For the second time this season, Miami Dolphins linebacker Jordyn Brooks has been honored with the AFC Defensive Player of the Week award — this time for his 20 tackle effort against the Washington Commanders in Madrid, Spain.
In addition to his 20 tackles, Brooks had a tackle for loss against Washington. His 20 take downs of opposing ball carriers last week is the highest amount of tackles any defender has had in an NFL game this season, and just four tackles shy of the all-time single game record of 24 (shared by David Harris of the New York Jets and Luke Kuechly of the Carolina Panthers).
Brooks leads the NFL in tackles this season with 125 — Bobby Wagner of the Commanders is a distant second with 107.
Miami’s 28-year-old linebacker won his first AFC Defensive Player of the Week award after Miami’s 34-10 victory over the Atlanta Falcons in week 8. That week, Cleveland Browns EDGE rusher Myles Garrett had 5 sacks, yet still didn’t take home the hardware. In week 11, Garrett had four sacks of Baltimore Ravens quarterback Lamar Jackson… and still lost the honor of Defensive Player of the Week to Brooks.
The Dolphins will continue to lean on Brooks’ leadership — both on and off the field — as the club inches closer to the conclusion of the regular season.
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