JetBlue Airways B68438 Flight Status Today

Domestic flight JetBlue Airways B68438 takes off from Jacksonville (VQQ) United States to New York (JFK) United States. It's operated by JetBlue Airways. The plane leaves Cecil Airport at 10:00 America/New_York. The flight is expected to land at John F. Kennedy International Airport at 12:11 America/New_York. The flight will last about 2 hours 11 minutes.

B68438 - JBU8438 JetBlue Airways (B6)
VQQ Jacksonville United States
Flight VQQ to JFK
JFK New York United States
Contact Airlines It may not operate on the date requested
Departure Cecil Airport
IATA: VQQICAO: KVQQ
Scheduled 10:00
None None
Arrival John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 12:11
5 20
B68438 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 11 minutes
  • Flight Distance: 1368 kms / 850 miles
Airline
  • JetBlue Airways
  • IATA: B6
  • ICAO: JBU
  • Operating Days: Sunday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320
  • Callsign: F-MDSD3
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Jacksonville: Tuesday 2026-01-20 18:04 PM
  • Current Time in New York: Tuesday 2026-01-20 18:04 PM

JetBlue Airways Flight B68438 FAQs

What is the scheduled flight duration for JetBlue Airways B68438 flight?

On average, nonstop flight takes 2 hour(s) 11 minutes, with the flight distance of 1368 km (850 miles).

What type of aircraft is used for the JetBlue Airways B68438 flight?

All JetBlue Airways B68438 flights are operated using Airbus A320 aircraft.

Which terminal the flight JetBlue Airways B68438 is arriving at?

Flight JetBlue Airways B68438 arrives in John F. Kennedy International Airport at Terminal 5.

How many JetBlue Airways B68438 flights are operated a week?

1 flights per week. The Flight JetBlue Airways B68438 is operated on Sunday.