Photo credit: Chris Leduc
Tom Brady was named NFC Offensive Player of the Week on Wednesday for his Week 5 performance against the Maimi Dolphins, in which Brady threw for 411 yards and five touchdowns while completing 73.2 percent of his passes en route to a 45-17 victory by the Tampa Bay Buccaneers.
This is Brady's 32nd Player of the Week award in his career, the most by any player in NFL history. It also marks his second such award during his time with the Buccaneers, after earning the honor for his 2020 Week 4 showing against the Los Angeles Chargers.
Brady is the third Buccaneers player to win a weekly award this season, following punter Bradley Pinion (Special Teams, Week 1) and safety Mike Edwards (Defensive, Week 2).
After a two-week slump for the Buccaneers' offense, not a fault of Brady's as he threw for over 700 yards in that stretch, the unit got back to form against a solid Dolphins' pass defense this past weekend and looks to carry the momentum into Philadelphia this Thursday against the Eagles.
Brady, certainly, has all of the momentum in the world right now. Brady leads the NFL in passing yards with 1,767, ranks second in passing touchdowns with 15, and fifth in passer rating at 108.5 through the first five weeks of the season.
Stay tuned to AllBucs for further coverage of the Tampa Bay Buccaneers, and other NFL news and analysis. Follow along on social media at @SIBuccaneers on Twitter and Tampa Bay Buccaneers on Sports Illustrated on Facebook.