{"id":76,"date":"2023-10-30T00:14:03","date_gmt":"2023-10-30T00:14:03","guid":{"rendered":"https:\/\/ahogan.org\/?page_id=76"},"modified":"2026-11-05T04:13:00","modified_gmt":"2026-11-05T04:13:00","slug":"car-lengths","status":"publish","type":"page","link":"https:\/\/ahogan.org\/index.php\/car-lengths\/","title":{"rendered":"Car Lengths"},"content":{"rendered":"

Exercise<\/span><\/h2>\n

You are driving on the highway at a speed of 60 mph. A stationary obstacle becomes visible ahead and you need to stop. Under the assumptions below, how far will you travel before coming to rest? How many car lengths is that? How does the answer change when there is less friction between the tires and the road?<\/p>\n

 <\/p>\n

Pre-requisites<\/span><\/h2>\n

This exercise uses speed, acceleration, force, and friction. A speed of 60 mph is 88 feet per second, and gravitational acceleration near Earth\u2019s surface is approximately 32 feet per second per second. Assume a level road and allow 2.5 seconds to perceive the obstacle and react. For this simplified calculation, assume constant braking deceleration begins immediately afterward, neglecting the time needed for braking force to build. Model the braking as a locked-wheel skid with a coefficient of friction of 0.4, then compare it with a lower-friction case of 0.2. These are illustrative assumptions, not fixed values for every vehicle or road surface.<\/p>\n

 <\/p>\n

Facts<\/span><\/h2>\n

When an obstacle becomes visible, the driver needs time to perceive it, decide what to do, and move a foot to the brake pedal. For this exercise, we allow 2.5 seconds for that process. At 88 feet per second, the car travels 220 feet before braking begins. Actual perception and reaction can take less or more time.<\/p>\n

Once braking begins, the car continues moving while friction reduces its speed. In our simplified model, a coefficient of friction of 0.4 produces a braking distance of 302.5 feet.<\/p>\n

Adding the 220 feet traveled before braking to the 302.5 feet traveled during braking gives a total stopping distance of 522.5 feet.<\/p>\n

For the lower-friction case, use a coefficient of 0.2. The braking distance doubles to 605 feet, while the distance traveled before braking remains 220 feet. The total stopping distance is therefore 825 feet.<\/p>\n

Taking one car length to be 18 feet, these totals are approximately 29 car lengths for the first case and 46 car lengths for the lower-friction case.<\/p>\n

Rain, snow, ice, tires, and other conditions affect the available friction. The two coefficients illustrate how the calculation changes; they do not assign one stopping distance to all dry roads or another to all wet or snowy roads.<\/p>\n

 <\/p>\n

 <\/p>\n

Thoughts and Opinions<\/span><\/h2>\n

An unspecified amount of time ago, I was involved in a single-car accident with some of these same conditions. What was different was I did not stop, and I did strike the object.<\/p>\n

Why might the gap needed behind another moving car be shorter than the distance needed to stop before a stationary object?<\/p>\n

The stationary object matters. It does not move farther down the road while you react and brake. The distances calculated above describe stopping before that object under our stated assumptions.<\/p>\n

When I was in my accident, I struck a large rock (about 3 feet tall, a few hundred pounds of demolished asphalt). It was road debris that fell out of the back of a truck. Because it was 5am and dark, it blended in with the road and I did not react at all, instead striking the stationary rock at a speed of at least 88 feet per second head-on. It tore a hole in the engine. I am blessed to be alive today to share these details.<\/p>\n

Had the rock become visible far enough ahead, there might have been time to stop. Under the first set of assumptions in this exercise, perceiving it, reacting, and braking to a stop would require about 523 feet, or 29 car lengths. That calculation illustrates the distance involved; it cannot establish exactly what would have happened in my accident.<\/p>\n

Now imagine two cars traveling at the same speed. Suppose they brake in exactly the same way, but the following driver begins braking 2.5 seconds after the leading driver. Because their braking distances are equal, the following car closes the gap by the distance it travels during that delay: 220 feet at 60 mph. This is an idealized comparison, with no additional margin.<\/p>\n

Real traffic does not guarantee equal braking performance or a clear road ahead. A leading vehicle may stop more abruptly, strike an obstacle, or move aside to reveal something stationary. The idealized comparison therefore cannot supply a universal following distance.<\/p>\n

For practical following guidance, the New Jersey Driver Manual<\/a> recommends the three-seconds-plus rule. Choose a fixed roadside object. When the vehicle ahead passes it, begin counting; your vehicle should reach it at least three seconds later in good conditions. Increase the interval to four or more seconds in bad weather, with additional space and reduced speed as conditions require. At 60 mph, three seconds corresponds to 264 feet\u2014about 15 of our assumed car lengths\u2014but measure the following gap in seconds rather than relying on a fixed number of car lengths. No single interval guarantees that every collision can be avoided.<\/p>\n

 <\/p>\n

Feel free to email me with your thoughts and opinions on this matter at\u00a0mr@ahogan.org<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Exercise You are driving on the highway at a speed of 60 mph. A stationary obstacle becomes visible ahead and you need to stop. Under the assumptions below, how far will you travel before coming to rest? How many car lengths is that? How does the answer change when there is less friction between the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-76","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/ahogan.org\/index.php\/wp-json\/wp\/v2\/pages\/76","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ahogan.org\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/ahogan.org\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/ahogan.org\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ahogan.org\/index.php\/wp-json\/wp\/v2\/comments?post=76"}],"version-history":[{"count":19,"href":"https:\/\/ahogan.org\/index.php\/wp-json\/wp\/v2\/pages\/76\/revisions"}],"predecessor-version":[{"id":243,"href":"https:\/\/ahogan.org\/index.php\/wp-json\/wp\/v2\/pages\/76\/revisions\/243"}],"wp:attachment":[{"href":"https:\/\/ahogan.org\/index.php\/wp-json\/wp\/v2\/media?parent=76"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}