document.write( "Question 648934: sue walks at a rate of 10 feet every 5 seconds while tom walks at a rate of 15 feet every 5 seconds. If sue has a head start of 25 feet, after how many seconds will they be at the same spot? \n" ); document.write( "
Algebra.Com's Answer #406888 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
Sue walks at a rate of 10 feet every 5 seconds while tom walks at a rate
\n" ); document.write( " of 15 feet every 5 seconds.
\n" ); document.write( " If sue has a head start of 25 feet, after how many seconds will they be at the
\n" ); document.write( " same spot?
\n" ); document.write( ":
\n" ); document.write( "Change the speeds to ft/sec
\n" ); document.write( "Sue: 10/5 = 2 ft/sec
\n" ); document.write( "Tom: 15/5 = 3 ft sec
\n" ); document.write( ":
\n" ); document.write( "Let t = time (in sec) for T to catch S
\n" ); document.write( ":
\n" ); document.write( "When T catches S, T will have walked 25 more ft than S
\n" ); document.write( "Write a distance equation; dist = speed * time
\n" ); document.write( ":
\n" ); document.write( "3t = 2t + 25
\n" ); document.write( "3t - 2t = 25
\n" ); document.write( "t = 25 sec for Tom to catch Sue
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Confirm this,
\n" ); document.write( "Tom: 3(25) = 75ft
\n" ); document.write( "Sue: 2(25) = 50ft; plus her 25 ft head start = 75 ft \n" ); document.write( "
\n" );