https://file.aiquickdraw.com/mathbot/file/1730984537786cpydwhy9.jpg

The problem presented is a kinematics question involving two segments of a journey.

**Problem Summary:**
1. A person travels from home to a bus station by motorbike at 36 km/h, westward, covering the distance in 10 minutes.
2. From the bus station, the person then boards a bus traveling southward at 48 km/h, for 30 minutes.

**Questions:**
a. Calculate the total distance and the average speed of the person over the entire journey.
b. Determine the overall displacement and resultant velocity, including direction.

### Solution:

1. **First Leg (Motorbike):**
   - **Speed, \( v_1 \) = 36 km/h**
   - **Time, \( t_1 \) = 10 minutes = \(\frac{10}{60}\) hours = \(\frac{1}{6}\) hours**
   - **Distance, \( d_1 = v_1 \times t_1 = 36 \times \frac{1}{6} = 6 \text{ km} \)** (westward)

2. **Second Leg (Bus):**
   - **Speed, \( v_2 \) = 48 km/h**
   - **Time, \( t_2 \) = 30 minutes = \(\frac{30}{60}\) hours = 0.5 hours**
   - **Distance, \( d_2 = v_2 \times t_2 = 48 \times 0.5 = 24 \text{ km} \)** (southward)

3. **Total Distance:**
   - Total distance traveled \( = d_1 + d_2 = 6 + 24 = 30 \text{ km} \)

4. **Average Speed:**
   - Total time \( = t_1 + t_2 = \frac{1}{6} + 0.5 = \frac{2}{3} \text{ hours} \)
   - Average speed \( = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{30}{\frac{2}{3}} = 45 \text{ km/h} \)

5. **Displacement (Resultant Distance):**
   - Using the Pythagorean theorem to find the displacement since the travel path forms a right triangle (westward and then southward):
     \[
     \text{Displacement} = \sqrt{d_1^2 + d_2^2} = \sqrt{6^2 + 24^2} = \sqrt{36 + 576} = \sqrt{612} \approx 24.74 \text{ km}
     \]

6. **Resultant Velocity (Magnitude and Direction):**
   - **Magnitude of resultant velocity:**
     \[
     \text{Resultant Velocity} = \frac{\text{Displacement}}{\text{Total Time}} = \frac{24.74}{\frac{2}{3}} \approx 37.11 \text{ km/h}
     \]
   - **Direction of resultant displacement:**
     - Angle \( \theta \) from the westward direction (using \(\tan^{-1}\)):
       \[
       \theta = \tan^{-1} \left( \frac{d_2}{d_1} \right) = \tan^{-1} \left( \frac{24}{6} \right) = \tan^{-1}(4) \approx 75.96^\circ
       \]
       So, the direction is approximately \( 75.96^\circ \) south of west.

### Answers:
- **(a)** Total distance: 30 km, Average speed: 45 km/h.
- **(b)** Displacement: 24.74 km at \( 75.96^\circ \) south of west, Resultant velocity: 37.11 km/h.

Would you like further details on any specific step?

### Related Questions:
1. How would the average speed change if the bus traveled at a different speed?
2. What if the motorbike trip lasted 15 minutes instead of 10 minutes?
3. How does changing directions (e.g., northwest) affect the resultant velocity calculation?
4. What is the difference between displacement and total distance traveled in such journeys?
5. How would the result change if the person reversed directions halfway?

#### Tip:
For multi-leg journeys, remember that total distance considers the entire path, while displacement only measures the straight-line distance from start to end.

Một người đi xe máy từ nhà đến bến xe bus cách nhà với tốc độ 36 km/h về phía Tây, trong 10 phút. Đến bến xe, người đó lên xe bus đi tiếp phía Nam với tốc độ 48 km/h, trong 30 phút.

a. Tính quãng đường và tốc độ của người đó đi được trong cả chuyến đi?
b. Xác định độ dịch chuyển tổng hợp và vận tốc của người đó trong cả chuyến đi (hướng và độ lớn)?

Solve a kinematics problem involving a two-part journey where a person travels west by motorbike and then south by bus. This problem covers calculating total distance, average speed, displacement, and resultant velocity using vector addition and the Pythagorean theorem. Suitable for high school students.

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