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To solve the problem, we need to model the motion of both cars mathematically and determine when and where they will meet.

### Key Information:
- Car A passes point X at 10:00, traveling to the right at 10 m/s.
- Car B passes point Y (3 km or 3000 m apart from X) at 10:02, traveling to the left at 15 m/s.
- We are to find when and where they pass each other.

### Let’s set up a coordinate system:
- Position of Car A at time \( t \) (after 10:00) can be represented as:
  \[
  \text{Position of Car A} = 10t \text{ meters} \quad (t \text{ in seconds, starting at } 10:00)
  \]
- Car B starts 120 seconds later (since 10:02 is 120 seconds after 10:00). So, the position of Car B at time \( t \) (after 10:00) can be written as:
  \[
  \text{Position of Car B} = 3000 - 15(t - 120) \text{ meters} \quad \text{(since it's moving left, hence the negative velocity)}
  \]
  
### To find the time when they meet, set the positions equal to each other:
\[
10t = 3000 - 15(t - 120)
\]

### Solve this equation:
1. Expand the terms on the right-hand side:
   \[
   10t = 3000 - 15t + 1800
   \]
   \[
   10t = 4800 - 15t
   \]

2. Combine like terms:
   \[
   10t + 15t = 4800
   \]
   \[
   25t = 4800
   \]

3. Solve for \( t \):
   \[
   t = \frac{4800}{25} = 192 \text{ seconds}
   \]

So, **192 seconds** after 10:00 is when the cars meet. Converting this to minutes, they meet at **10:03:12** (192 seconds is 3 minutes and 12 seconds).

### Position where they meet:
Substitute \( t = 192 \) into Car A’s position equation:
\[
\text{Position of Car A} = 10 \times 192 = 1920 \text{ meters}
\]
Thus, they will meet **1920 meters to the right of point X**.

---

The equation could not be written as \( t + 120 \) because Car B starts moving at 10:02, 120 seconds later than Car A. In the equation, \( t - 120 \) accounts for that delay, meaning Car B is only moving after the 120-second mark, hence adjusting the time accordingly.

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

### Related Questions:
1. What would happen if both cars started at the same time?
2. How would the result change if Car B moved faster than Car A?
3. Can you derive the meeting time if the distance between the cars was 5 km instead of 3 km?
4. How do we model the scenario if Car B started before Car A?
5. What if Car A accelerated after passing point X?

#### Tip:
Always check the time references carefully when modeling delayed motion in physics problems!

Calculate the time when and the position where Car A and Car B will pass each other. Round the time off to the nearest minute and state the position in metres to the right of X.

Learn how to calculate the time and position where Car A and Car B, moving at different speeds, pass each other on a straight track. This problem involves kinematics and relative motion, with a step-by-step solution using linear equations. Suitable for high school students in Grades 10-12.

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