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Aeromexico AM4270 Flight Status Today
Domestic flight Aeromexico AM4270 takes off from Atlanta (ATL) United States to New York (JFK) United States. It's operated by Aeromexico. The plane leaves Hartsfield-Jackson Atlanta International Airport at 17:27 America/New_York. The flight is expected to land at John F. Kennedy International Airport at 20:00 America/New_York. The flight will last about 2 hours 7 minutes.
- Type: Domestic Flight
- Flight Duration: 2 hours 7 minutes
- Flight Distance: 1223 kms / 760 miles
- Aeromexico
- IATA: AM
- ICAO: AMX
- Operating Days: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday
- Service type: Normal passenger
- Seats: 149
- Freight capacity: 5.7 tons
- Passenger classes: Economy, Business Class, Shuttle, Premium Economy
- Aircraft: Boeing 767-300 (winglets) Passenger
- Callsign: AMX4270
- Departure Timezone: America/New_York
- Arrival Timezone: America/New_York
- Current Time in Atlanta: Friday 2026-02-06 18:02 PM
- Current Time in New York: Friday 2026-02-06 18:02 PM
Flight routes similar to Aeromexico AM4270
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LA6146 LATAM Airlines |
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DL2386 Delta |
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KE7081 Korean Air |
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UX3432 Air Europa |
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EY8240 Etihad Airways |
06/02/2026 13:45 |
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|
LO5736 Polish Airlines LOT |
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FI7904 Icelandair |
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Aeromexico Flight AM4270 FAQs
What is the scheduled flight duration for Aeromexico AM4270 flight?
On average, nonstop flight takes 2 hour(s) 7 minutes, with the flight distance of 1223 km (760 miles).
What type of aircraft is used for the Aeromexico AM4270 flight?
All Aeromexico AM4270 flights are operated using Boeing 767-300 (winglets) Passenger aircraft.
Which terminal the flight Aeromexico AM4270 is arriving at?
Flight Aeromexico AM4270 arrives in John F. Kennedy International Airport at Terminal 4.
How many Aeromexico AM4270 flights are operated a week?
7 flights per week. The Flight Aeromexico AM4270 is operated on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday .
