document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1091781>Question 1091781</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A car is involved in a single vehicle accident and the tyre marks were more than 75 metres long. Under road conditions the distance d ( in metres). it takes a car travelling at v m/s to stop is approximated by the equation d= 0.05v^2 + v . The accident occurred in a 100 km/h speed zone. was there a likelihood of the driver exceeding this speed limit ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #706324 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=706324\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If d=0.05vē+v, then:<BR>\n" );
document.write( "75=0.05vē+v<BR>\n" );
document.write( "0.05vē+v-75=0<BR>\n" );
document.write( "5vē+100v-7500=0<BR>\n" );
document.write( "vē+20v-1500=0<BR>\n" );
document.write( "(v+50)(v-30)=0<BR>\n" );
document.write( "v=30 or -50<BR>\n" );
document.write( "The car was going 30 m/s, or 108 kmh<BR>\n" );
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