Air Canada AC4738 Flight Status Today

Domestic flight Air Canada AC4738 takes off from Jacksonville (JAX) United States to Newark (EWR) United States. It's operated by Air Canada. The plane leaves Jacksonville International Airport at 12:55 America/New_York. The flight is expected to land at Newark Liberty International Airport at 15:29 America/New_York. The flight will last about 2 hours 35 minutes.

AC4738 - ACA4738 Air Canada (AC)
JAX Jacksonville United States
Flight JAX to EWR
EWR Newark United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 15-April-2020
Departure Jacksonville International Airport
IATA: JAXICAO: KJAX
Scheduled 12:55
None A10
Arrival Newark Liberty International Airport
IATA: EWRICAO: KEWR
Scheduled 15:29
A 24
AC4738 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 35 minutes
  • Flight Distance: 1319 kms / 820 miles
Airline
  • Air Canada
  • IATA: AC
  • ICAO: ACA
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 50
  • Freight capacity: 1 tons
  • Passenger classes: Economy, Shuttle, Premium Economy
More Detail
  • Aircraft: E145
  • Callsign: ACA4738
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Jacksonville: Thursday 2026-04-09 17:12 PM
  • Current Time in Newark: Thursday 2026-04-09 17:12 PM

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Air Canada Flight AC4738 FAQs

What is the scheduled flight duration for Air Canada AC4738 flight?

On average, nonstop flight takes 2 hour(s) 35 minutes, with the flight distance of 1319 km (820 miles).

What type of aircraft is used for the Air Canada AC4738 flight?

All Air Canada AC4738 flights are operated using E145 aircraft.

Which terminal the flight Air Canada AC4738 is arriving at?

Flight Air Canada AC4738 arrives in Newark Liberty International Airport at Terminal A.