document.write( "Question 7758: Runner A is initially 1.0 mi west of a flagpole and is running with a constant velocity of 2.0 mi/h due east. Runner B is initially 3.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far are the runners from the flagpole when they meet?\r
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document.write( "My work:
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document.write( "4mi=7t\r
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document.write( "t=4/7\r
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document.write( "t=.5714 (not right answer) \n" );
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Algebra.Com's Answer #4295 by prince_abubu(198)![]() ![]() ![]() You can put this solution on YOUR website! If runner A is 1.0 mi west of the flagpole and runner B is 3.0 mi east of the flagpole, they would be 4 miles apart when the clock starts ticking. This means that the sum of the distances they travel must add up to 4 miles once they meet. They will also spend the same exact time running from clock start to the very moment they meet.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "OK. Distance that runner A travels at time t is 2t. Distance that runner B travels is 5t. 2t + 5t = 4 miles. so 7t = 4, tthen t = 4/7 of an hour. They will meet after 4/7 of an hour, or roughly after 34 minutes. But let's stick with 4/7 hour.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We're not done yet. How far from where runner A started did they meet? We'll have to plug in the 4/7 to the 2t. That gives us 8/7, or 1 1/7 of a mile. If runner A was 1 mile from the flagpole to begin with, and meets runner B 1 1/7 mile after he passed the flagpole, then they would be 1/7 of a mile away from the flagpole at the exact time they meet. \n" ); document.write( " |