5. As shown in Figure 3, there is an insect A at vertex $C_{1}$ of a cube box with an edge length of $10 \mathrm{~cm}$, and there is an insect B at vertex $A$ inside the box (the thickness of the box is negligible). Assuming insect A starts from vertex $C_{1}$ and crawls downward along edge $C_{1} C$ at a speed of $1 \mathrm{~cm} / \mathrm{s}$, and at the same time, insect B starts from vertex $A$ and crawls on the inner walls of the box at a speed of $2 \mathrm{~cm} / \mathrm{s}$, then insect B needs at least ( ) to catch insect A (rounded to $1 \mathrm{~s}$).
(A) $10 \mathrm{~s}$
(B) $8 \mathrm{~s}$
(C) $6 \mathrm{~s}$
(D) $4 \mathrm{~s}$
5. B.

When insect A crawls down from vertex $C_{1}$ along edge $C_{1} C$, insect B crawls along any one of the four paths shown in Figure 6.
Figure 6
In Figure 6(a) and Figure 6(b), if insect B takes $x \mathrm{~s}$ to catch insect A along the path $A \rightarrow E \rightarrow F$, then in the right triangle $\triangle A C F$, we have $(2 x)^{2}=(10-x)^{2}+20^{2}$, solving which gives $x=10$;

In Figure 6(c) and Figure 6(d), if insect B takes $y \mathrm{~s}$ to catch insect A along the path $A \rightarrow E \rightarrow F$, then in the right triangle $\triangle A D F$ or $\operatorname{Rt} \triangle A B F$, we have $(2 y)^{2}=10^{2}+(20-y)^{2}$, solving which gives $y \approx 8$. Therefore, the minimum time required is $8 \mathrm{~s}$.