Distance from Warrensburg to Clayton

Distance between Warrensburg and Clayton is 296 kilometers (184 miles).
Driving distance from Warrensburg to Clayton is 340 kilometers (211 miles).

296 km

184 miles

340 km

211 miles

Distance Map Between Warrensburg and Clayton

Warrensburg, Jefferson City, United States ↔ Clayton, Jefferson City, United States = 184 miles = 296 km.

How far is it between Warrensburg and Clayton

Warrensburg is located in United States with (38.7628,-93.7361) coordinates and Clayton is located in United States with (38.6426,-90.3237) coordinates. The calculated flying distance from Warrensburg to Clayton is equal to 184 miles which is equal to 296 km.

If you want to go by car, the driving distance between Warrensburg and Clayton is 340.15 km. If you ride your car with an average speed of 112 kilometers/hour (70 miles/h), travel time will be 03 hours 02 minutes. Please check the avg. speed travel time table on the right for various options.
Difference between fly and go by a car is 44 km.

City/Place	Latitude and Longitude	GPS Coordinates
Warrensburg	38.7628, -93.7361	38° 45´ 46.0440'' N 93° 44´ 9.7800'' W
Clayton	38.6426, -90.3237	38° 38´ 33.1800'' N 90° 19´ 25.4280'' W

Estimated Travel Time Between Warrensburg and Clayton

Average Speed	Travel Time
30 mph (48 km/h)	07 hours 05 minutes
40 mph (64 km/h)	05 hours 18 minutes
50 mph (80 km/h)	04 hours 15 minutes
60 mph (97 km/h)	03 hours 30 minutes
70 mph (112 km/h)	03 hours 02 minutes
75 mph (120 km/h)	02 hours 50 minutes

Warrensburg, Jefferson City, United States

Related Distances from Warrensburg

Cities	Distance
Warrensburg to Arnold 3	356 km
Warrensburg to Saint Peters	305 km
Warrensburg to Kirkwood	333 km
Warrensburg to Poplar Bluff	496 km
Warrensburg to University City	342 km

Clayton, Jefferson City, United States

Related Distances to Clayton

Cities	Distance
Oakville 2 to Clayton 2	33 km
Independence to Clayton 2	378 km
Kirkwood to Clayton 2	16 km
Springfield to Clayton 2	339 km
Saint Joseph to Clayton 2	478 km