Step 1: Use the fact that the time before meeting is the geometric mean of the two after-meeting times.
For this type of problem, the time taken to meet, \( t \), satisfies \( t = \sqrt{t_1 \times t_2} \), where \( t_1 = 4 \) hours and \( t_2 = 9 \) hours are the times taken after meeting. So \[ t = \sqrt{4 \times 9} = \sqrt{36} = 6 \text{ hours} \]
Step 2: Write the distance covered by each train up to the meeting point.
In 6 hours before meeting, train 1 covers \( 6v_1 \) and train 2 covers \( 6v_2 \), where \( v_1 \) and \( v_2 \) are their speeds.
Step 3: Use the after-meeting distances to link the speeds.
After meeting, train 1 covers the distance train 2 had already covered, \( 6v_2 \), in 4 hours, so \[ v_1 = \frac{6v_2}{4} = 1.5\, v_2 \]
Step 4: Write this as a ratio.
\[ \frac{v_1}{v_2} = 1.5 = \frac{3}{2} \] So the speed ratio is \[ \boxed{3:2} \]