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An aircraft flies at a TAS of 380 kt. It flies from A to B and back to A. Distance AB is 480 NM. When going from A to B, it experiences a headwind componentof 60 kt. The wind remains constant. The duration of the flight will be
  • A
    3 h 00 min.
  • B
    2 h 32 min.
  • C
    2 h 35 min.
  • D
    2 h 10 min.

On the outbound leg (from A to B), the Ground Speed will be 320 kt (True Airspeed 380 kt with a 60 kt headwind). Covering a distance of 480 NM will take 1 hour 30 minutes (480 NM ÷ 320 kt).

Upon making a 180° turn and returning from B to A, the original 60 kt headwind transforms into a 60 kt tailwind. This results in a Ground Speed of 440 kt (True Airspeed 380 kt + 60 kt tailwind). Covering the same 480 NM at a Ground Speed of 440 kt will take 1 hour 05 minutes (480 NM ÷ 440 kt).

Therefore, the total flight time from A to B and back to A will be 2 hours 35 minutes.

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