For the second time in his career, Tampa Bay Buccaneers quarterback Jameis Winston has been named the NFC Offensive Player of the Week.
Winston received that award on Wednesday following his extremely productive outing in the Buccaneers' 55-40 Week Four win over the Los Angeles Rams. He had first earned that honor in Week One of the 2016 campaign, after a four-touchdown performance in a 31-24 road win at Atlanta. He is the second Buccaneer to win a Player of the Week award in 2019, as outside linebacker Shaq Barrett captured defensive honors after his three-sack game in a Week Two win at Carolina.
Winston is the fifth Buccaneer quarterback to win multiple player of the week awards, including Ryan Fitzpatrick, who was the recipient in each of the first two weeks last season. Also on that list: Steve DeBerg (1984, Week 3; 1987, Week 1), Vinny Testaverde (1989, Week 1; 1992, Week 2) and Brad Johnson (2002, Weeks 9 and 14; 2003, Week 6).
On Sunday, Winston led the Buccaneers to its highest single-game point total ever in knocking off the previously-unbeaten Rams. He completed 28 of 41 passes for 385 yards, four touchdowns, one interception and a passer rating of 120.5. He completed 68.3% of his passes against the league's fourth-ranked pass defense and averaged 9.4 yards per attempt. Winston and Patrick Mahomes (Week Two) are the only NFL quarterbacks to throw at least 40 passes in a game this year and average more than nine yards per attempt.
In addition to touchdown passes of nine and three yards to Chris Godwin and of 13 yards to Cameron Brate, Winston also hit Mike Evans on a 67-yard bomb that gave the Bucs an 18-point lead in the fourth quarter. Over his last three games, he has thrown for 973 yards and eight touchdowns against just two interceptions.
Winston is also a candidate for this week's FedEx Air Player of the Week, which is chosen by fan voting on NFL.com.
