Record crowds expected for Independence Day travel

AAA says 2.3 million Georgians are likely to take trips, most of them 50-plus miles by car.
Fireworks burst into air during the July 4th celebration in Marietta. (Michael Blackshire/Michael.blackshire@ajc.com)

Credit: Michael Blackshire

Credit: Michael Blackshire

Fireworks burst into air during the July 4th celebration in Marietta. (Michael Blackshire/Michael.blackshire@ajc.com)

A record number of people are expected to travel during the Fourth of July holiday period, according to the latest forecast from AAA.

The auto club forecasts that more than 70.9 million people in the United States will travel at least 50 miles for the Independence Day period from June 29 through July 7. That includes 2.3 million people in Georgia taking trips for the holiday.

Across the country, “There will be 3.5 million more travelers than last year. That means even more people at airports and popular attractions like beaches, lakes, and theme parks,” said AAA Vice President of Travel Debbie Haas in a written statement.

A record number of people are expected to fly — more than 5.7 million nationwide, up nearly 7% from last year, and 12% higher than 2019 levels. The air traffic is expected to include more than 161,700 people in Georgia. Travelers should prepare for crowded airports and planes and the possibility of long lines and airport parking shortages.

But the vast majority of people traveling at least 50 miles are going by car. More than 2 million people in Georgia are expected to take a road trip, setting a record for the third consecutive year.

“Road travelers should prepare for congestion in the afternoon and evening hours, particularly near larger metro areas, theme parks, and popular attractions,” said AAA spokeswoman Montrae Waiters in a written statement.

On the highways, the worst traffic delays are expected on Wednesday, July 3, and Sunday, July 7, according to data firm INRIX.