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02 October 2015


The National Football League announced today that Tampa Bay Buccaneers linebacker Kwon Alexander has been voted Pepsi NFL Rookie of the Week for Week 3.
“Kwon has done a great job for us in the middle of our defense this year and we are excited to see him getting recognized for his level of play,” said Buccaneers Head Coach Lovie Smith. “Defensively, we focus on making big plays and forcing takeaways, and that is exactly what Kwon did for us last week in Houston. He is getting more comfortable in his role each week and we look forward to seeing him continue to develop.”
Facing the Houston Texans, Alexander registered 10 tackles, including one for loss, along with two passes defensed and one interception, the first of his career. Alexander is tied for the lead in passes defensed (4) among all linebackers and is one of seven rookies to have an interception. His 23 tackles lead all rookies. Alexander is the first defensive player in Buccaneers history to win Pepsi Rookie of the Week. In Week 2, Buccaneers quarterback Jameis Winston won the award.
Since 2002, no teammates on opposite sides of the ball have won the award in consecutive weeks. The last team to have different players win the award in consecutive weeks were the San Diego Chargers in 2013 (WR Keenan Allen, Week 6 & T D.J. Fluker, Week 7).
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