After a sterling performance against the Giants, New York Jets safety Jamal Adams was honored by the league.
Jamal Adams swiped one more big score away before his New York Jets moved on to Washington. The strong safety was named the AFC’s Defensive Player of the Week by the NFL on Wednesday, the first such award of his career.
Adams played perhaps the finest game of his young career last Sunday afternoon against the New York Giants. Facing the Jets’ in-state interconference rival, Adams earned nine tackles, including a career-best two sacks of quarterback Daniel Jones in what became a 34-27 Jets victory.
Both of Adams’s sacks forced the ball from Jones’ grasp and the latter was perhaps the most electrifying defensive play from Week 10.
The Jets led 14-13 at halftime and gave the ball to the Giants to begin the second half. On the third play of the opening drive, Adams stole the ball away from Jones and took it all the way back for the lead-extending score.
It was Adams’s second touchdown of the season, having previously taken a Jarrett Stidham interception back for a score against New England in Week 3.
Adams is the second Jet to win a Player of the Week Award, with quarterback Sam Darnold winning Offensive honors after Week 6’s win over Dallas. The previous Jets Defensive Player of the Week honor came in November 2017, when linebacker Jordan Jenkins had two sacks and a forced fumble in a victory against Buffalo.
Victorious AFC brethren alongside Adams are Baltimore quarterback Lamar Jackson and Miami kicker Jason Sanders. Jackson, the winner of the Offensive award for the second week in a row and third time this season, tallied 288 total yards and four total touchdowns in the Ravens’ 49-13 win in Cincinnati.
Sanders accounted for all but six of the Dolphins’ 16 points in Indianapolis via three field goals and an extra point in a 16-12 triumph.
Adams and the Jets return to action on Sunday afternoon against the Washington Redskins (1:00 p.m. ET, FOX).
