National Football League

Jaguars' Telvin Smith named AFC Defensive Player of the Week

Published Oct. 22, 2014 10:54 a.m. ET

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Jacksonville Jaguars rookie linebacker Telvin Smith has been named AFC Defensive Player of the Week. The announcement was made Wednesday by the NFL.

Smith, a fifth-round pick out of Florida State, registered a sack, forced fumble, interception and two passes defensed as the Jaguars defeated the Cleveland Browns 24-6 for their first win of the season.

Smith's forced fumble came in the third quarter when he sacked Browns quarterback Brian Hoyer. The ball was recovered by linebacker Paul Posluszny, and the takeaway led to a field goal which increased the Jaguars' lead to 10-6.

In the fourth quarter, Smith picked off Hoyer and returned it 15 yards to set up the game-clinching touchdown. It was the first interception by the Jaguars since the second quarter of their season opener Sept. 7.

Smith became the first AFC rookie to win a player of the week award in 2014 and second overall (cornerback Kyle Fuller, Chicago, Week 2).

You can follow Ken Hornack on Twitter @HornackFSFla or email him at khornack32176@gmail.com.

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