document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=942534>Question 942534</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a 4 x 100 relay race (each leg 100 meters long), the first place runner has a 0.550 second lead and is running at a constant speed of 8.40 m/s.  What minimum speed must the second runner have in order to catch up with the first runner by the end of their leg of the race?  \r<BR>\n" );
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document.write( "I know that v=d/t and that the formulas will be equal to each other because they finish the leg of the race together.  but I'm confused if I add the 0.55 second lead to the constant speed of the first runner or adjust the distance?  I drew a picture and I can visually see the problem, just confused about how to deal with the .55 second lead.  Thanks in advance for your help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #574630 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=574630\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> In a 4 x 100 relay race (each leg 100 meters long), the first place runner has a 0.550 second lead and is running at a constant speed of 8.40 m/s. What minimum speed must the second runner have in order to catch up with the first runner by the end of their leg of the race? <BR>\n" );
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document.write( "let x=increase in speed of second runner<BR>\n" );
document.write( "8.40+x=catch-up speed of second runner<BR>\n" );
document.write( "travel time=distance/speed<BR>\n" );
document.write( "..<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=100%2F8.40-100%2F%288.40%2Bx%29=0.55 ALIGN=MIDDLE  ALT=\"100%2F8.40-100%2F%288.40%2Bx%29=0.55\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "lcd:8.4(x+8.4)<BR>\n" );
document.write( "840+100x-840=.55(8.4x+70.56)<BR>\n" );
document.write( "100x=4.62x+38.808<BR>\n" );
document.write( "95.38x=38.808<BR>\n" );
document.write( "x=0.41<BR>\n" );
document.write( "8.4+x=8.81<BR>\n" );
document.write( "What minimum speed must the second runner have? 8.81 m/s </SPAN>\n" );
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