Let's solve Part C by calculating the distance the plane will be from its intended position after 10 minutes if no corrective action is taken.

Given:

1. Wind velocity (from the southwest) = 110 km/h
2. Airplane's speed (due south) = 700 km/h
3. Time = 10 minutes = $\frac{10}{60}$ hours = $\frac{1}{6}$ hours

Approach:

1. The wind blows from the southwest, which means it has a component towards the east and a component towards the north. Since the wind is coming from 45 degrees to the south and west directions, we can resolve the wind's velocity into two perpendicular components:

   • $v_{\text{wind east}} = 110 \times \cos(45^{\circ})$
   • $v_{\text{wind north}} = 110 \times \sin(45^{\circ})$

2. The displacement caused by the wind in 10 minutes will be the distance the wind pushes the plane in that time:

   • $\Delta x = v_{\text{wind east}} \times t$
   • $\Delta y = v_{\text{wind north}} \times t$

3. Calculate the resultant displacement from the intended position:

   • $\Delta r = \sqrt{(\Delta x)^2 + (\Delta y)^2}$

Calculation:

Let's calculate this step-by-step.The plane will be approximately 18.3 km away from its intended position after 10 minutes if no corrective action is taken.

Would you like more details on the calculations, or do you have any other questions?

Here are 5 related questions you might find interesting:

1. How would the displacement change if the wind speed were doubled?
2. What if the wind came from the northwest instead of the southwest?
3. How would the airplane's heading need to change to counteract the wind?
4. How does the wind's angle affect the resulting displacement?
5. What would be the time to reach a specific displacement if the wind speed changed?

Tip: Remember to convert all time units consistently when solving physics problems!