We can solve this problem by setting up equations based on the distances traveled by both trains.

Step 1: Define the Variables

Let:

• $v_1$ be the speed of the first train (leaving Union Station).
• $v_2$ be the speed of the second train (leaving Central Station).
• The total distance between Union Station and Central Station is $216 \, \text{km}$.

Step 2: Scenario 1 - Actual Departure Times

In the actual scenario, the first train departs at 9:00 AM from Union Station, and the second train departs at 10:00 AM from Central Station. They meet at 12:00 PM.

Time Traveled by the Trains:

• The first train travels for 3 hours (from 9:00 AM to 12:00 PM).
• The second train travels for 2 hours (from 10:00 AM to 12:00 PM).

Distances Traveled:

• The first train travels a distance of $3v_1$.
• The second train travels a distance of $2v_2$.

Since they meet, the sum of the distances traveled by both trains must equal the total distance between the stations: $3v_1 + 2v_2 = 216 \, \text{km}$

Step 3: Scenario 2 - Hypothetical Departure Times

In the second scenario, the second train starts at 9:00 AM, and the first train starts at 10:30 AM. They still meet at noon.

Time Traveled by the Trains:

• The second train travels for 3 hours (from 9:00 AM to 12:00 PM).
• The first train travels for 1.5 hours (from 10:30 AM to 12:00 PM).

Distances Traveled:

• The second train travels a distance of $3v_2$.
• The first train travels a distance of $1.5v_1$.

Again, since they meet, the sum of the distances traveled must be 216 km: $1.5v_1 + 3v_2 = 216 \, \text{km}$

Step 4: Solve the System of Equations

Now, we have two equations:

1. $3v_1 + 2v_2 = 216$
2. $1.5v_1 + 3v_2 = 216$

We can solve this system of equations to find $v_1$ and $v_2$.

First, let's solve the first equation for $v_1$: $v_1 = \frac{216 - 2v_2}{3}$

Now substitute this into the second equation: $1.5 \left(\frac{216 - 2v_2}{3}\right) + 3v_2 = 216$ Simplify the equation: $0.5(216 - 2v_2) + 3v_2 = 216$ $108 - v_2 + 3v_2 = 216$ $2v_2 = 108$ $v_2 = 54 \, \text{km/h}$

Now substitute $v_2 = 54$ into the first equation: $3v_1 + 2(54) = 216$ $3v_1 + 108 = 216$ $3v_1 = 108$ $v_1 = 36 \, \text{km/h}$

Final Answer:

• The speed of the first train is $36 \, \text{km/h}$.
• The speed of the second train is $54 \, \text{km/h}$.

Would you like further details or clarification?

---

Here are 5 related questions to expand your understanding:

1. How would the meeting time change if both trains started at 9:30 AM?
2. What if the distance between the stations was 300 km? How would the speeds change?
3. What would happen if one train traveled 10 km/h faster than the other?
4. Could you find the time they meet if both trains had the same speed?
5. What if a third train left 1 hour later at a different speed—where would it meet the others?

Tip: In distance-speed-time problems, always break down each part into distance and time traveled to set up a system of equations.