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J.J. Watt named Defensive Player of the Week for 7th time

J.J. Watt reacts after Brett Maher of the Dallas Cowboys kicks a 45 yard field goal to tie the game in the fourth quarter at NRG Stadium on October 7, 2018 in Houston, Texas.
J.J. Watt reacts after Brett Maher of the Dallas Cowboys kicks a 45 yard field goal to tie the game in the fourth quarter at NRG Stadium on October 7, 2018 in Houston, Texas. (2018 Getty Images)

HOUSTON – Houston Texans defensive end J.J. Watt was named the American Football Conference’s Defensive Player of the Week on Wednesday.

The Texans thumped the Tennessee Titans at NRG Stadium on Monday with a 34-17 win.

Watt had nine total tackles, a solo sack for a loss of eight yards, a forced fumble, two quarterback hits and one tackle for loss. He split another sack with defensive end Christian Covington for a loss of four yards.

Watt now leads the AFC with 11.5 sacks this season and is tied for second in the NFL.

Since entering the NFL in 2011, Watt has netted a total of 87.5 sacks. He has also recorded more sacks against the Titans than any opponent since 1982.

This is Watt’s seventh Defensive Player of the Week award in his NFL career and his first since 2015.


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