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Let's break down the problem step-by-step.

### 1. Average Speed for the Whole Trip
**Average speed** is the total distance traveled divided by the total time taken. To calculate this:
- The trip involves several segments. Let's determine the total distance by summing the absolute values of displacement in each segment:
  - From \( t = 0 \) to \( t = 10 \) seconds (point A): 
    - Displacement = 100 m
  - From \( t = 10 \) to \( t = 20 \) seconds (point B): 
    - Additional displacement = 200 - 100 = 100 m
  - From \( t = 20 \) to \( t = 30 \) seconds (point C): 
    - Additional displacement = 0 - 200 = 200 m
  - From \( t = 30 \) to \( t = 60 \) seconds: 
    - Additional displacement = 200 m downward to -200 m (300 m in total).

  Total distance traveled = \( 100 + 100 + 200 + 300 = 700 \) meters.

- Total time = 60 seconds.

  Average speed = Total distance / Total time = \( \frac{700 \text{ m}}{60 \text{ s}} \approx 11.67 \) m/s.

### 2. Average Velocity for the Whole Trip
**Average velocity** is the total displacement divided by the total time taken. Here:
- Initial displacement = 0 m.
- Final displacement = -200 m (at \( t = 60 \) seconds).

  Total displacement = Final displacement - Initial displacement = \( -200 - 0 = -200 \) m.

- Total time = 60 seconds.

  Average velocity = Total displacement / Total time = \( \frac{-200 \text{ m}}{60 \text{ s}} \approx -3.33 \) m/s.

### 3. Instantaneous Velocity at Points A, B, and C
**Instantaneous velocity** at any point on a displacement-time graph is the slope of the tangent at that point.

- **Point A (t = 10 s)**:
  - The slope of the line from \( t = 0 \) to \( t = 10 \):
    - Velocity = \( \frac{\Delta d}{\Delta t} = \frac{100 \text{ m} - 0 \text{ m}}{10 \text{ s} - 0 \text{ s}} = 10 \) m/s.

- **Point B (t = 20 s)**:
  - The slope of the line from \( t = 10 \) to \( t = 20 \):
    - Velocity = \( \frac{200 \text{ m} - 100 \text{ m}}{20 \text{ s} - 10 \text{ s}} = \frac{100}{10} = 10 \) m/s.

- **Point C (t = 30 s)**:
  - The slope of the line from \( t = 20 \) to \( t = 30 \):
    - Velocity = \( \frac{0 \text{ m} - 200 \text{ m}}{30 \text{ s} - 20 \text{ s}} = \frac{-200}{10} = -20 \) m/s.

### Summary of the Answers:
1. **Average Speed**: Approximately \( 11.67 \) m/s.
2. **Average Velocity**: Approximately \( -3.33 \) m/s.
3. **Instantaneous Velocities**:
   - At **A**: 10 m/s
   - At **B**: 10 m/s
   - At **C**: -20 m/s

Would you like further details on any of these calculations or the concepts involved?

### Follow-up Questions:
1. How does the direction of travel affect the average velocity but not the average speed?
2. Why is average velocity a vector quantity while average speed is scalar?
3. Can you explain the significance of instantaneous velocity at a specific point?
4. How would the average velocity change if the messenger returned to the starting point?
5. How can you identify if the object is accelerating or decelerating from the graph?

### Tip:
Instantaneous velocity can be thought of as the object's speed and direction at any given moment, which is why it corresponds to the slope of the tangent line on a displacement-time graph.

A messenger started to bike westward along a road near his office. A displacement vs. time graph of his trip is shown. Find the (a) average speed for the whole trip, (b) average velocity for the whole trip, and (c) instantaneous velocity at points A, B, and C.

Analyze a messenger's trip using a displacement vs. time graph. Calculate average speed, average velocity, and instantaneous velocity at key points (A, B, C). Explore concepts in kinematics such as displacement and velocity, ideal for students in Grades 9-12.

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