Devin Bush wins AFC defensive player of the week
Our Summary
The NFL has named Cleveland Browns linebacker Devin Bush as the AFC Defensive Player of the Week. Bush achieved this recognition following a standout performance where he returned an interception 97 yards for a touchdown and recorded 14 total tackles. This feat marks Bush as the first player since 2000 to have at least 14 tackles and an interception of at least 95 yards in a single game. This is Bush's second career Defensive Player of the Week award, with the first occurring while he was with the Pittsburgh Steelers in Week 6 of the 2019 season. The acknowledgment comes amid significant changes for the Browns, including the end of head coach Kevin Stefanski’s tenure with the team.
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