To determine the average velocity of the train during the trip, we need to follow these steps:

(1) Convert time intervals to hours

Since the speed is given in km/h, we must express the time intervals in hours:

• 28.0 minutes = $\frac{28.0}{60} = 0.467 \, \text{hours}$
• 24.0 minutes = $\frac{24.0}{60} = 0.400 \, \text{hours}$
• 38.0 minutes = $\frac{38.0}{60} = 0.633 \, \text{hours}$

(2) Calculate the displacement in each segment

The train moves at a constant speed of 42.0 km/h, so we can calculate the displacement for each leg of the journey using $\text{displacement} = \text{speed} \times \text{time}$.

Segment 1: Eastward motion

• Displacement = $42.0 \, \text{km/h} \times 0.467 \, \text{hours} = 19.614 \, \text{km}$
• Direction = east

Segment 2: 54.0° east of north

• Displacement = $42.0 \, \text{km/h} \times 0.400 \, \text{hours} = 16.8 \, \text{km}$
• Direction = 54.0° east of north
  • We can break this into components:
    • $x_2 = 16.8 \times \cos(54.0^\circ) = 9.88 \, \text{km}$ (eastward component)
    • $y_2 = 16.8 \times \sin(54.0^\circ) = 13.57 \, \text{km}$ (northward component)

Segment 3: Westward motion

• Displacement = $42.0 \, \text{km/h} \times 0.633 \, \text{hours} = 26.586 \, \text{km}$
• Direction = west (opposite to east)

(3) Calculate net displacement components

Now we calculate the total displacement in the east-west (x-axis) and north-south (y-axis) directions by summing the components.

• Total east-west displacement: $x_{\text{total}} = 19.614 \, \text{km} + 9.88 \, \text{km} - 26.586 \, \text{km} = 2.908 \, \text{km}$ (positive means eastward)

• Total north-south displacement: $y_{\text{total}} = 0 \, \text{km} + 13.57 \, \text{km} = 13.57 \, \text{km}$ (positive means northward)

(4) Calculate the magnitude of the total displacement

The magnitude of the total displacement is the length of the vector formed by these components: $\text{magnitude} = \sqrt{(x_{\text{total}})^2 + (y_{\text{total}})^2} = \sqrt{(2.908)^2 + (13.57)^2} = 13.88 \, \text{km}$

(5) Calculate the total time

The total time for the trip is the sum of the time intervals: $t_{\text{total}} = 0.467 \, \text{hours} + 0.400 \, \text{hours} + 0.633 \, \text{hours} = 1.500 \, \text{hours}$

(6) Calculate the average velocity

The magnitude of the average velocity is: $v_{\text{avg}} = \frac{\text{total displacement}}{\text{total time}} = \frac{13.88 \, \text{km}}{1.500 \, \text{hours}} = 9.25 \, \text{km/h}$

(7) Calculate the direction of the average velocity

The direction (relative to east) is given by the angle of the displacement vector: $\theta = \tan^{-1} \left( \frac{y_{\text{total}}}{x_{\text{total}}} \right) = \tan^{-1} \left( \frac{13.57}{2.908} \right) = 77.9^\circ \, \text{north of east}$

Final answers:

• (a) The magnitude of the average velocity is 9.25 km/h.
• (b) The direction of the average velocity is 77.9° north of east.

Would you like more details on any of the steps, or do you have any questions?

Here are some related questions:

1. How do we calculate displacement when a path includes angles?
2. What is the difference between speed and velocity?
3. Why do we use components to solve vector problems?
4. How can trigonometry help in calculating vector components?
5. How would the average velocity change if the train moved faster?

Tip: When solving vector problems, always break vectors into components to simplify calculations.