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The problem states that a train moves at a constant speed of 61.0 km/h with the following movements:
- **First**: East for 41.0 minutes,
- **Second**: 49.0° east of due north for 23.0 minutes,
- **Third**: West for 51.0 minutes.

We are asked to determine:
1. The **magnitude** of the average velocity.
2. The **angle** (relative to east) of the average velocity.

### Steps to Solve:

1. **Convert time to hours**:
   - \( t_1 = 41.0 \, \text{min} = \frac{41.0}{60} \, \text{hours} \)
   - \( t_2 = 23.0 \, \text{min} = \frac{23.0}{60} \, \text{hours} \)
   - \( t_3 = 51.0 \, \text{min} = \frac{51.0}{60} \, \text{hours} \)

2. **Determine distances traveled in each segment**:
   - Distance in the first segment (eastward): \( d_1 = 61.0 \times t_1 \)
   - Distance in the second segment (at 49.0° east of due north): \( d_2 = 61.0 \times t_2 \)
   - Distance in the third segment (westward): \( d_3 = 61.0 \times t_3 \)

3. **Resolve the second segment into components**:
   - Eastward component: \( d_{2x} = d_2 \times \sin(49.0^\circ) \)
   - Northward component: \( d_{2y} = d_2 \times \cos(49.0^\circ) \)

4. **Calculate the total displacement**:
   - Total eastward displacement: \( D_x = d_1 - d_3 + d_{2x} \)
   - Total northward displacement: \( D_y = d_{2y} \)

5. **Find the magnitude and direction of the average velocity**:
   - Magnitude: \( D = \sqrt{D_x^2 + D_y^2} \)
   - Angle relative to east: \( \theta = \tan^{-1}\left(\frac{D_y}{D_x}\right) \)

6. **Average velocity**:
   - Average velocity magnitude: \( v_{avg} = \frac{D}{t_{\text{total}}} \), where \( t_{\text{total}} = t_1 + t_2 + t_3 \)

Let's calculate this now.### Results:
1. The **magnitude** of the average velocity during the trip is approximately **8.90 km/h**.
2. The **angle** of the average velocity relative to east is approximately **64.00°** north of east.

Would you like any further clarifications or additional details on these calculations?

Here are 5 related questions for further exploration:
1. How would the average velocity change if the speed of the train was increased?
2. How can you calculate the displacement if the angles of movement change?
3. How would the total displacement change if the train moved west first and then east?
4. What would the outcome be if the time spent moving in each direction was different?
5. How is the average speed different from average velocity in this context?

**Tip:** When breaking a vector into components, always pay attention to the direction and angles involved, especially when resolving them into horizontal and vertical components.

A train at a constant 61.0 km/h moves east for 41.0 min, then in a direction 49.0° east of due north for 23.0 min, and then west for 51.0 min. What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip?

Learn how to calculate the average velocity of a train moving in multiple directions at a constant speed. This problem involves vector addition, trigonometry, and kinematics. It covers how to resolve vectors into components and use trigonometric functions to find the resultant magnitude and direction.

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