D Given $\triangle A B C$, if for any $t \in \mathbf{R},|\overrightarrow{B A}-t \overrightarrow{B C}| \geqslant|\overrightarrow{A C}|$, then $\triangle A B C$ must be ( ).
(A) Acute triangle
(B) Obtuse triangle
(C) Right triangle
(D) The answer is uncertain
Let $\angle ABC = \alpha$, and draw $AD \perp BC$ at $D$. From $|\overrightarrow{BA} - t \overrightarrow{BC}| \geqslant |\overrightarrow{AC}|$, we get
$$
|\overrightarrow{BA}|^{2} - 2t \overrightarrow{BA} \cdot \overrightarrow{BC} + t^{2}|\overrightarrow{BC}|^{2} \geqslant |\overrightarrow{AC}|^{2}.
$$

Let $t = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BC}|^{2}}$, substituting into the above equation, we have
$$
|\overrightarrow{BA}|^{2} - 2|\overrightarrow{BA}|^{2} \cos^{2} \alpha + \cos^{2} \alpha |\overrightarrow{BA}|^{2} \geqslant |\overrightarrow{AC}|^{2},
$$

which simplifies to $|\overrightarrow{BA}|^{2} \sin^{2} \alpha \geqslant |\overrightarrow{AC}|^{2}$, or $|\overrightarrow{BA}| \sin \alpha \geqslant |\overrightarrow{AC}|$. Therefore, $|\overrightarrow{AD}| \geqslant |\overrightarrow{AC}|$. This implies $\angle ACB = \frac{\pi}{2}$. Hence, the answer is C.