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Myles Garrett is AFC Defensive Player of the Week

Browns defensive end honored for impact plays in Week 4 win over the Dallas Cowboys.

NFL: Washington Football Team at Cleveland Browns Ken Blaze-USA TODAY Sports

Cleveland Browns defensive end Myles Garrett is the AFC Defensive Player of the Week.

Garrett earned the honor after posting two sacks, a forced fumble and three tackles in Cleveland’s win over the Dallas Cowboys.

Garrett got to work early against the Cowboys with an 11-yard sack of quarterback Dak Prescott on the second play of the game. He later added a strip-sack of Prescott and the Browns converted that turnover into a one-yard touchdown pass from quarterback Baker Mayfield to tight end Austin Hooper. That began a run that saw the Browns go from a 14-14 tie to a 41-14 lead.

During the Browns three-game winning streak, Garrett has posted one sack and one forced fumble in each game. He is tied for the league lead with five sacks, three forced fumbles and two fumble recoveries, according to clevelandbrowns.com.

Through the season’s first four games, Garrett has forced six turnovers, which is as many as Bud Dupree of the Pittsburgh Steelers and Shaquil Barrett of the Tampa Bay Buccaneers had in 2019 when they lead the NFL in that category, according to Next Gen Stats:

Not bad for a player that some still claim “does not make impact plays,” no?