5. In the rectangular coordinate plane, given points $A(0,2)$, $B(0,1)$, $D(t, 0)(t>0)$, and $M$ is a moving point on line segment $AD$. If $|AM| \leqslant 2|BM|$ always holds, then the minimum value of the positive real number $t$ is $\qquad$
5. $\frac{2 \sqrt{3}}{3}$.

Let $M(x, y)$.
From $|A M| \leqslant 2|B M|$, we get $x^{2}+\left(y-\frac{2}{3}\right)^{2} \geqslant \frac{4}{9}$.
Therefore, the line segment $A D$ is always outside the Apollonius circle $x^{2}+\left(y-\frac{2}{3}\right)^{2}=\frac{4}{9}$.
When $t$ is at its minimum, the line segment $A D$ is tangent to the circle, thus, $l_{A D}: \frac{x}{t}+\frac{y}{2}=1$.
Therefore, $\frac{\left|\frac{2}{3} t-2 t\right|}{\sqrt{4+t^{2}}}=\frac{2}{3} \Rightarrow t=\frac{2 \sqrt{3}}{3}$.