An aircraft flies at an altitude where the ambient atmospheric density is $\rho = 0.30625\text{ kg/m}^3$. Standard sea-level density is $\rho_0 = 1.225\text{ kg/m}^3$. If the equivalent airspeed indicated by flight instruments is $V_e = 120\text{ m/s}$, the true airspeed $V$ (in $\text{m/s}$) of the aircraft is:

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles Equivalent airspeed ($V_e$) and True airspeed ($V$) are related by dynamic pressure equivalence: $$\frac{1}{2} \rho_0 V_e^2 = \frac{1}{2} \rho V^2 \implies V = \frac{V_e}{\sqrt{\sigma}}$$ where $\sigma = \frac{\rho}{\rho_0}$ is the relative atmospheric density ratio. ### 2. Analytical Formulation & Calculation Given: - $\rho = 0.30625\text{ kg/m}^3$ - $\rho_0 = 1.225\text{ kg/m}^3$ $$\sigma = \frac{\rho}{\rho_0} = \frac{0.30625}{1.225} = \frac{1}{4} = 0.25$$ $$\sqrt{\sigma} = \sqrt{0.25} = 0.50$$ ### 3. Step-by-Step Calculation $$V = \frac{V_e}{\sqrt{\sigma}} = \frac{120}{0.50} = 240\text{ m/s}$$ ### 4. Distractor Forensics - Candidate multiplying by $\sqrt{\sigma}$ instead of dividing gets $120 \times 0.5 = 60\text{ m/s}$. ### 5. Exam Trap & Standard Reference - **Trap Alert:** True airspeed is always *higher* than equivalent airspeed at high altitudes ($V > V_e$). - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 4.

← Prev Question Next Question →