Air Canada AC4735 Flight Status Today

Domestic flight Air Canada AC4735 takes off from Newark (EWR) United States to Jacksonville (JAX) United States. It's operated by Air Canada. The plane leaves Newark Liberty International Airport at 09:00 America/New_York. The flight is expected to land at Jacksonville International Airport at 11:52 America/New_York. The flight will last about 2 hours 52 minutes.

AC4735 - ACA4735 Air Canada (AC)
EWR Newark United States
Flight EWR to JAX
JAX Jacksonville United States
Contact Airlines It may not operate on the date requested
Departure Newark Liberty International Airport
IATA: EWRICAO: KEWR
Scheduled 09:00
C 130
Arrival Jacksonville International Airport
IATA: JAXICAO: KJAX
Scheduled 11:52
None None
AC4735 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 52 minutes
  • Flight Distance: 1319 kms / 820 miles
Airline
  • Air Canada
  • IATA: AC
  • ICAO: ACA
  • Operating Days: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday
Passenger Services
  • Service type: Normal passenger
  • Seats: 64
  • Freight capacity: 0 tons
  • Passenger classes: First Class, Economy, Business Class, Shuttle
More Detail
  • Aircraft: Embraer 175
  • Callsign: ACA4735
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Newark: Tuesday 2026-01-20 15:34 PM
  • Current Time in Jacksonville: Tuesday 2026-01-20 15:34 PM

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United Airlines UA2335
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Air Canada Flight AC4735 FAQs

What is the scheduled flight duration for Air Canada AC4735 flight?

On average, nonstop flight takes 2 hour(s) 52 minutes, with the flight distance of 1319 km (820 miles).

What type of aircraft is used for the Air Canada AC4735 flight?

All Air Canada AC4735 flights are operated using Embraer 175 aircraft.

How many Air Canada AC4735 flights are operated a week?

7 flights per week. The Flight Air Canada AC4735 is operated on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday .