A jet aircraft executes a steady, coordinated level turn at a true airspeed $V = 180\text{ m/s}$ with a bank angle $\phi = 45^\circ$. Taking $g = 9.81\text{ m/s}^2$, the rate of turn $\omega$ (in degrees per second, $^\circ/\text{s}$) is closest to:

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### 1. Strategy & Theoretical Principles In a coordinated, level turn with bank angle $\phi$: 1. Centripetal acceleration balance: $$g \tan \phi = \frac{V^2}{R} = V \omega$$ where $\omega = \frac{V}{R}$ is the turn rate in radians per second. 2. Turn rate in $\text{rad/s}$: $$\omega = \frac{g \tan \phi}{V}$$ 3. Convert to degrees per second: $$\omega (^{\circ}/\text{s}) = \omega (\text{rad/s}) \times \frac{180}{\pi}$$ ### 2. Analytical Formulation & Calculation Given: - $V = 180\text{ m/s}$ - $\phi = 45^\circ \implies \tan(45^\circ) = 1.0$ - $g = 9.81\text{ m/s}^2$ $$\omega = \frac{9.81 \times 1.0}{180} = \frac{9.81}{180} = 0.0545\text{ rad/s}$$ Convert to degrees per second: $$\omega (^{\circ}/\text{s}) = 0.0545 \times \frac{180}{\pi} = \frac{9.81}{\pi} \approx \frac{9.81}{3.14159} \approx 3.1226^{\circ}/\text{s} \approx 3.12^{\circ}/\text{s}$$ ### 3. Distractor Forensics - Candidate reporting in $\text{rad/s}$ gets $0.0545$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Always verify whether the question asks for turn rate in $\text{rad/s}$ or $^{\circ}/\text{s}$. - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 6.

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