Joe Mixon Named AFC Offensive Player of the Week

Joe Mixon Named AFC Offensive Player of the Week

CINCINNATI — Bengals running back Joe Mixon was named AFC Offensive Player of the Week for his Week 4 performance in Sunday's 33-25 win over the Jacksonville Jaguars. 

He posted season-highs in rushing yards (151) and yards from scrimmage (181). His 181 total yards were the most by an AFC player this season. 

Mixon also scored a career-high three touchdowns. His nine-yard touchdown reception in the first half helped the Bengals tie the game. Then, he had second half scoring runs of 34 and 23 yards, which sealed the deal for Cincinnati. 

"I think with the football in my hands, it’s liable to go [for a touchdown] on any play," Mixon said after the game. "These past couple of weeks have been a little tough. But the thing was, nobody was noticing we were so close. It’s always one block away, one play away. We were able to execute those things today. I’m just so grateful to be able to play here each and every day and I never take nothing for granted, especially my teammates. It’s just been great. We just gotta keep on building week in, week out and keep on executing, keep trying to get those wins week by week."

Mixon has ran the ball 77 times for 315 yards this season. His per-carry average went up from 3.2 to 4.1 following his stellar effort on Sunday. 

He's the first Bengals player to win the award since A.J. Green won it in Week 3 of the 2015 season. Green had 10 receptions for 227 yards and 2 touchdowns on the road in Baltimore—which is exactly where the Bengals are headed this weekend. 

Watch some of Mixon's highlights from Week 4 below. For more on the Bengals, including the latest NFL news, go here!

181 total yards, 3 touchdowns, a WIN and AFC Offensive Player of the Week for @Joe_MainMixon

📺: #CINvsBAL - 10/11 at 1:00 p.m. ET on CBS pic.twitter.com/ad2vJ85D36

— Cincinnati Bengals (@Bengals) October 7, 2020