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.