Philadelphia Eagles running back Saquon Barkley added to his historic 2024 season on Thursday night, as it was announced at the NFL Honors that he won the AP's Offensive Player of the Year Award.
Over 16 regular-season appearances, Barkley rushed for 2,005 yards and 13 touchdowns on 345 carries—becoming the ninth player in NFL history to rush for at least 2,000 yards in a season. His performance helped the Eagles' offense score 27.2 points per game in 2024, which ranked seventh in the NFL.
The 27-year-old rushed for 100+ yards in 11 of 16 games while finishing just 101 yards short of breaking Eric Dickerson's single-season rushing yards record. Barkley did not suit up for the Eagles in Week 18 as they had the No. 2 seed in the NFL playoffs locked up.
Speaking of the playoffs, though his performance doesn't count towards the AP voting, Barkley has been incredible through Philly's run to the Super Bowl thus far. Through three games, he's rushed for 442 yards (100+ in each) and five touchdowns while averaging 6.7 yards per carry. Philly has averaged 35 points per game this postseason.
Barkley beat out finalists Cincinnati Bengals quarterback Joe Burrow, Bengals wide receiver Ja'Marr Chase, Baltimore Ravens running back Derrick Henry, and Ravens quarterback Lamar Jackson for the award.
His Eagles will take on the Kansas City Chiefs this coming Sunday in Super Bowl LIX. Kickoff is scheduled for 6:30 p.m. ET from New Orleans' Caesars Superdome.
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This article was originally published on www.si.com as Eagles RB Saquon Barkley Wins Offensive Player of the Year Award Ahead of Super Bowl LIX.
