FOXBOROUGH, Mass. – Linebacker Kyle Van Noy was named AFC Defensive Player of the Week by the National Football League for his performance in the Patriots 16-10 win over Buffalo on Sunday.
Van Noy led the team with eight tackles, two sacks, three quarterback hits and two forced fumbles. Late in the game with the Patriots holding a 16-10 lead and the Buffalo offense at the New England 39-yard line, Van Noy hit Buffalo QB Matt Barkley, forcing a pass that was intercepted by LB Jamie Collins with 1:27 left to play to secure the victory.
It is the first time that Van Noy has earned the NFL award and marks the Patriots first AFC Defensive Player of the Week award since LB Dont’a Hightower was honored after Week 6 of the 2016 season.
This award marks the second consecutive week that a Patriots player has been honored by the NFL. Rookie P Jake Bailey was named AFC Special Teams Player of the Week for his performance in Week 3 vs. the New York Jets.