5. As shown in Figure 2, three points $P$, $S$, and $R$ move at a uniform speed on the sides of $\triangle A B C$. When $t=0$, they start from $A$, $B$, and $C$ respectively, and when $t=1 \mathrm{~s}$, they simultaneously reach $B$, $C$, and $A$. Then, the fixed point in this motion process is the ( ) of $\triangle P Q R$.
(A) incenter
(B) circumcenter
(C) orthocenter
(D) centroid
5.D.

According to the problem, $\frac{A P}{A B}=\frac{B S}{B C}=\frac{C R}{C A}=\lambda$.
Let $G$ be the centroid of $\triangle P S R$, then
$$
\begin{array}{l}
A G=\frac{1}{3}(A P+A S+A R) \\
=\frac{1}{3}[\lambda A B+A B+\lambda B C+(1-\lambda) A C] \\
=\frac{1}{3}(A B+A C) .
\end{array}
$$

Therefore, $G$ is the centroid of $\triangle A B C$.