Joe Pagano

Why Study Math? - A Lesson in Predation - Does the Cheetah Catch the Gazelle?



Posted: Thursday, May 05, 2011

by Joe Pagano
Math by Joe

Imagine that you are a hungry cheetah grazing out on the savannah in Africa. The sun is blazing overhead at a cool 110 degrees Fahrenheit and because of the recent heat you have been unable to locate any prey to eat because they have all been hiding out in some cool shade somewhere. Suddenly off in the distance you espy a delicious looking gazelle. You know it is hot, but you have one last burst of energy. You know you can run up to 60 miles per hour. The catch looks possible. The question is, do your calculations make sense? because if they don't, you don't get to eat once again.

How can we use basic algebra to solve the problem of whether the cheetah catches the gazelle? This is a classic example of how an understanding of basic math affords one the ability to solve all kinds of problems related to the real world. After all, it may be a problem in which you are not determining whether the cheetah in fact eats, but whether you yourself do!

To solve the problem at hand, we use some basic information and some basic algebra. A gazelle can run 73 feet per second, and it can sustain this speed for several minutes before shutting down. The cheetah, the fastest land animal on earth, is simply built for speed, and can run up to 88 feet per second. Reaching this top speed, however, is physiologically taxing to the cheetah; for this reason, this blazing speed can only be maintained for 20 seconds, after which, the cheetah must shut down and rest.

Given the above information, the question becomes, how close must the gazelle be to the cheetah in order that dinner be served? Let's set up the problem as follows. Let x represent the distance between the cheetah and gazelle when the chase begins. Since distance is equal to rate times time, in formula, d = rt, we have in 20 seconds the cheetah will run 88*20 or 1760 feet; and the gazelle will run 73*20 or 1460 feet. To solve for x, we have 88*20 = 73*20 + x. Solving, we have that x = 300 feet. We can also see that the cheetah will run 300 more feet during the 20 seconds so that the safe distance away for the gazelle will be at least 300 feet.

What is interesting about the given problem is that gazelles know instinctively what their safety net is and will not start running unless they know they need to widen their safety distance. Who said gazelles weren't smart? I guess it goes to show that when survival is on the line, we all become smarter.

Once again, we see how a little bit of algebra can help determine the answer to a real world problem; in this case, whether dinner is or is not served-for the cheetah that is. Now wouldn't it be nice if all dinners were that simple!

Joe is a prolific writer of self-help and educational material and is the creator and author of over a dozen books and ebooks which have been read throughout the world. He is a former teacher of high school and college mathematics and has recently returned as a professor of mathematics at a local community college in New Jersey.
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