document.write( "Question 132952This question is from textbook Applied College Algebra
\n" ); document.write( ": Running at an average rate of 6 meters per second, a sprinter ran to the end of a track. The sprinter then jogged back to the starting point at an average rate of 2 meters per second. The total time for the sprint and the jog back was 2 minutes 40 seconds. Find the length of the track. \n" ); document.write( "
Algebra.Com's Answer #97150 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
\n" ); document.write( "Let L=length of track
\n" ); document.write( "Time spent running to end of track= L/6
\n" ); document.write( "Time spent jogging back=L/2\r
\n" ); document.write( "\n" ); document.write( "And we are told that these two times add up to 2min 40sec or 160sec (let's deal in seconds so we don't get confused). So:\r
\n" ); document.write( "\n" ); document.write( "L/6 + L/2=160 multiply each term by 6
\n" ); document.write( "L+3L=960
\n" ); document.write( "4L=960 divide both sides by 4
\n" ); document.write( "L=240 meters------------------------length of track\r
\n" ); document.write( "\n" ); document.write( "(Note in the equation above, both L/6 and L/2 are expressed in (meters/meters/sec)or sec). \r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "240/6+240/2=160 sec
\n" ); document.write( "40+120=160
\n" ); document.write( "160=160\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor
\n" ); document.write( " \n" ); document.write( "
\n" );