document.write( "Question 150990: suzy can run 4m/sec and tim can run 6m/sec. How far ahead of tim must suzy be to not to fall behind tim in the first 10 secondsof running? Use a graph to check ou answer? \n" ); document.write( "

Algebra.Com's Answer #110991 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let x=distance ahead of Tim that Suzy must be to not to fall behind tim in the first 10 secondsof running\r
\n" ); document.write( "\n" ); document.write( "Distance Tim runs in the first 10 seconds=(6m/s)*10 sec=60 meters\r
\n" ); document.write( "\n" ); document.write( "Distance Suzy runs in the first 10 seconds plus her head start of x meters equals 4m/s*10 sec + x=40 meters plus x
\n" ); document.write( "Now we know that when the above two distances are equal, Tim will have caught up and in the next instance, Suzy will have fallen behind, so our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "60=40+x subtract 40 meters from each side\r
\n" ); document.write( "\n" ); document.write( "60-40=x
\n" ); document.write( "x=20 meters--------------------distance ahead that Suzy must be so as not to fall behind in the first 10 seconds\r
\n" ); document.write( "\n" ); document.write( "MAYBE YOU CAN DO THE GRAPH \r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "20+4*10=60
\n" ); document.write( "60=60\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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