3. As shown in Figure 3, given two perpendicular lines intersecting at point $O$, particle 甲 moves from point $A$ to point $C$ from west to east at a speed of 1.5 cm/s, and particle 乙 moves from point $B$ to point $D$ from south to north at a speed of 2.5 cm/s. If $A O=3$ cm, $B O=4$ cm, then after $\qquad$ seconds, the triangle formed by points $C$, $D$, and $O$ will be similar to $\triangle A O B$.
3. $\frac{14}{11}$ or $\frac{50}{29}$ or $\frac{16}{9}$.

As shown in Figure 5, let the particle 甲 reach position $C_{1}$ and particle 乙 reach position $D_{1}$ after $t$ seconds. At this time, $\triangle C_{1} O D_{1} \backsim \triangle B O A$, so $\frac{C_{1} O}{B O}=\frac{D_{1} O}{A O}$, which gives $\frac{3-1.5 t}{4} = \frac{4-2.5 t}{3} \Rightarrow t_{1}=\frac{14}{11}$.

If particle 甲 continues to move forward to position $C_{2}$ and particle 乙 continues to move forward to position $D_{2}$, at this time, $\triangle C_{2} O D_{2} \backsim \triangle B O A$, so $\frac{C_{2} O}{B O} = \frac{D_{2} O}{A O}$, which gives $\frac{3-1.5 t}{4} = \frac{2.5 t-4}{3} \Rightarrow t_{2}=\frac{50}{29}$.
Similarly, if $\triangle C_{3} O D_{3} \backsim \triangle A O B$, then $\frac{C_{3} O}{A O} = \frac{O D_{3}}{O B}$, which gives $\frac{3-1.5 t}{3} = \frac{2.5 t-4}{4} \Rightarrow t_{3}=\frac{16}{9}$.
If $\triangle C_{4} O D_{4} \backsim \triangle A O B$, then $\frac{O C_{4}}{O A} = \frac{O D_{4}}{O B}$, which gives $\frac{1.5 t-3}{3} = \frac{2.5 t-4}{4} \Rightarrow t=0$ (not valid, discard).