Aeromexico AM4968 Flight Status Today

Domestic flight Aeromexico AM4968 takes off from New York (JFK) United States to Syracuse (SYR) United States. It's operated by Aeromexico. The plane leaves John F. Kennedy International Airport at 20:13 America/New_York. The flight is expected to land at Hancock International Airport at 21:38 America/New_York. The flight will last about 1 hours 18 minutes.

AM4968 - AMX4968 Aeromexico (AM)
JFK New York United States
Flight JFK to SYR
SYR Syracuse United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 02-March-2021
Departure John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 20:13
4 B55
Arrival Hancock International Airport
IATA: SYRICAO: KSYR
Scheduled 21:38
B 23
AM4968 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 18 minutes
  • Flight Distance: 336 kms / 208 miles
Airline
  • Aeromexico
  • IATA: AM
  • ICAO: AMX
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 69
  • Freight capacity: 1.9 tons
  • Passenger classes: Economy, Shuttle, Premium Economy
More Detail
  • Aircraft: Canadair (Bombardier) Regional Jet 700 and Challenger 870
  • Callsign: AMX4968
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in New York: Sunday 2026-05-03 21:34 PM
  • Current Time in Syracuse: Sunday 2026-05-03 21:34 PM

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Aeromexico Flight AM4968 FAQs

What is the scheduled flight duration for Aeromexico AM4968 flight?

On average, nonstop flight takes 1 hour(s) 18 minutes, with the flight distance of 336 km (208 miles).

What type of aircraft is used for the Aeromexico AM4968 flight?

All Aeromexico AM4968 flights are operated using Canadair (Bombardier) Regional Jet 700 and Challenger 870 aircraft.

Which terminal the flight Aeromexico AM4968 is arriving at?

Flight Aeromexico AM4968 arrives in Hancock International Airport at Terminal B.