Consider an airplane in an accelerated climb. At a given instant in this climb, the specific excess power is 120 ft/s
Consider an airplane in an accelerated climb. At a given instant in this climb, the specific excess power is 120 ft/s, the instantaneous velocity is 500 ft/s, and the instantaneous rate of climb is 3,000 ft/min. Calculate the instantaneous acceleration.
\[\frac{dh}{dt} = 3,000 ft/min = \frac{3000}{60} = 50\,ft/s\]\[\frac{dh}{dt} = P_{s} – \frac{V_{\infty }}{g}A\]\[\Rightarrow A = -\frac{\left ( 50 – 120 \right )32.2}{500} = 4.508 \,ft/s^{2}\]
