MIAMI GARDENS, Fla. – The NFL today announced that Miami Dolphins quarterback Tua Tagovailoa has been named AFC Offensive Player of the Week for his performance in Sunday’s win at Baltimore.
It’s Tagovailoa’s first AFC Offensive Player of the Week award and Miami’s first AFC Offensive Player of the Week since Ryan Fitzpatrick won the award in Week 16 of the 2019 season vs. Cincinnati.
Tagovailoa completed 36-of-50 passes (72.0 pct.) for 469 yards and six touchdowns for a 124.1 quarterback rating in the Week 2 victory over Baltimore. The 36 completions, 469 passing yards and six passing touchdowns were all career highs.
His six passing touchdowns tied a team record while his 36 completions rank third and 469 passing yards rank fourth in single-game team history. Tagovailoa helped lead the Dolphins to their largest fourth-quarter comeback (21 points) and largest road comeback in team history. In the fourth quarter, he completed 13-of-17 passes (76.5 pct.) for 199 yards and four touchdowns. His four passing touchdowns were the most by any NFL quarterback in the fourth quarter of a game since Oct. 21, 2007 (Sage Rosenfels vs. Tennessee). Tagovailoa became the second-youngest NFL player (Patrick Mahomes) since 1950 to throw for at least 450 yards and six touchdowns in game.
