Miami Dolphins running back Reggie Bush was named the AFC Offensive Player of the Week on Wednesday. He rushed for 172 yards and two touchdowns in Miami's 35-13 victory over the Oakland Raiders.

Bush showed toughness and versatility by making big plays and grinding out tough yards to wear the Raiders down. Bush also took a lot of pressure off rookie quarterback Ryan Tannehill, who picked up his first victory as a pro.

This is the second time Bush has won the award with the Dolphins. He received the same honor in Week 15 last season.