Jaguars linebacker Foyesade Oluokun had a big day to help Jacksonville defeat Carolina to open the 2025 season.
He’s now been named AFC defensive player of the week.
Oluokun recorded 10 total tackles, a forced fumble, and an interception in the 26-10 victory.
He was the only player in the league to force multiple turnovers in Week 1.
This is Oluokun’s third career player of the week award — his first with Jacksonville. He previously won two with Atlanta.
