CLEVELAND — Upon signing a 5-year, $125 million contract extension that made him the highest-paid defensive player in football at the time, Cleveland Browns defensive end Myles Garrett set his expectation at becoming the NFL's Defensive Player of the Year.
Through the first four weeks of the 2020 season, it would be tough to argue that the No. 1 pick of the 2017 NFL Draft isn't in a position to accomplish just that.
On Sunday, Garrett recorded a strip-sack for the third time in as many weeks, with his second-quarter takeaway playing a pivotal role in the Browns' 49-38 win over his hometown Dallas Cowboys. For his efforts -- which included three tackles, two sacks and a forced fumble -- Garrett was named the AFC Defensive Player of the Week.
Garrett's honor -- which marks the first time he's been named the AFC Defensive Player of the Week -- comes amid a season in which the Texas A&M product appears to be on pace for one a career year. In four games, Garrett has tallied an NFL-high five sacks, to go along with three forced fumbles and two tackles for a loss as Cleveland has amassed a 3-1 record -- its best start since the 2001 season.
This is the type of season that Garrett envisioned having when he discussed his goals for the season after signing his mega extension in July. After a suspension for striking Pittsburgh Steelers quarterback Mason Rudolph in the head with his helmet brought his 2019 season to a premature end, Garrett has reignited his Defensive Player of the Year campaign in a big way.
"I don’t want to be the same player I was last year; I want to be better in all aspects,‘' Garrett said in July. "Even on [last season's] trajectory, I was known to be in the [Defensive] Player of the Year conversation. So I don't want to make it a conversation anymore. This next year, I want to ball out and win that award, but I want to take my team to the playoffs and even higher than that."