Answer
$14$ cm
Work Step by Step
First, define the coordinate system.
Let the clock center be the origin and
when the hand points to 3 o'clock, we take it to be the +x direction,
when it points to 12 oclock, this is +y.
(a)
We interpret the initial position of the hand: $\vec{r_{1}}=(10\mathrm{c}\mathrm{m})\hat{\mathrm{i}}$
and the final position: $\vec{r_{2}}=(-10\mathrm{c}\mathrm{m})\hat{\mathrm{j}}$.
We have some formulas at hand,
$\left[\begin{array}{lll}
\vec{r}=x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+z\hat{\mathrm{k}} & (4- 1) & \text{Position vector} \\
\Delta\vec{r}=\vec{r}_{2}-\vec{r}_{1} & (4- 2) & \text{Displacement} \\
\Delta\vec{r}=(x_{2}-x_{1})\hat{\mathrm{i}}+(y_{2}-y_{1})\hat{\mathrm{j}}+(z_{2}-z_{1})\hat{\mathrm{k}} & (4- 3) & \text{Displacement}
\end{array}\right]$
Using 4-2,
$\Delta\vec{r}=\vec{r}_{2}-\vec{r}_{1}=(-10cm)\hat{j}-(10cm)\hat{i} =(-10cm)\hat{i}+(-10cm)\hat{j}.$
Its magnitude is
$|\Delta\vec{r}|=\sqrt{(-10\mathrm{c}\mathrm{m})^{2}+(-10\mathrm{c}\mathrm{m})^{2}}=14$ cm
