FOXBOROUGH, Mass. — Running back Ezekiel Elliott is officially heading to the AFC, signing a free agent deal with the New England Patriots.
The team announced the addition of the three-time Pro Bowler on Wednesday. A person familiar with the terms told The Associated Press that the seven-year veteran with the Dallas Cowboys is joining the Patriots on a one-year deal worth $4 million and incentives could boost his compensation to $6 million.
Coach Bill Belichick told reporters in Green Bay, Wisconsin, on Wednesday morning that Elliott was with the team for its two days of joint practices with the Packers in advance of its preseason game. His participation is still to be determined.
The 28-year-old ran for 68 touchdowns and more than 8,000 yards with the Cowboys. He led the NFL in rushing yards in 2016, when he was the runner-up for the AP Rookie of the Year award, and again in 2018. He also has more than 2,300 receiving yards, and 12 touchdown catches.
Elliott’s yards per game have decreased every season of his career, a trend that partly reflects the decreased usage of running backs in the NFL since the Cowboys used the No. 4 overall draft pick to select him in 2016.
He is expected to serve as a compliment to running back Rhamondre Stevenson, who is projected to be New England's lead option out of the backfield.
The Patriots suffered in short-yardage situations last season, an area where Elliott excels despite showing diminished overall output at the end of his Dallas tenure.
AP Pro Football Writer Rob Maaddi contributed to this report.