Travis Kelce Named AFC Offensive Player of the Week

Kelce racked up a career-best 191 yards in last week’s win over Los Angeles

The best players always seem to step up in the biggest of moments, and that's exactly what tight end Travis Kelce did last Thursday in what turned out to be a critical victory for the Kansas City Chiefs.

Kelce racked up a career-best 191 yards and two touchdowns in the win, helping Kansas City defeat Los Angeles and seize a two-game lead in the AFC West. The All-Pro tight end was right in the middle of a furious rally that featured three touchdown drives – each of which covered 75 yards – between the fourth quarter and overtime.

He hauled in a 69-yard reception that set up a 1-yard, game-tying scoring strike between quarterback Patrick Mahomes and wide receiver Tyreek Hill midway through the fourth quarter before catching a 7-yard touchdown with just over a minute remaining to help force overtime. He then caught a 34-yard strike from Mahomes during the extra period that ended the game and secured the victory.

It was simply a herculean effort in a huge moment, and it caught the attention of the NFL. Kelce was named the AFC "Offensive Player of the Week" on Wednesday for his performance, marking the first time that a Chiefs' tight end has ever won the award. Additionally, Kelce is the first tight end to earn "Player of the Week" honors since Rob Gronkowski did so in 2017.

Kelce's big performance also helped him surpass 1,000 receiving yards for a sixth-consecutive campaign, making him just the 19th player – and first tight end – to record that many 1,000-yard seasons in a row.

The man is epitomizing greatness before our eyes, and Thursday's performance was just another example of it.

Advertising