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06 November 2015

BUCCANEERS LB KWON ALEXANDER VOTED PEPSI ROOKIE OF THE WEEK

The National Football League announced today that Tampa Bay Buccaneers linebacker Kwon Alexander has been voted Pepsi NFL Rookie of the Week for Week 8.
 
Alexander, who also won NFC Defensive Player of the Week, set a career-high with 11 tackles in Tampa Bay’s overtime win at Atlanta, while adding a pass defensed, an interception, a forced fumble and a fumble recovery. He is the first rookie linebacker to record an interception, a forced fumble and a fumble recovery in the same game since Demeco Ryans (2006) and the first Buccaneers player to accomplish the feat since John Lynch (1996). His two takeaways led to 10 Buccaneers points, while his 11 tackles were tied for the eighth-most in the NFL in Week 8.
 
This is Alexander’s second Pepsi Rookie of the Week Award (also Week 3), and the fourth the team has earned overall (QB Jameis Winston, Weeks 2 & 5). Tampa Bay’s four Pepsi Rookie of the Week awards are the most in single season in team history. Alexander and Winston are the first teammates to earn the award multiple times since 2012 (Washington QB Robert Griffin III & RB Alfred Morris). They are the first ever teammates on opposite side of the ball to win the award multiple times.
 
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