Distance Calculator

Distance from Secaucus to Sayreville

Distance between Secaucus and Sayreville is 45 kilometers (28 miles).
Driving distance from Secaucus to Sayreville is 52 kilometers (32 miles).

air 45 km
air 28 miles
car 52 km
car 32 miles

Distance Map Between Secaucus and Sayreville

Secaucus, Trenton, United StatesSayreville, Trenton, United States = 28 miles = 45 km.

How far is it between Secaucus and Sayreville

Secaucus is located in United States with (40.7896,-74.0565) coordinates and Sayreville is located in United States with (40.4593,-74.361) coordinates. The calculated flying distance from Secaucus to Sayreville is equal to 28 miles which is equal to 45 km.

If you want to go by car, the driving distance between Secaucus and Sayreville is 51.59 km. If you ride your car with an average speed of 112 kilometers/hour (70 miles/h), travel time will be 00 hours 27 minutes. Please check the avg. speed travel time table on the right for various options.
Difference between fly and go by a car is 7 km.

City/PlaceLatitude and LongitudeGPS Coordinates
Secaucus 40.7896, -74.0565 40° 47´ 22.3800'' N
74° 3´ 23.5080'' W
Sayreville 40.4593, -74.361 40° 27´ 33.3720'' N
74° 21´ 39.5280'' W

Estimated Travel Time Between Secaucus and Sayreville

Average SpeedTravel Time
30 mph (48 km/h) 01 hours 04 minutes
40 mph (64 km/h) 00 hours 48 minutes
50 mph (80 km/h) 00 hours 38 minutes
60 mph (97 km/h) 00 hours 31 minutes
70 mph (112 km/h) 00 hours 27 minutes
75 mph (120 km/h) 00 hours 25 minutes
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