LANDOVER, MD - DECEMBER 22: Saquon Barkley #26 of the New York Giants is shoved out of bounds by Kayvon Webster #38 of the Washington Redskins during overtime at FedExField on December 22, 2019 in Landover, Maryland.
Scott Taetsch/Getty Images

Saquon Barkley has been named the NFC Offensive Player of the Week for his efforts in the New York Giants’ Week 16 win over Washington.

This past Sunday, Saquon Barkley played his best game of the year and arguably the best game of his entire career. In a 41-35 overtime win over the Washington Redskins, the New York Giants running back carried the ball 22 times for 189 yards and one touchdown. He additionally caught four balls for 90 yards and one score through the air.

For his efforts, the NFL has named Barkley the NFC Offensive Player of the Week for Week 16.

Barkley’s 189 rushing yards set a new career-high for him in the pros. His previous career-high was 170 rushing yards set in a Week 14 win over the Redskins last season. His 279 yards from scrimmage additionally set a new career-high.

For the majority of the year, Barkley had been dealing with an ankle injury. He suffered a high ankle sprain in the Week 3 win over Tampa Bay, which caused him to miss three games. Despite the ankle healing enough to which he could play, it was noticeable that the injury was slowing him down.

But the last two weeks, Barkley’s shown he’s back to his old self. In the Week 15 win over the Miami Dolphins, the second-year pro carried the ball 24 times for 112 yards and a pair of touchdowns. He also caught four balls for 31 yards.

This is just the second time the league has named Saquon the NFC Offensive Player of the Week. They previously provided that honor to him after his performance during Week 11 of 2018. In a win over the Buccaneers, Barkley carried the ball 27 times for 142 yards and two touchdowns. He also racked up two receptions for 10 yards and one score through the air.

NYY

NYM

NYG

NYJ

NYK

BKN

NYR

NYI

NJD

SJU