To calculate the average velocity for the girl's walk to school, we need to find the displacement vector and the total time.

Step 1: Displacement Vector

Displacement is the straight-line distance from the starting point to the ending point, considering the direction.

1. Northward Displacement:
   The girl walks 1 km North (1000 m).

2. Southwest Displacement:
   The girl walks 200 m Southwest. Southwest means $45^\circ$ between South and West. The components of this displacement are:

   • South: $200 \cos(45^\circ) = \frac{200}{\sqrt{2}} = 141.42 \, \text{m}$
   • West: $200 \sin(45^\circ) = \frac{200}{\sqrt{2}} = 141.42 \, \text{m}$

   Combining these, the overall displacement is:

   • $\Delta x = -141.42 \, \text{m}$ (West)
   • $\Delta y = 1000 - 141.42 = 858.58 \, \text{m}$ (North)

   The resultant displacement vector is: $\text{Resultant Displacement} = \sqrt{\Delta x^2 + \Delta y^2} = \sqrt{(-141.42)^2 + (858.58)^2}$ $= \sqrt{20000 + 737204.7364} \approx 870.85 \, \text{m}$

3. Direction of Displacement (Optional):
   The angle $\theta$ of the resultant displacement is: $\tan^{-1}\left(\frac{\Delta y}{-\Delta x}\right) = \tan^{-1}\left(\frac{858.58}{-141.42}\right) \approx -80.74^\circ$ So, the direction is slightly west of north.

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Step 2: Total Time

1. First Segment:
   The girl walks 1 km in 15 minutes: $t_1 = 15 \times 60 = 900 \, \text{seconds}$

2. Second Segment:
   The girl walks 200 m Southwest in 160 seconds: $t_2 = 160 \, \text{seconds}$

   Total time is: $t_{\text{total}} = t_1 + t_2 = 900 + 160 = 1060 \, \text{seconds}$

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Step 3: Average Velocity

Average velocity is the total displacement divided by the total time: $\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$ $\text{Average Velocity} = \frac{870.85}{1060} \approx 0.821 \, \text{m/s}$

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Final Answer:

The girl's average velocity is approximately: $\boxed{0.82 \, \text{m/s}} \, \text{towards } -80.74^\circ \, \text{(slightly west of north)}.$

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Would you like me to provide more details or clarify any part of this calculation?

Related Questions:

1. What is the difference between speed and velocity in this scenario?
2. How would the average velocity change if the southwest displacement were longer?
3. How does direction affect the calculation of displacement versus distance?
4. What would her average speed be for the trip?
5. What is the significance of components when calculating displacement in 2D motion?

Tip:

When working with displacement vectors, always resolve components accurately and use Pythagoras’ theorem to find the resultant.