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To solve this problem, let's analyze the movement of both cars and set up equations to determine when and where they will meet.

### Given Information:
1. **Distance between X and Y**: \(3 \text{ km} = 3000 \text{ m}\).
2. **Car A**:
   - Speed: \(10 \text{ m/s}\) (moving to the right from point X).
   - Starts at \(10:00\).
3. **Car B**:
   - Speed: \(15 \text{ m/s}\) (moving to the left from point Y).
   - Starts at \(10:02\).

### Steps:

1. **Define Variables:**
   - Let \(t\) be the time in seconds after \(10:00\) when Car A and Car B will meet.

2. **Position Equations:**
   - **Car A's Position** (relative to X): Since Car A starts at X and moves right, its position after \(t\) seconds is:
     \[
     \text{Position of Car A} = 10 \times t
     \]
   - **Car B's Position** (relative to X): Car B starts from Y, which is 3000 m away from X, and moves left. Since Car B starts 2 minutes (120 seconds) after Car A, the time \(t\) for Car B will be \(t - 120\) seconds. Its position is:
     \[
     \text{Position of Car B} = 3000 - 15 \times (t - 120)
     \]

3. **Set Up Equation for Meeting Point:**
   - At the meeting point, the positions of Car A and Car B will be equal:
     \[
     10t = 3000 - 15(t - 120)
     \]

4. **Solve for \(t\):**
   Expanding and simplifying the equation:
   \[
   10t = 3000 - 15t + 1800
   \]
   \[
   25t = 4800
   \]
   \[
   t = 192 \text{ seconds}
   \]

5. **Calculate the Meeting Position:**
   Substitute \(t = 192\) back into Car A’s position equation:
   \[
   \text{Position} = 10 \times 192 = 1920 \text{ m (to the right of X)}
   \]

6. **Convert Time to Minutes:**
   Since Car A started at 10:00, 192 seconds is approximately \(3 \text{ minutes and 12 seconds}\). Thus, the meeting time is around **10:03**.

### Answer:
- **Time of meeting**: 10:03.
- **Position of meeting**: 1920 meters to the right of X.

Would you like more details on this solution, or do you have any questions?

### Related Questions:
1. What would be the meeting position if Car A were traveling at 20 m/s?
2. How would the answer change if Car B started at 10:01 instead of 10:02?
3. What is the impact of both cars moving in the same direction on the meeting point?
4. How much time would it take for Car A alone to reach point Y?
5. If Car A had an acceleration instead of a constant speed, how would we solve the problem?

### Tip:
Always ensure units are consistent across all calculations to avoid errors.

Calculate the time and position where Car A and Car B will pass each other, given that Car A passes point X at 10:00 traveling to the right at 10 m/s and Car B passes point Y at 10:02 traveling to the left at 15 m/s. Points X and Y are 3 km apart.

Determine the time and position where two cars moving towards each other on a straight track will meet. Car A travels at 10 m/s and Car B at 15 m/s. Points X and Y are 3 km apart, with Car A starting at 10:00 and Car B at 10:02. This problem involves concepts of linear motion and algebra, suitable for Grades 9-12.

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