The Patriots beat the Dolphins to wrap up the No. 2 seed in Week 18 and one of their running backs was a big reason why.
Rhamondre Stevenson has been named AFC offensive player of the week, the NFL announced on Wednesday.
Stevenson finished the contest with 153 yards from scrimmage (131 rushing, 22 receiving) and three touchdowns (two rushing, one receiving).
He was the only player in Week 18 to have three scrimmage touchdowns.
This is the first offensive player of the week award for Stevenson. He finished the season with 603 yards and seven touchdowns while catching 32 passes for 345 yards with two TDs in 14 games.
