The National Football League (NFL) recently announced that Buffalo Bills running back Devin Singletary has been named the NFL Player of the Week for Week 10. This honor is awarded to the player who has the most outstanding performance during a given week.
Singletary, a rookie out of Florida Atlantic University, had a breakout performance in the Bills’ 20-3 win over the Cleveland Browns. He rushed for 95 yards on 15 carries and scored two touchdowns. He also caught three passes for 45 yards. His performance was instrumental in helping the Bills to their sixth consecutive victory.
Singletary is the first Bills player to be named NFL Player of the Week since LeSean McCoy in Week 14 of the 2017 season. He is also the first rookie running back to receive the honor since Kareem Hunt in Week 3 of the 2017 season.
Singletary’s performance has been a major factor in the Bills’ success this season. He has rushed for 568 yards and four touchdowns on the season, and he has also caught 18 passes for 139 yards and one touchdown. He has been a reliable and consistent presence in the backfield for the Bills, and his performance against the Browns was a reminder of why he was a third-round pick in this year’s NFL Draft.
Devin Singletary’s selection as NFL Player of the Week is a testament to his hard work and dedication. He is quickly becoming one of the league’s top running backs, and he is sure to have many more outstanding performances in the future. Congratulations to Devin Singletary on this well-deserved honor!