Discussion
A flight path vector is drawn 4 degrees below the horizon at 150 knots true airspeed. The rate of descent is approximately
Answer: B
Correct answer: 1060 feet per minute
Convert the speed to feet per minute and take the vertical component:
\[ 150\ \mathrm{kt} = 150 \times \frac{6076}{60} = 15\,190\ \mathrm{ft/min} \]
\[ ROD = 15\,190 \times \sin 4^\circ = 15\,190 \times 0.0698 = 1060\ \mathrm{ft/min} \]
This is why the flight path vector is so useful: its position relative to the horizon is the flight path angle, so the pilot reads the actual trajectory directly instead of inferring it from pitch attitude and vertical speed separately.
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