MIAMI GARDENS, Fla. — Miami Dolphins linebacker Jordyn Brooks was named AFC Defensive Player of the Week Wednesday following a standout performance in this past Sunday’s 34-10 win at Atlanta, the NFL announced.
It’s the first time Brooks has earned the award in his career and marks the Dolphins’ first AFC Defensive Player of the Week selection since Tyrel Dodson in Week 17 of the 2024 season against Cleveland.
Brooks led Miami with 10 tackles (seven solo), 1 sack and three tackles for loss, including six tackles (five solo), the sack and all three tackles for loss in the first half alone.
He became the only NFL player this season to post that stat line in a single half and the first Dolphins player to do so since Ndamukong Suh against Philadelphia in 2015.
His performance helped the Dolphins defense contain the Falcons’ fourth-ranked rushing attack to just 45 yards, the fewest allowed by Miami since holding the Jets to 23 yards in 2023.
In the first half, Atlanta managed only 10 rushing yards, 58 total yards, and failed to convert any of five third-down attempts.
It was an impressive feat considering that Bijan Robinson and Tyler Allgeier are one of the NFL’s best running back tandems.
Brooks, who earned his first career team captaincy this season after joining Miami as an unrestricted free agent in 2024, leads the NFL with 85 tackles (53 solo) entering Week 9.
He also has a fumble recovery and 2.5 sacks this season, and is the only player in the NFL to record 80-plus tackles, a fumble recovery and multiple sacks in each of the last two seasons.
Up Next:
Miami (2-6) will host the Baltimore Ravens (2-5) at Hard Rock Stadium on Thursday night. Kickoff is set for 8:15 p.m.
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