3. In rectangle $A B C D$, $A B=10$ cm, $B C=$ 20 cm, a moving point $M$ starts from point $B$ along side $A B$ towards endpoint $A$, with a speed of 1 cm per second, and a moving point $N$ starts from point $C$ along side $B C$ towards point $B$, with a speed of 1 cm per second. Then, by the 10th second, the distance traveled by the midpoint $P$ of the moving segment $M N$ is
3. $5 \sqrt{2}$ cm.

As shown in Figure 7, take the midpoints $E$ and $F$ of $AD$ and $BC$ respectively, and connect $EF$, $EM$, $EN$, $EP$, and $PF$.
It is easy to prove that quadrilateral $ABFE$ is a square.
Then $\triangle MAE \cong \triangle NFE$
$\Rightarrow \triangle EMN$ is an isosceles right triangle
$\Rightarrow PE \perp MN$.
Also, $\angle EPN = \angle EFC = 90^{\circ}$
$\Rightarrow E, P, F, N$ are concyclic
$\Rightarrow \angle EFP = \angle ENP = 45^{\circ}$
$\Rightarrow$ point $P$ is always on the diagonal $AF$ of square $ABFE$.
Initially, point $P$ is at point $F$, and at the 10th second, point $P$ is at the midpoint of $AF$, thus, by the 10th second, the distance traveled by the midpoint $P$ of the moving segment $MN$ is $\frac{1}{2} AF = 5 \sqrt{2}$ cm.