Virgin Atlantic VS945 Flight Status Today

Domestic flight Virgin Atlantic VS945 takes off from New York (JFK) United States to Milwaukee (MKE) United States. It's operated by Virgin Atlantic. The plane leaves John F. Kennedy International Airport at 18:10 America/New_York. The flight is expected to land at General Mitchell International Airport at None America/Chicago. The flight will last about 2 hours 42 minutes.

VS945 - VIR945 Virgin Atlantic (VS)
JFK New York United States
Flight JFK to MKE
MKE Milwaukee United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 18:10
None None
Arrival General Mitchell International Airport
IATA: MKEICAO: KMKE
Scheduled None
None None
VS945 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 42 minutes
  • Flight Distance: 1199 kms / 745 miles
Airline
  • Virgin Atlantic
  • IATA: VS
  • ICAO: VIR
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A350
  • Callsign: None
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/Chicago
  • Current Time in New York: Wednesday 2026-01-21 20:58 PM
  • Current Time in Milwaukee: Wednesday 2026-01-21 19:58 PM

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Virgin Atlantic Flight VS945 FAQs

What is the scheduled flight duration for Virgin Atlantic VS945 flight?

On average, nonstop flight takes 2 hour(s) 42 minutes, with the flight distance of 1199 km (745 miles).

What type of aircraft is used for the Virgin Atlantic VS945 flight?

All Virgin Atlantic VS945 flights are operated using Airbus A350 aircraft.