Discussion
An aeroplane's roll damping gives \( L_p = -1.6 \ \mathrm{s^{-1}} \). The time for the roll rate to fall to half its initial value after the ailerons are centred is
Answer: A
The mode is first order with time constant \( \tau = 1/1.6 = 0.625 \ \mathrm{s} \), and the half-amplitude time is
\[ \begin{aligned} t_{1/2} &= 0.693\,\tau \\ &= 0.693 \times 0.625 = 0.43 \ \mathrm{s} \end{aligned} \]
so a little over four tenths of a second. The 0.63 s option is the time constant itself, which is the 63 per cent time rather than the 50 per cent one. 1.60 s is the derivative read back as a time. 0.87 s doubles the correct answer, which would follow from using \( 2\ln 2 \) in the numerator.
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