Question 1210546
To determine which plane returns first, we need to calculate the total travel time for each plane, accounting for how the wind affects their ground speed.

### Plane B: Traveling East and West (The "Wind-Aligned" Case)

Plane B travels to point Q (100 miles East) and back. The 40 mi/hr wind is blowing from the West (moving East).

* **Outbound (Eastward):** The wind is a tailwind.
* Ground Speed = 
* Time = 


* **Return (Westward):** The wind is a headwind.
* Ground Speed = 
* Time = 


* **Total Time for B:** 

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### Plane A: Traveling North and South (The "Crosswind" Case)

Plane A travels to point P (100 miles North) and back. To maintain a straight North-South path, the plane must angle itself into the 40 mi/hr westerly wind. This creates a right triangle where the airspeed is the hypotenuse and the wind is one side.

* **Calculating Ground Speed ():**
Using the Pythagorean theorem:




* **Total Time for A:** Since the ground speed is the same for both the Northbound and Southbound legs (the wind is a constant crosswind):
* Total Time = 



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### Conclusion

Comparing the results:

* **Plane A:** ~2.18 hours
* **Plane B:** ~2.38 hours

**Plane A (traveling North/South) returns to O first.** This is a classic physics result: a crosswind slows a plane down less than a headwind/tailwind combination of the same magnitude over a round trip.

Would you like me to show you the mathematical proof for why the crosswind is always faster, regardless of the specific wind speed?