determine critical point for an engine failure scenario: TRIP DISTANCE 1080NM, GROUNDSPEED OUT 320KTS, REDUCED GROUNDSPEED OUT 295KTS, REDUCED GROUNDSPEED HOME 255KTS

• A. 540 NM
• B. 501 NM
• C. 724 NM
• D. 682 NM

Answer: (B)

In engine-out planning, the critical point is where the time to continue to the destination equals the time to return to the origin, given the reduced speeds on the outbound and return legs.

Let x be the distance from the origin at the moment of engine failure. Time to reach the destination from there is (1080 − x) / 295 hours, since the reduced outbound speed is 295 knots. Time to return to the origin is x / 255 hours, with the reduced return speed of 255 knots. Set them equal: (1080 − x)/295 = x/255. Solving gives 255(1080 − x) = 295x → 275,400 − 255x = 295x → 275,400 = 550x → x ≈ 500.7 NM, about 501 NM.

So the critical point is approximately 501 nautical miles from the origin along the outbound leg. The other speeds describe normal vs. reduced performance, but the calculation uses the reduced outbound and return speeds.