document.write( "Question 939886: Two persons are to run a race but one can run 12 meters per second, whereas the other can run 6 meters per second. If the slower runner has a 93- meter head start,how long will it be before the faster runner catches the slower runner, if they begin at the same time? \n" ); document.write( "

Algebra.Com's Answer #572756 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Start a stop watch when they both start running
\n" ); document.write( "Let $\"+t+\"$ = the time in seconds on the stopwatch
\n" ); document.write( "when the faster runner catches the slower runner
\n" ); document.write( "Let $\"+d+\"$ = the distance the faster runner
\n" ); document.write( "travels until they meet
\n" ); document.write( "---------------------------------------
\n" ); document.write( "Slower runner's equation:
\n" ); document.write( "(1) $\"+d+-+93+=+6t+\"$
\n" ); document.write( "Faster runner's equation:
\n" ); document.write( "(2) $\"+d+=+12t+\"$
\n" ); document.write( "----------------------
\n" ); document.write( "Substitute (2) into (1)
\n" ); document.write( "(1) $\"+12t+-+93+=+6t+\"$
\n" ); document.write( "(1) $\"+6t+=+93+\"$
\n" ); document.write( "(1) $\"+t+=+15.5+\"$
\n" ); document.write( "The faster runner catches the slower runner in 15.5 sec
\n" ); document.write( "check:
\n" ); document.write( "(1) $\"+d+-+93+=+6%2A15.5+\"$
\n" ); document.write( "(1) $\"+d+=+93+%2B+93+\"$
\n" ); document.write( "(1) $\"+d+=+186+\"$ m
\n" ); document.write( "and
\n" ); document.write( "(2) $\"+d+=+12%2A15.5+\"$
\n" ); document.write( "(2) $\"+d+=+186+\"$ m
\n" ); document.write( "OK\r
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