Traveling BusesDate: 06/01/97 at 03:26:21 From: Richard Ramsey Subject: Uniform motion both equal and unequal distances Here is my problem. A southbound bus left Fort Walton Beach at 9 AM. Two hours later a northbound bus left the same station. If the buses traveled at the same rate and were 352 kilometers apart at 2 PM, find the rate of the buses. Here is a sample problem where they travel unequal distances - please help me! At 8 PM Achilles left camp and headed south at 20 kilometers per hour. At 10 PM, Patrocles headed south from the same camp. If Patrocles was 50 kilometers ahead by 3 AM, what was his speed? Date: 06/01/97 at 05:30:27 From: Doctor Mitteldorf Subject: Re: uniform motion both equal and unequal distances Dear Richard, The key step in any word problem is to take the words in the problem and write them as an equation. It takes understanding and experience. Practice and lots of examples are the only ways I know to get good at this. In the first problem, it says the two rates were the same. We'll let that be our variable, and we'll call it v for velocity. This problem happens over the course of 5 hours (9 to 2) and it's divided into two parts: in the first 2 hours, there's one bus moving and one standing still; in the last 3 hours, both buses are moving but in opposite directions. During the first 2 hours, the separation increases by 2*v: speed x time = distance. During the last 3 hours, each bus covers a distance of 3*v. We also know that the buses are 352 km apart after five hours, so the sum of the distances traveled should be equal to 352 km, right? The equation that comes out of all this is: 2*v + 3*v + 3*v = 352 8*v = 352 v = 352/8 v = 44 km/hr The second problem sounds similar, but you really have to start over again and think it through. What is it you want to know here? Achilles's path is completely known from the start. He travels for a total of 7 hours at a speed of 20 kph, so he goes 140 miles all together. Patrocles goes 50 km further, or 190 km all together. He makes the 190 km trip in only 5 hours, since he started at 10 PM, so he must be traveling at 190/5 = 38 kph. For more about this kind of word problem, see "Two Trains" in the Dr. Math FAQ: http://mathforum.org/dr.math/faq/faq.two.trains.html -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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