document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1124495>Question 1124495</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The stopping distance of a car after the brakes have been applied varies directly as the square of the speed.\r<BR>\n" );
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document.write( "A.If a car traveling 60mph can stop in 200ft.,how far will it take a car to stop if it is traveling at 50mph\r<BR>\n" );
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document.write( "B.Suppose the car was able to stop in 50ft. At what rate was its traveling?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #740797 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=740797\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> d=k*s^2<BR>\n" );
document.write( "200=k*60^2<BR>\n" );
document.write( "200/3600=k<BR>\n" );
document.write( "k=1/18 or 0.057\r<BR>\n" );
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document.write( "50=k*s^2<BR>\n" );
document.write( "50/(1/18)=s^2<BR>\n" );
document.write( "900=s^2<BR>\n" );
document.write( "s=30 mph\r<BR>\n" );
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document.write( "This makes sense, because as the speed doubles (30 to 60) the stopping distance should increase by a factor of 4 (50 to 200). </SPAN>\n" );
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