## Statement

## Solution

**Data**

ω = 6 π rad/s

v = 3 m/s

**Resolution**

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

$$v=\omega \xb7R\Rightarrow \phantom{\rule{0ex}{0ex}}R=\frac{v}{\omega}\Rightarrow \phantom{\rule{0ex}{0ex}}R=\hspace{0.17em}\frac{3}{6\pi}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)R=0.1m}\phantom{\rule{0ex}{0ex}}$$

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

$$s=\phi \xb7R\Rightarrow \phantom{\rule{0ex}{0ex}}\phi =\frac{s}{R}\Rightarrow \phantom{\rule{0ex}{0ex}}\phi =\frac{1}{0.1}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)\phi =10rad}$$

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:

$$\overrightarrow{r}=R\xb7\mathrm{cos}\phi \xb7\overrightarrow{i}+R\xb7\mathrm{sin}\varphi \xb7\overrightarrow{j}\Rightarrow \phantom{\rule{0ex}{0ex}}\overrightarrow{r}=0.1\xb7\mathrm{cos}\left(10rad\right)\xb7\overrightarrow{i}+0.1\xb7\mathrm{sin}(10rad)\xb7\overrightarrow{j}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)\overrightarrow{r}=-0.08\xb7\overrightarrow{i}-0.05\overrightarrow{j}}$$