MIAMI GARDENS, Fla. – The NFL announced that Miami Dolphins running back Raheem Mostert has been named AFC Offensive Player of the Week for his performance in Sunday's win vs. Carolina.
It's Mostert's first AFC Offensive Player of the Week honor and he becomes the second Dolphins running back to win the award this season, joining De'Von Achane, who won it in Week 3 vs. Denver. The last time an NFL team had two different running backs win Offensive Player of the Week honors in the same season was in 2016, when Pittsburgh's DeAngelo Williams (Week 1) and Le'Veon Bell (Week 14) each earned the award.
Mostert had three touchdowns in Miami's 42-21 victory over Carolina. He totaled 17 carries for 115 yards (6.8 avg.) and two scores on the ground while catching three passes for 17 yards (5.7 avg.) and a touchdown. He was the only player in the NFL to score three touchdowns this week and the only one to record both a rushing and a receiving score. Of Mostert's 17 carries, six were for first downs and three went for 10-plus yards. He helped Miami tally 162 rushing yards, marking the fifth game in a row the Dolphins have had a 100-yard rusher and rushed for more than 140 yards as a team.
Mostert has now scored a touchdown in five of Miami's six games this season and his 11 overall touchdowns and 66 points both lead the league. He has already set career highs with nine rushing touchdowns and 11 overall touchdowns with 11 games still remaining in the season.
This is the third AFC Offensive Player of the Week award for the Dolphins this season after quarterback Tua Tagovailoa earned the honor in Week 1 at the L.A. Chargers and Achane won it in Week 3 vs. Denver. The last time the Dolphins won three AFC Offensive Player of the Week Awards in the same season was in 2016, when running back Jay Ajayi earned the award three times and quarterback Matt Moore earned the honor once.
