The problem describes a train moving at a constant speed of 49.0 km/h through three segments:

1. First segment: 30.0 minutes east.
2. Second segment: 27.0 minutes at a direction 51.0° east of due north.
3. Third segment: 40.0 minutes west.

We need to find:

• (a) The magnitude of its average velocity.
• (b) The angle of its average velocity (relative to east).

Step 1: Convert the time from minutes to hours

Since the speed is in km/h, convert the times into hours:

• 30.0 min = $\frac{30.0}{60}$ hours = 0.5 hours.
• 27.0 min = $\frac{27.0}{60}$ hours = 0.45 hours.
• 40.0 min = $\frac{40.0}{60}$ hours = 0.6667 hours.

Step 2: Calculate the distances traveled in each segment

Using the formula: $\text{distance} = \text{speed} \times \text{time}$ The distances are:

• East (Segment 1): $d_1 = 49.0 \times 0.5 = 24.5 \, \text{km}$ east.
• Direction 51.0° east of north (Segment 2): $d_2 = 49.0 \times 0.45 = 22.05 \, \text{km}$ at 51.0° east of north.
• West (Segment 3): $d_3 = 49.0 \times 0.6667 = 32.67 \, \text{km}$ west.

Step 3: Resolve the distances into components

First segment (East):

• $x_1 = 24.5 \, \text{km}$ (East).
• $y_1 = 0 \, \text{km}$ (North).

Second segment (51° east of north):

• $x_2 = 22.05 \times \sin(51.0^\circ) = 22.05 \times 0.777 = 17.13 \, \text{km}$ (East component).
• $y_2 = 22.05 \times \cos(51.0^\circ) = 22.05 \times 0.629 = 13.87 \, \text{km}$ (North component).

Third segment (West):

• $x_3 = -32.67 \, \text{km}$ (West).
• $y_3 = 0 \, \text{km}$ (North).

Step 4: Calculate the total displacement

Now sum up the components:

• Total $x$-component: $x_{\text{total}} = 24.5 + 17.13 - 32.67 = 8.96 \, \text{km} \, \text{(East)}.$

• Total $y$-component: $y_{\text{total}} = 0 + 13.87 + 0 = 13.87 \, \text{km} \, \text{(North)}.$

Step 5: Find the magnitude of the displacement

Using the Pythagorean theorem: $\text{magnitude of displacement} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} = \sqrt{(8.96)^2 + (13.87)^2} = \sqrt{80.3 + 192.4} = \sqrt{272.7} = 16.51 \, \text{km}.$

Step 6: Find the average velocity

The total time of travel: $t_{\text{total}} = 0.5 + 0.45 + 0.6667 = 1.6167 \, \text{hours}.$

The magnitude of the average velocity is: $v_{\text{avg}} = \frac{\text{total displacement}}{\text{total time}} = \frac{16.51}{1.6167} = 10.21 \, \text{km/h}.$

Step 7: Find the direction of the average velocity

The angle $\theta$ relative to east is given by: $\theta = \tan^{-1}\left( \frac{y_{\text{total}}}{x_{\text{total}}} \right) = \tan^{-1}\left( \frac{13.87}{8.96} \right) = \tan^{-1}(1.548) = 57.36^\circ.$

So, the angle of the average velocity is approximately 57.4° north of east.

Final Answers:

• (a) Magnitude of the average velocity: 10.21 km/h.
• (b) Angle of the average velocity: 57.4° north of east.

Do you want more details or have any questions about these steps?
Here are 5 related questions you can explore:

1. How would the result change if the speed of the train were different?
2. What is the significance of resolving vectors into components?
3. How does the concept of displacement differ from distance?
4. Could we solve this problem using polar coordinates directly?
5. How does average velocity differ from average speed?

Tip: When solving vector problems, always make sure to break them into components to simplify calculations.