Rhamondre Stevenson named AFC offensive player of the week
Our Summary
Rhamondre Stevenson of the New England Patriots has been named the AFC Offensive Player of the Week following his standout performance against the Miami Dolphins in Week 18. Stevenson contributed significantly to the Patriots securing the No. 2 seed in the playoffs by amassing 153 yards from scrimmage, including 131 rushing yards and 22 receiving yards, along with three touchdowns. His three scrimmage touchdowns were unmatched by any other player in Week 18. This marks Stevenson's first time receiving the Offensive Player of the Week honor. Over the season, he recorded 603 rushing yards, seven rushing touchdowns, and caught 32 passes for 345 yards and two touchdowns in 14 games.
Storyline
See how this story developed over time. Our AI tracks related coverage to show you the full context—from initial reports to latest updates.
Start your mornings informed
Everything important from the NFL, delivered daily at 8 AM ET. No ads. No clickbait. Just 2 minutes.
Monday Morning - Chiefs Trade + 4 More Stories
Good morning! Here's what matters in the NFL today:
🔥 Chiefs Trade for Elite Pass Rusher
Kansas City addresses defensive needs with blockbuster move. Why it matters: Changes playoff dynamics in AFC.
📊 BY THE NUMBERS
47% - Increase in Chiefs' pass rush win rate after trade
⏰ LOOKING AHEAD
Watch for Ravens' response at trade deadline tomorrow...
Continue reading for complete coverage...
What you get every morning
- All 32 teams covered
- 2-minute read guaranteed
- Human-verified news only
- Zero ads, zero tracking
More Stories Await
Dive deeper into NFL coverage and discover what makes The Daily Handoff different.