Aeromexico AM4972 Flight Status Today

Domestic flight Aeromexico AM4972 takes off from Knoxville (TYS) United States to Atlanta (ATL) United States. It's operated by Aeromexico. The plane leaves McGhee Tyson Airport at 17:25 America/New_York. The flight is expected to land at Hartsfield-Jackson Atlanta International Airport at 18:33 America/New_York. The flight will last about 1 hours 4 minutes.

AM4972 - AMX4972 Aeromexico (AM)
TYS Knoxville United States
Flight TYS to ATL
ATL Atlanta United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 01-February-2020
Departure McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 17:25
None 2
Arrival Hartsfield-Jackson Atlanta International Airport
IATA: ATLICAO: KATL
Scheduled 18:33
S C40
AM4972 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 4 minutes
  • Flight Distance: 244 kms / 152 miles
Airline
  • Aeromexico
  • IATA: AM
  • ICAO: AMX
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 76
  • Freight capacity: 1.9 tons
  • Passenger classes: Economy, Shuttle, Premium Economy
More Detail
  • Aircraft: CRJ9
  • Callsign: AMX4972
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Knoxville: Sunday 2025-12-28 21:35 PM
  • Current Time in Atlanta: Sunday 2025-12-28 21:35 PM

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Aeromexico Flight AM4972 FAQs

What is the scheduled flight duration for Aeromexico AM4972 flight?

On average, nonstop flight takes 1 hour(s) 4 minutes, with the flight distance of 244 km (152 miles).

What type of aircraft is used for the Aeromexico AM4972 flight?

All Aeromexico AM4972 flights are operated using CRJ9 aircraft.

Which terminal the flight Aeromexico AM4972 is arriving at?

Flight Aeromexico AM4972 arrives in Hartsfield-Jackson Atlanta International Airport at Terminal S.