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The problem provided is in Vietnamese and discusses a motorcycle's motion in different directions with varying speeds over a 6-minute time frame. The goal is to find the total displacement vector, the average velocity vector, and the average speed of the motorcycle, considering east as the positive x-direction. 

Let's break down the problem:

1. **Phase 1 (Southward motion)**:
   - Velocity: \( v_1 = 20 \, \text{m/s} \)
   - Time: \( t_1 = 3 \, \text{minutes} = 180 \, \text{seconds} \)
   - Direction: South (negative y-direction)
   - Displacement: \( d_1 = v_1 \times t_1 = 20 \times 180 = 3600 \, \text{m} \)

2. **Phase 2 (Westward motion)**:
   - Velocity: \( v_2 = 25 \, \text{m/s} \)
   - Time: \( t_2 = 2 \, \text{minutes} = 120 \, \text{seconds} \)
   - Direction: West (negative x-direction)
   - Displacement: \( d_2 = v_2 \times t_2 = 25 \times 120 = 3000 \, \text{m} \)

3. **Phase 3 (Northwestward motion)**:
   - Velocity: \( v_3 = 30 \, \text{m/s} \)
   - Time: \( t_3 = 1 \, \text{minute} = 60 \, \text{seconds} \)
   - Direction: Northwest (45° from the negative x and positive y)
   - Displacement: 
     - Along the x-axis: \( d_{3x} = v_3 \times t_3 \times \cos(45^\circ) = 30 \times 60 \times \frac{1}{\sqrt{2}} = 1273 \, \text{m} \) (negative x)
     - Along the y-axis: \( d_{3y} = v_3 \times t_3 \times \sin(45^\circ) = 30 \times 60 \times \frac{1}{\sqrt{2}} = 1273 \, \text{m} \) (positive y)

### Total Displacement:
The total displacement vector has components in both x and y directions:
- \( d_x = -3000 - 1273 = -4273 \, \text{m} \)
- \( d_y = -3600 + 1273 = -2327 \, \text{m} \)

Thus, the total displacement vector \( \vec{d} \) is:
\[
\vec{d} = (-4273 \, \text{m}, -2327 \, \text{m})
\]
The magnitude of the total displacement is:
\[
|\vec{d}| = \sqrt{(-4273)^2 + (-2327)^2} \approx 4874 \, \text{m}
\]

### Average Velocity:
The total time is \( t = 6 \, \text{minutes} = 360 \, \text{seconds} \). The average velocity vector \( \vec{v}_{avg} \) is:
\[
\vec{v}_{avg} = \frac{\vec{d}}{t} = \left(\frac{-4273}{360}, \frac{-2327}{360}\right) \, \text{m/s}
\]
\[
\vec{v}_{avg} = (-11.87 \, \text{m/s}, -6.46 \, \text{m/s})
\]

The magnitude of the average velocity is:
\[
|\vec{v}_{avg}| = \frac{4874}{360} \approx 13.54 \, \text{m/s}
\]

### Average Speed:
The total distance traveled is the sum of all the displacements:
\[
\text{Total distance} = 3600 + 3000 + 1800 = 8400 \, \text{m}
\]
The average speed is:
\[
\text{Average speed} = \frac{\text{Total distance}}{t} = \frac{8400}{360} \approx 23.33 \, \text{m/s}
\]

### Summary:
1. **Total displacement**: 4874 meters
2. **Average velocity**: \( (-11.87 \, \text{m/s}, -6.46 \, \text{m/s}) \), magnitude: 13.54 m/s
3. **Average speed**: 23.33 m/s

Would you like any further details or explanations?

Here are 5 related questions for practice:
1. How would the total displacement change if the motorcycle headed east in phase 2 instead of west?
2. What would be the effect on the average velocity if phase 3 was directed northeast instead of northwest?
3. How does the direction of motion influence the average speed calculation?
4. If the velocity in phase 1 was increased to 30 m/s, how would that impact the total displacement?
5. How can the displacement vector be visualized on a coordinate plane for clearer understanding?

**Tip**: Always pay attention to the directions of motion when calculating vectors, as they can drastically affect the final result.

Một tài xế mô-tô đi về phía nam với tốc độ 20 m/s mất 3 phút, sau đó vòng qua phía tây với tốc độ 25 m/s mất 2 phút và cuối cùng anh ta đi theo hướng tây bắc với tốc độ 30 m/s trong 1 phút. Trong 6 phút di chuyển trên, hãy tìm vector độ dời tổng hợp, tốc độ trung bình và vận tốc trung bình của xe. Chọn chiều dương trục x là hướng đông.

Solve a kinematics problem involving a motorcycle's motion in different directions. Calculate the total displacement vector, average velocity, and speed using vector analysis over a 6-minute period. This problem covers concepts like displacement, velocity, and the Pythagorean theorem, suitable for Grades 10-12.

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