Fins’ Bush Named AFC Player Of The Week

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Reggie Bush

MIAMI GARDENS, FL – SEPTEMBER 16: Reggie Bush #22 of the Miami Dolphins rushes during a game against the Oakland Raiders at Sun Life Stadium on September 16, 2012 in Miami Gardens, Florida. (Photo by Mike Ehrmann/Getty Images)

Miami Dolphins

MIAMI (CBSMiami) – After running for a career high 26 times for 172 yards and 2 touchdowns, Miami Dolphins running back Reggie Bush has been named the AFC Offensive Player of the Week.

Bush was key to the Dolphins picking up a victory over the Raiders and took much of the offensive burden off the shoulders of rookie quarterback Ryan Tannehill. Bush had two spectacular touchdown runs of 65 yards and 23 yards.

Bush’s rush of 23 yards featured him keeping his balance after a hit and bouncing to the goal line. His rush of 65 yards featured him avoiding several missed tackles on the way to the end zone. With the 65 yard run, he became the first Dolphin to ever run for two touchdowns of over 65 yards.

It’s the second time Bush has been named player of the week as a member of the Miami Dolphins and the third in his career. He was previously selected for his performance against Buffalo in Week 15 in 2011 and after a Week 13 game in 2006 as a member of the New Orleans Saints.

Bush is the first Dolphin to be named a player of the week on the young 2012 NFL season.

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