A train leaves union station for central station, 216km away, at 9am.  One hour later, a train leaves central station for union station.  They meet at noon.  If the second train had started at 9am and the first train at 10:30am, they would still have met at noon.  Find the speed of each train.

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?

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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.

A train leaves union station for central station, 216km away, at 9am. One hour later, a train leaves central station for union station. They meet at noon. If the second train had started at 9am and the first train at 10:30am, they would still have met at noon. Find the speed of each train.

Learn how to solve a distance-speed-time problem involving two trains traveling between Union and Central Station, with detailed step-by-step explanations. This example focuses on solving a system of linear equations and understanding the relationships between time, speed, and distance. Suitable for Grades 8-10 students.

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