Cincinnati Bengals running back Jeremy Hill has been named the AFC Offensive Player of the Week for his performance in Sunday’s 30-0 win at Cleveland, where he rushed 148 yards and two touchdowns.
The rookie from LSU is the first Bengals player to win a weekly award this year, and first since Andy Dalton earned the award in Week 14 last year against Indianapolis.
Hill also is nominated for two other weekly awards.
He is one of three nominees up for the ground portion of the FedEx Air and Ground NFL Players of the Week, along with Seattle’s Marshawn Lynch and Green Bay’s Eddie Lacy.
Fans can vote for that award online through 3 p.m. Thursday at nfl.com/FedEx.
And Hill is one of five nominees for the Pepsi NFL Player of the Week award, joining the New York Giants’ Odell Beckham, Arizona’s Chandler Catanzaro, Baltimore’s C.J. Mosley and Kansas City’s De’Anthony Thomas.
Votes are accepted online through 3 p.m. Friday at nfl.com/rookies.
Monday night Hill will go head to head with the reigning AFC Defensive Player of the Week, Denver’s Aqib Talib, who led the Broncos with a season-high eight tackles (seven solo) to go along with one interception and three passes defensed in a 22-10 victory against San Diego.