document.write( "Question 1077070: CAR A TRAVELING AT 35 MILES PER HOUR. CAR B IS 200 FEET BEHIND CAR A. AT WHAT SPEED MUST CAR B TRAVEL TO CATCH UP TO CAR A?
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #691631 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
If car B travels at \"+35+\" mi/hr, it will never catch up
\n" ); document.write( "to car A -it will always be \"+200+\" ft behind, so any
\n" ); document.write( "speed greater than \"+35+\" mi/hr will allow it to catch up
\n" ); document.write( "----------------------
\n" ); document.write( "Suppose B's speed = \"+45+\" mi/hr
\n" ); document.write( "Let \"+d+\" = distance that B travels until it catches A
\n" ); document.write( "Let \"+t+\" = time in hrs for B to catch A
\n" ); document.write( "Convert \"+200+\" ft to miles: \"+200%2F5280+\"
\n" ); document.write( "---------------------------------------
\n" ); document.write( "Equation for A:
\n" ); document.write( "(1) \"+d+-+200%2F5280+=+35t+\"
\n" ); document.write( "Equation for B:
\n" ); document.write( "(2) \"+d+=+45t+\"
\n" ); document.write( "-------------------------
\n" ); document.write( "Plug (2) into (1)
\n" ); document.write( "(1) \"+d+-+200%2F5280+=+35t+\"
\n" ); document.write( "(1) \"+45t+-+200%2F5280+=+35t+\"
\n" ); document.write( "(1) \"+10t+=+200%2F5280+\"
\n" ); document.write( "(1) \"+t+=+20%2F5280+\"
\n" ); document.write( "(1) \"+t+=+.003788+\" hrs
\n" ); document.write( "Convert hrs to sec
\n" ); document.write( "\"+.003788%2A3600+=+13.64+\"
\n" ); document.write( "B will catch A in 13.64 sec
\n" ); document.write( "--------------------------
\n" ); document.write( "Plug back into (2) to find \"+d+\"
\n" ); document.write( "Hope this helps
\n" ); document.write( "
\n" ); document.write( "
\n" );