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To solve this problem, we need to find the magnitude and angle (relative to east) of the train's **average velocity**. The train moves in three segments, and we need to determine the resultant displacement for the trip and divide by the total time.

### Given information:
1. Speed of the train: \( 56.0 \, \text{km/h} \)
2. Directions:
   - First: East (0° relative to east) for 35.0 min.
   - Second: 51.0° east of north for 23.0 min.
   - Third: West (180° relative to east) for 45.0 min.

### Converting time to hours (since speed is in km/h):
- \( t_1 = 35.0 \, \text{min} = \frac{35.0}{60} \, \text{hours} \)
- \( t_2 = 23.0 \, \text{min} = \frac{23.0}{60} \, \text{hours} \)
- \( t_3 = 45.0 \, \text{min} = \frac{45.0}{60} \, \text{hours} \)

### Step 1: Find the displacement vectors for each segment.

- **Displacement 1 (East):**
  \[
  \Delta \mathbf{r}_1 = 56.0 \times \frac{35.0}{60} \, \text{km} \quad \text{(along the x-axis)}
  \]

- **Displacement 2 (51° east of north):**
  This vector has both x and y components. Using trigonometry:
  \[
  \Delta \mathbf{r}_2 = 56.0 \times \frac{23.0}{60} \, \text{km}
  \]
  - x-component: \( \Delta x_2 = \Delta r_2 \times \sin(51.0^\circ) \)
  - y-component: \( \Delta y_2 = \Delta r_2 \times \cos(51.0^\circ) \)

- **Displacement 3 (West):**
  \[
  \Delta \mathbf{r}_3 = 56.0 \times \frac{45.0}{60} \, \text{km} \quad \text{(along the negative x-axis)}
  \]

### Step 2: Compute the total displacement vector by summing all components.

### Step 3: Calculate the average velocity by dividing the total displacement by the total time.

I'll perform the calculations now.The average velocity of the train during the trip is:

1. **Magnitude**: \( 8.96 \, \text{km/h} \)
2. **Angle (relative to east)**: \( 61.45^\circ \) north of east.

Would you like further details on the calculation or any other clarifications?

Here are some related questions to consider:
1. How would the result change if the speed of the train were different?
2. What if the second segment had a different angle relative to east?
3. How do you calculate average velocity for more complex paths?
4. Can you use this method to calculate displacement in 3D space?
5. How does this problem relate to real-world navigation systems?

**Tip**: To find an angle in a right triangle, you can use the inverse trigonometric functions like \(\tan^{-1}\), \(\sin^{-1}\), or \(\cos^{-1}\).

A train at a constant 56.0 km/h moves east for 35.0 min, then in a direction 51.0° east of due north for 23.0 min, and then west for 45.0 min. What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip?

Learn how to calculate the magnitude and direction of a train's average velocity as it moves east, in a direction 51° east of north, and west. Using vector addition, trigonometry, and the displacement formula, this solution breaks down the calculations step by step. Suitable for high school students in Grades 11-12.

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