5. Given planar vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ satisfy:
$$
\begin{array}{l}
|\boldsymbol{a}|=|\boldsymbol{b}|=|\boldsymbol{c}|=2, \boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}=\mathbf{0} . \\
\text { If } 0 \leqslant x \leqslant \frac{1}{2} \leqslant y \leqslant 1 \text {, then } \\
|x(\boldsymbol{a}-\boldsymbol{c})+y(\boldsymbol{b}-\boldsymbol{c})+\boldsymbol{c}|
\end{array}
$$

the minimum value is $\qquad$
5. $\frac{1}{2}$.

Since $|a|=|b|=|c|, a+b+c=0$, therefore, the angles between $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ are all $120^{\circ}$.
Thus, $\boldsymbol{a} \cdot \boldsymbol{b}=\boldsymbol{b} \cdot \boldsymbol{c}=\boldsymbol{a} \cdot \boldsymbol{c}=-2$.
Let $z=1-x-y$. Then
$$
\begin{array}{l}
|x(\boldsymbol{a}-\boldsymbol{c})+y(\boldsymbol{b}-\boldsymbol{c})+\boldsymbol{c}|^{2} \\
=|x \boldsymbol{a}+y \boldsymbol{b}+(1-x-y) \boldsymbol{c}|^{2} \\
=2\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right) \\
\geqslant 2\left((x-y)^{2}+(y-z)^{2}\right) \\
=2\left((y-x)^{2}+(2 y+x-1)^{2}\right) \\
\geqslant((y-x)+(2 y+x-1))^{2} \\
=(3 y-1)^{2} \geqslant \frac{1}{4},
\end{array}
$$

Equality holds if and only if $y=\frac{1}{2}, z=x=\frac{1}{4}$. 
Therefore, the minimum value sought is $\frac{1}{2}$.