After having a phenomenal day against the Indianapolis Colts and propelling the Jacksonville Jaguars to a Week 17 victory, rookie quarterback Gardner Minshew II has found himself earning another award that he’s all too familiar with. Thursday afternoon, it was announced that he won his seventh Pepsi Rookie of the Week award, giving him the most of any rookie in 2019.
Minshew was able to go 27-of-39 (69.2%) for 295 yards, and three touchdowns, which aided the Jags to a 38-20 victory to close the season. The victory also gave him a 6-6 record on the season as a starter.
Minshew’s nomination for the award closes what has been an interesting season for the rookie, who was thrust into the starting role Week 1 against the Kansas City Chiefs and renamed the starter Week 14 against the Los Angeles Chargers. Ultimately, he was able to finish the season 285-of-470 (60.6%) for 3,271 passing yards, 21 touchdowns and six picks, making a strong case to start in 2020.
Minshew’s success on the 2019 season may not be over just yet, as there is a chance he could win Offensive Rookie of the Year. Per FanDuel Sportsbook, he has the fifth-best odds (+3,300) to win the award, with rookie running back Josh Jacobs leading the way with odds of -160.
