document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=487931>Question 487931</A>: <I><SPAN STYLE=\"word-break: break-all;\"> one car travels 50 mph. if they start from the same place going the same direction, and the second car leaves an hour after the first car, how long will it take the second car to catch up? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #333193 by </B><A HREF=/tutors/aboutme.mpl?userid=chessace><B>chessace(471)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=chessace><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=chessace><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=333193\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2nd car speed S MPH, then meet at time T (start of 2nd car).<BR>\n" );
document.write( "S*T = 50*(T+1) is the distance they each go.<BR>\n" );
document.write( "T(S-50)=50<BR>\n" );
document.write( "T=50/(S-50) hours.<BR>\n" );
document.write( "Another way of looking at it is that the initial distance apart, 50m<BR>\n" );
document.write( "is decreasing by their relative speed (S-50):<BR>\n" );
document.write( "T=50/(S-50) hours. </SPAN>\n" );
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