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The deadliest holiday is the 4th of July

On Behalf of | Jul 2, 2015 | Car Accidents

A lot of holidays, for all their fun and festivity, turn out to be deadly for those in Illinois. New Year’s Eve is famous for this, but it’s not even the deadliest holiday in the United States. That distinction falls to Independence Day, on the 4th of July.

When looking at an entire decade, starting in 2000 and running to 2009, no day ended up with more deaths on the road than the 4th. It is true that more drunk driving deaths happen on New Year’s Day, but overall deaths happen more in the summer.

Alcohol still played a large part in those accidents, as half of the accidents involved at least one drunk driver. The majority of the accidents also take place during the night, making it one of the worse times to be on the road.

The problem, to a large degree, is that people spend all day celebrating the 4th, having barbeques and often drinking alcohol. By the end of the night, when everyone heads home, the roads become very hazardous.

The police have been stepping up their efforts to make things safer. They have even been putting checkpoints in place to catch drivers who are under the influence before they are involved in accidents. Despite this, the number of DUI accidents climbs each and every year, and the 4th continues to have the most accidents of the year.

Have you been hurt in an accident or have you lost a loved one while celebrating the 4th? If so, you must know your potential rights to financial compensation after the accident, whether alcohol was involved or not.

Source: Huliq, “4th of July Independence Day is Deadliest American Holiday,” Anissa Ford, accessed July 02, 2015