Air France AF8968 Flight Status Today

Domestic flight Air France AF8968 takes off from New York (JFK) United States to Columbus (CMH) United States. It's operated by Air France. The plane leaves John F. Kennedy International Airport at 14:56 America/New_York. The flight is expected to land at John Glenn Columbus International Airport at 16:59 America/New_York. The flight will last about 1 hours 22 minutes.

AF8968 - AFR8968 Air France (AF)
JFK New York United States
Flight JFK to CMH
CMH Columbus United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 01-April-2020
Departure John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 14:56
Arrival John Glenn Columbus International Airport
IATA: CMHICAO: KCMH
Scheduled 16:59
C
AF8968 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 22 minutes
  • Flight Distance: 776 kms / 482 miles
Airline
  • Air France
  • IATA: AF
  • ICAO: AFR
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 50
  • Freight capacity: 1.5 tons
  • Passenger classes: First Class, Economy, Shuttle, Premium Economy
More Detail
  • Aircraft: E75
  • Callsign: AFR8968
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in New York: Friday 2026-03-20 18:37 PM
  • Current Time in Columbus: Friday 2026-03-20 18:37 PM

Flight routes similar to Air France AF8968

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British Airways BA6626
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11:59

Air France Flight AF8968 FAQs

What is the scheduled flight duration for Air France AF8968 flight?

On average, nonstop flight takes 1 hour(s) 22 minutes, with the flight distance of 776 km (482 miles).

What type of aircraft is used for the Air France AF8968 flight?

All Air France AF8968 flights are operated using E75 aircraft.

Which terminal the flight Air France AF8968 is arriving at?

Flight Air France AF8968 arrives in John Glenn Columbus International Airport at Terminal C.