New York Giants RB Saquon Barkley was named NFC Offensive Player of the Week award, the NFL announced Wednesday.
The fifth-year running back rushed for 164 yards on 18 carries (9.1-yard avg.), including a 68-yard run that set up his four-yarder for the Giants’ first touchdown of the season. He added a team-high six receptions for 30 yards. Barkley caught a pass from Daniel Jones for the two-point conversion that accounted for the game-deciding points in the Giants’ season-opening 21-20 victory against the Tennessee Titans in Nissan Stadium.
Barkley’s 164 rushing yards and 194 yards from scrimmage both topped all NFL players on Kickoff Weekend.
“It’s great, but any individual award is really a team award,” he said about winning the award. “I don’t get that without the help of my offensive linemen, wide receivers blocking down the field, (Quarterback) Daniel (Jones) making great calls, and (Offensive Coordinator Mike) Kafka making great calls. At the end of the day, the most important thing last week was getting the win, and we’ve got to shift to Carolina.”
The 164 rushing yards was the third-highest total of his career and the most he’s had in a game outside FedExField. Barkley ran for a career-high 189 yards on Dec. 22, 2019, and 170 yards on Dec. 9, 2018. Both of those games were in Washington.
Barkley’s 68-yard run was on the first play of the third quarter. It is his fourth 68-yard run, tied for the second-longest of his career.
This is Barkley’s third Player of the Week award; he is the only Giants running back to be honored since Ahmad Bradshaw in 2011.
Barkley and the Giants are hoping to get win number two as New York(1-0) battles the Panthers on Sunday in their home-opener.