Distance Calculator

Distance from Sayreville Junction to Jackson

Distance between Sayreville Junction and Jackson is 89 kilometers (55 miles).
Driving distance from Sayreville Junction to Jackson is 106 kilometers (66 miles).

air 89 km
air 55 miles
car 106 km
car 66 miles

Distance Map Between Sayreville Junction and Jackson

Sayreville Junction, Trenton, United StatesJackson, Trenton, United States = 55 miles = 89 km.

How far is it between Sayreville Junction and Jackson

Sayreville Junction is located in United States with (40.4654,-74.3304) coordinates and Jackson is located in United States with (39.7765,-74.8624) coordinates. The calculated flying distance from Sayreville Junction to Jackson is equal to 55 miles which is equal to 89 km.

If you want to go by car, the driving distance between Sayreville Junction and Jackson is 106.22 km. If you ride your car with an average speed of 112 kilometers/hour (70 miles/h), travel time will be 00 hours 56 minutes. Please check the avg. speed travel time table on the right for various options.
Difference between fly and go by a car is 17 km.

City/PlaceLatitude and LongitudeGPS Coordinates
Sayreville Junction 40.4654, -74.3304 40° 27´ 55.3680'' N
74° 19´ 49.5480'' W
Jackson 39.7765, -74.8624 39° 46´ 35.4000'' N
74° 51´ 44.5680'' W

Estimated Travel Time Between Sayreville Junction and Jackson

Average SpeedTravel Time
30 mph (48 km/h) 02 hours 12 minutes
40 mph (64 km/h) 01 hours 39 minutes
50 mph (80 km/h) 01 hours 19 minutes
60 mph (97 km/h) 01 hours 05 minutes
70 mph (112 km/h) 00 hours 56 minutes
75 mph (120 km/h) 00 hours 53 minutes
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