document.write( "Question 227082: A car leaves a town traveling at 40 miles per hour. Two hours later a second car leaves traveling the same route and overtakes the first car in 5 hours. How fast was the second traveling? \n" ); document.write( "
Algebra.Com's Answer #168841 by josmiceli(19441)\"\" \"About 
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Assume I have a stopwatch, and I start it
\n" ); document.write( "when the 2nd car leaves
\n" ); document.write( "How far has the 1st car gone when I start
\n" ); document.write( "the stopwatch?
\n" ); document.write( "\"40%2A2+=+80\" mi
\n" ); document.write( "The key to the problem is that, after the
\n" ); document.write( "stopwatch starts, both cars will travel
\n" ); document.write( "for the same amount of time
\n" ); document.write( "Let \"d\" = distance 2nd car must go to
\n" ); document.write( "catch 1st car in mi
\n" ); document.write( "Let \"s\" = speed of 2nd car in mi/hr
\n" ); document.write( "For 1st car:
\n" ); document.write( "(1) \"d+-+80+=+40%2A5\"
\n" ); document.write( "For 2nd car:
\n" ); document.write( "(2) \"d+=+s%2A5\"
\n" ); document.write( "This is 2 equations and 2 unknowns, so it's solvable\r
\n" ); document.write( "\n" ); document.write( "(1) \"d+-+80+=+40%2A5\"
\n" ); document.write( "\"d+=+80+%2B+200\"
\n" ); document.write( "\"d+=+280\" mi
\n" ); document.write( "And, substituting,
\n" ); document.write( "(2) \"d+=+s%2A5\"
\n" ); document.write( "\"280+=+5s\"
\n" ); document.write( "\"s+=+56\"
\n" ); document.write( "The speed of the 2nd car was 56 mi/hr\r
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