document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1176850>Question 1176850</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The data given below provides the distance required for a particular type of car to stop when traveling at a variety of speeds. A reasonable model for braking distance is d=as^k, where d is distance, s is speed, and a and k are constants. Use the data in the table to find values for a and k, and also type out the equation that models the breaking distance.\r<BR>\n" );
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document.write( "Speed (MPH)     10  20   30   40<BR>\n" );
document.write( "Distance (Feed)5.5  22  49.5  88\r<BR>\n" );
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document.write( "a= <BR>\n" );
document.write( "k=<BR>\n" );
document.write( "Model: </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #804796 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=804796\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 5.5=a10^k<BR>\n" );
document.write( "a=5.5/10^k<BR>\n" );
document.write( "22=a20^k<BR>\n" );
document.write( "a=22/20^k<BR>\n" );
document.write( "set those two equal<BR>\n" );
document.write( "(5.5/10^k)=(22/20^k)<BR>\n" );
document.write( "1/10^k=4/20^k<BR>\n" );
document.write( "(2^k)=4<BR>\n" );
document.write( "k=2<BR>\n" );
document.write( "a=0.055<BR>\n" );
document.write( "-<BR>\n" );
document.write( "check in third where 49.5=0.055^30^2,  which is true, as is the fourth.<BR>\n" );
document.write( "a=0.055<BR>\n" );
document.write( "k=2<BR>\n" );
document.write( "Model is distance = 0.055*speed^2<BR>\n" );
document.write( "d=0.55*s^2<BR>\n" );
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