JetBlue Airways B6440 Flight Status Today

Domestic flight JetBlue Airways B6440 takes off from Cleveland (CLE) United States to Boston (BOS) United States. It's operated by JetBlue Airways. The plane leaves Cleveland Hopkins International Airport at 07:00 America/New_York. The flight is expected to land at Logan International Airport at 08:40 America/New_York. The flight will last about 1 hours 33 minutes.

B6440 - JBU440 JetBlue Airways (B6)
CLE Cleveland United States
Flight CLE to BOS
BOS Boston United States
Contact Airlines It may not operate on the date requested
Departure Cleveland Hopkins International Airport
IATA: CLEICAO: KCLE
Scheduled 07:00
None C6
Arrival Logan International Airport
IATA: BOSICAO: KBOS
Scheduled 08:40
C C28
B6440 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 33 minutes
  • Flight Distance: 903 kms / 561 miles
Airline
  • JetBlue Airways
  • IATA: B6
  • ICAO: JBU
  • Operating Days: Thursday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Embraer 190
  • Callsign: F-KNEL1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Cleveland: Thursday 2026-04-30 06:06 AM
  • Current Time in Boston: Thursday 2026-04-30 06:06 AM

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LATAM Airlines LA6250
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SAS SK3310
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JetBlue Airways Flight B6440 FAQs

What is the scheduled flight duration for JetBlue Airways B6440 flight?

On average, nonstop flight takes 1 hour(s) 33 minutes, with the flight distance of 903 km (561 miles).

What type of aircraft is used for the JetBlue Airways B6440 flight?

All JetBlue Airways B6440 flights are operated using Embraer 190 aircraft.

Which terminal the flight JetBlue Airways B6440 is arriving at?

Flight JetBlue Airways B6440 arrives in Logan International Airport at Terminal C.

How many JetBlue Airways B6440 flights are operated a week?

1 flights per week. The Flight JetBlue Airways B6440 is operated on Thursday.