Position vector from distance traveled in u.c.m.

Data

ω = 6 π rad/s
v = 3 m/s

Resolution

If we apply the definitions of the angular magnitudes, we know that:

$$\begin{gathered} v=\omega ·R \Rightarrow \\ R = \frac{v}{\omega } \Rightarrow \\ R =\hspace{0.17em}\frac{3}{6\pi } \Rightarrow \\ \boxed{R =0.1 m} \end{gathered}$$

Now applying the definition of distance traveled based on the angular position and the radius:

$$\begin{gathered} s=\phi ·R \Rightarrow \\ \phi =\frac{s}{R} \Rightarrow \\ \phi =\frac{1}{0.1}\Rightarrow \\ \boxed{\phi =10 rad} \end{gathered}$$

Now that we know the radius and the angular position when the distance traveled is 1 m, we can determine the position vector using the following expression:

$$\begin{gathered} \overrightarrow{r}=R·\cos \phi · \overrightarrow{i }+ R · \sin \varphi ·\overrightarrow{ j} \Rightarrow \\ \overrightarrow{r}=0.1·\cos \left(10 rad\right) · \overrightarrow{i} + 0.1 · \sin (10 rad) · \overrightarrow{j} \Rightarrow \\ \boxed{\overrightarrow{r}=-0.08 ·\overrightarrow{ i }- 0.05 \overrightarrow{j}} \end{gathered}$$