3. The law of motion of the first tourist: $S=\sqrt{1+6 t}-1$, and the second tourist: $-S=$ $6\left(t-\frac{1}{6}\right)$ for $t \geq \frac{1}{6} ; S=0$ for $t<\frac{1}{6}$. The required condition is met if and only if one of the tourists is on one side of the road sign, and the other is on the other side (one or the other may also be at the boundary point - on the line of the sign). Since the first tourist reaches the sign at time $t_{1}: \sqrt{1+6 t_{1}}-1=2$, i.e., at $t_{1}=\frac{4}{3}$, and the second tourist reaches the sign at time $t_{2}: 6\left(t_{2}-\frac{1}{6}\right)=2$, i.e., at $t_{2}=\frac{1}{2}$, the suitable moments of time are $t \in\left[\frac{1}{2}, \frac{4}{3}\right]$, i.e., from half an hour to 1 hour 20 minutes after the first tourist's departure.
Answer: from half an hour to 1 hour and 20 minutes after the first tourist exits.