Aeromexico AM5208 Flight Status Today

Domestic flight Aeromexico AM5208 takes off from Syracuse (SYR) United States to New York (JFK) United States. It's operated by Aeromexico. The plane leaves Hancock International Airport at 11:40 America/New_York. The flight is expected to land at John F. Kennedy International Airport at 12:55 America/New_York. The flight will last about 1 hours 15 minutes.

AM5208 - AMX5208 Aeromexico (AM)
SYR Syracuse United States
Flight SYR to JFK
JFK New York United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 06-April-2019
Departure Hancock International Airport
IATA: SYRICAO: KSYR
Scheduled 11:40
4
Arrival John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 12:55
EM B1
AM5208 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 15 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: 50
  • Freight capacity: 1.5 tons
  • Passenger classes: Economy, Shuttle, Premium Economy
More Detail
  • Aircraft: CRJ2
  • Callsign: AMX5208
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Syracuse: Monday 2025-12-29 05:47 AM
  • Current Time in New York: Monday 2025-12-29 05:47 AM

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Aeromexico Flight AM5208 FAQs

What is the scheduled flight duration for Aeromexico AM5208 flight?

On average, nonstop flight takes 1 hour(s) 15 minutes, with the flight distance of 336 km (208 miles).

What type of aircraft is used for the Aeromexico AM5208 flight?

All Aeromexico AM5208 flights are operated using CRJ2 aircraft.

Which terminal the flight Aeromexico AM5208 is arriving at?

Flight Aeromexico AM5208 arrives in John F. Kennedy International Airport at Terminal EM.