document.write( "Question 433975: A long distance runner started on a course running at an average speed of 6 mph. one-half hour later, a second runner began the same course at an average speed of 7mph. How long after the second runner started did the second runner overtake the first runner? \n" ); document.write( "

Algebra.Com's Answer #853261 by MathTherapy(10587) $\"\"$ $\"About$

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document.write( "A long distance runner started on a course running at an average speed of 6 mph. one-half hour later, a second runner began the\r\n" );
document.write( "same course at an average speed of 7mph.  How long after the second runner started did the second runner overtake the first runner?\r\n" );
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document.write( "Let time taken by FASTER runner to get to meet up point, be T\r\n" );
document.write( "With FASTER runner's speed being 7 mph, DISTANCE FASTER runner covers, to get to meet-up point = 7T \r\n" );
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document.write( "Since SLOWER runner started one-half ()hour before FASTER runner, then time slower runner takes to get to meet-up point = T + \r\n" );
document.write( "With SLOWER runner's speed being 6 mph, DISTANCE SLOWER runner covers, to get to meet-up point = 6(T + ) \r\n" );
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document.write( "When both got to meet-up point, they'd covered the same distance\r\n" );
document.write( "As such, we get the following DISTANCE equation: 7T = 6(T + )\r\n" );
document.write( "                                                 7T = 6T + 3\r\n" );
document.write( "                                            7T - 6T = 3\r\n" );
document.write( "Time FASTER runner takes to get to meet-up point/catch up with SLOWER runner, or T = 3 hours

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