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).

45 km

28 miles

52 km

32 miles

Distance Map Between Secaucus and Sayreville

Secaucus, Trenton, United States ↔ Sayreville, 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/Place	Latitude and Longitude	GPS 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 Speed	Travel 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

Secaucus, Trenton, United States

Related Distances from Secaucus

Cities	Distance
Secaucus to Vineland	195 km
Secaucus to Rahway	32 km
Secaucus to Dover 4	58 km
Secaucus to Roselle 2	32 km
Secaucus to Orange 3	20 km

Sayreville, Trenton, United States

Related Distances to Sayreville

Cities	Distance
Wayne to Sayreville	66 km
South Plainfield to Sayreville	25 km
Summit to Sayreville Junction	46 km
Tinton Falls to Sayreville Junction	38 km
Secaucus to Sayreville Junction	49 km