JetBlue Airways B66495 Flight Status Today

Domestic flight JetBlue Airways B66495 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 08:00 America/New_York. The flight is expected to land at John F. Kennedy International Airport at 09:51 America/New_York. The flight will last about 4 hours 43 minutes.

B66495 - JBU6495 JetBlue Airways (B6)
VQQ Jacksonville United States
Flight VQQ to JFK
JFK New York United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 22-April-2021
Departure Cecil Airport
IATA: VQQICAO: KVQQ
Scheduled 08:00
A 8
Arrival John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 09:51
None None
B66495 Detail
  • Type: Domestic Flight
  • Flight Duration: 4 hours 43 minutes
  • Flight Distance: 1368 kms / 850 miles
Airline
  • JetBlue Airways
  • IATA: B6
  • ICAO: JBU
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 148
  • Freight capacity: 9.4 tons
  • Passenger classes: Economy, Shuttle, Premium Economy
More Detail
  • Aircraft: Airbus A320
  • Callsign: JBU495
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Jacksonville: Wednesday 2026-01-21 08:59 AM
  • Current Time in New York: Wednesday 2026-01-21 08:59 AM

Flight routes similar to JetBlue Airways B66495

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JetBlue Airways B68438
JetBlue Airways 		21/01/2026
10:00 		2 hours 11 minutes
12:11

JetBlue Airways Flight B66495 FAQs

What is the scheduled flight duration for JetBlue Airways B66495 flight?

On average, nonstop flight takes 4 hour(s) 43 minutes, with the flight distance of 1368 km (850 miles).

What type of aircraft is used for the JetBlue Airways B66495 flight?

All JetBlue Airways B66495 flights are operated using Airbus A320 aircraft.