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" );
document.write( "\n" );
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 #574615 by </B><A HREF=/tutors/aboutme.mpl?userid=macston><B>macston(5194)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=macston><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=macston><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=574615\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> At 8.4 m/s, the first runner will finish the 100 m leg in (100/8.4)sec. Since he has a 0.550 second start, he has (100/8.4)-0.550 sec left to finish. The second runner must run 100 m /((100/8.4)-.550) to tie the first runner. So the second runner must run at 8.81 m/s   </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );