KANSAS CITY, Mo. – Days after a record-setting performance, the NFL named Chiefs running back Jamaal Charles as the AFC Offensive Player of the Week.
Charles paced the Chiefs in Week 15 with a five-touchdown performance to go along with 215 total yards (195 receiving) in a 56-31 win against the Oakland Raiders.
He became the only player in NFL history to record four receiving touchdowns and one rushing touchdown in a single game. Charles also became the first running back in NFL history to record four receiving touchdowns in a single game.
“He obviously is very valuable to this team,” Chiefs coach Andy Reid said of Charles during Wednesday’s media session. “I appreciate probably most how he comes to work every day and the attitude he brings. He wants to get better every day.”
Wednesday’s recognition marks the third career AFC Offensive Player of the Week for Charles. The two-time Pro Bowler previously won the award in Week 3 of the 2012 season.
“I’m happy to get it, but I know they’re happy for me,” Charles said of teammates’ reactions to Wednesday’s announcement. “They know I work, I bust my tail in practice every week and I go out there and play so hard for my teammates.”
While winning awards is enjoyable, Charles told reporters he gets more gratification from what the Chiefs have done 11 times this season.
“The most fun is winning,” Charles said “That puts a smile on everybody’s face.”
Charles is the first running back to be recognized in 2013 with the AFC Offensive Player of the Week honor.
He joins teammates Justin Houston, Tamba Hali, Dexter McCluster and Ryan Succop as Chiefs recognized by the NFL with the weekly award during the 2013 regular season.