FAA Commercial Pilot Airplane (CAX)Aerodynamics and PerformanceEasy
A pilot departs with a gross weight of 3,000 pounds and a stall speed of 60 KCAS. After burning off fuel during the flight, the airplane's weight decreases to 2,700 pounds. Assuming all other factors remain constant, what is the new stall speed?
- A54.0 KCAS
- B60.0 KCAS
- C56.9 KCAS
- D63.2 KCAS
Show answer & explanationAnswer & explanation
Correct answer: C. 56.9 KCAS
Stall speed varies with the square root of the weight ratio: Vs2 = Vs1 × √(W2/W1) = 60 × √(2,700/3,000) = 60 × √0.9 = 60 × 0.9487 ≈ 56.9 KCAS. As weight decreases, stall speed decreases proportionally to the square root of the weight ratio, not linearly.
Why the other options are wrong
- A. Too low; would require a much greater weight reduction than actually occurred.
- B. Assumes no change in stall speed, ignoring the weight reduction.
- D. Incorrectly shows an increase in stall speed when weight decreased.
Stall Speed vs. Weight
Stall speed changes with the square root of the ratio of new weight to original weight, since lift required to sustain flight is proportional to weight.
- Formula: Vs2 = Vs1 × √(W2/W1)
- Lighter weight = lower stall speed
- Relationship is square-root, not linear
Memory trick: Lighter plane, lazier stall — square root of the scale.