega t \sin\omega t)\\\cos(\widehat{\vec {av}})= \vec {a}\cdot\vec {v}/ |\vec {a}|\cdot|\vec {v}|\\=(a_xv_x+a_yv_y)/(a_x^2+a_y^2)^{1/2}(v_x^2+v_y^2)^{1/2}$
Second method: polar coordinates.
$\vec {r}=r\hat{r}, \quad \hat{r}=\cos\varphi \hat{x}+ \sin\varphi \hat{y} \\\vec {v}=\dot{\vec {r}}=\dot{r}\hat{r}+r\dot{\hat{r}}\\ =\dot{r}\hat{r}+ r(-\sin\varphi\hat{x}+\cos\varphi\hat{y}) \dot{\varphi}\\=\dot{r}\hat{r}+ r\dot{\varphi}\hat{\varphi}\\\dot{\hat{\varphi}}=(-\cos\varphi\hat{x}-\sin\varphi\hat{y})\dot{\varphi}\\=-\dot{\varphi}\hat{r}\\\dot{\hat{r}}=\dot{\varphi}\hat{\varphi}\\\vec {a}=\dot{\vec {v}}=\ddot{r}\hat{r} + \dot{r}\dot{\hat{r}}+\dot{r}\dot{\varphi}\hat{\varphi}+r\ddot{\varphi}\hat{\varphi} + r\dot{\varphi}\dot{\hat{\varphi}}\\=(\ddot{r}-\dot{\varphi}^2r)\hat{r} + (r\ddot{\varphi} + 2\dot{r}\dot{\varphi})\hat{\varphi}$
We have $\varphi=\omega t$ so that $\dot{\varphi}=\omega$, $\ddot{\varphi}=0$. We have $r=k\omega t$, so that $\dot{r}=k\omega$, $\ddot{r}=0$. Now
$\vec {v}=k\omega \hat{r}+ \omega r\hat{\varphi}\\\vec {a}=-\omega^2 r \hat{r}+ 2k\omega^2\hat{\varphi}$