Let's analyze the graphs in the context of the given problem again to determine if Graph 1 or Graph 2 is correct:

### Summary of the Situation:
- **Initial Velocity**: -24.0 m/s (south, taken as negative).
- **Final Velocity**: -6.0 m/s (south, taken as negative).
- **Duration of Deceleration**: 9.0 seconds.
- **Time Interval**: from \(t = -2\) to \(t = 10\) seconds.

### Graph Analysis for Graph 1 and Graph 2:

1. **Graph 1**:
   - This graph shows an initial constant velocity (horizontal line) before \(t = 0\) and after \(t = 9\), indicating two periods of constant velocity.
   - The slope changes between \(t = 0\) and \(t = 9\), suggesting constant acceleration.
   - The graph starts at approximately -24.0 m/s at \(t = 0\) and rises to approximately -6.0 m/s at \(t = 9\).
   - This graph appears to follow the exact change from -24.0 m/s to -6.0 m/s over 9 seconds and a constant velocity period before \(t = 0\) and after \(t = 9\).

2. **Graph 2**:
   - Graph 2 also shows an initial constant velocity but with different magnitudes and timings.
   - This graph suggests that the velocity before \(t = 0\) is different from the expected -24.0 m/s and the slope change occurs after \(t = 0\) as it should.
   - This graph doesn't seem to match the expected linear deceleration from -24.0 m/s to -6.0 m/s as clearly as Graph 1.

### Conclusion:

**Graph 1** accurately represents the situation:
- It shows a constant velocity of -24.0 m/s before the brakes are applied (from \(t = -2\) to \(t = 0\)).
- It shows a linear increase in velocity (deceleration) from -24.0 m/s to -6.0 m/s from \(t = 0\) to \(t = 9\).
- It shows a constant velocity of -6.0 m/s after \(t = 9\).

**Graph 1** is the correct graph for the given scenario.

### Do you have any other questions, or would you like further clarification?

### Related Questions:
1. How can we calculate the displacement of the train using Graph 1?
2. How would the graph change if the final speed were different (e.g., -12.0 m/s)?
3. What would happen to the graph if the acceleration were not constant?
4. Can we determine the total distance traveled by the train from the graph?
5. What does the area under the velocity-time graph represent in this context?

### Tip:
When dealing with velocity-time graphs, the slope of the graph indicates acceleration, and the area under the graph represents displacement.

Explore the analysis of velocity-time graphs to determine whether Graph 1 or Graph 2 correctly represents the given scenario. Detailed comparison of initial velocity, acceleration, and constant velocity periods from -24.0 m/s to -6.0 m/s over 9 seconds. Suitable for high school students studying kinematics and velocity-time graphs.

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