Tom Brady Earns AFC Player Of The Week For Performance In Hometown San Francisco
BOSTON (CBS) -- For the first time in his 17-year career, Tom Brady got to make a trip to play the San Francisco 49ers, the team he cheered for while growing up. He came away not only with a victory, but also with an accolade from the NFL.
Brady was named on Wednesday as the AFC Offensive Player of the Week for his 280-yard, four-touchdown, zero-interception performance against the 49ers.
For Brady, it's the 27th time he's earned player of the week honors, which ties him with Peyton Manning for the most in NFL history.
Brady grew up just down the road in San Mateo, famously attending the NFC Championship Game when Dwight Clark made "The Catch." In a quirk of scheduling circumstance, he never got the chance to play in San Francisco, as the Patriots' only trip there during his tenure came in 2008, when he was injured. He did previously play the 49ers twice at home, winning in 2005 and losing in 2012. He also played in nearby Oakland in 2002 and 2011.
It was the 21st time in Brady's career that he finished a game with at least four touchdowns and zero interceptions.
Previously this season, Brady earned player of the week honors in his first game of the year, which was Week 5 in Cleveland. He also earned AFC Offensive Player of the Month for his work in October.
Brady and the 8-2 Patriots will travel to New Jersey to take on the Jets this Sunday afternoon.
