document.write( "Question 659306: A truck leaves the dock at 10:00 am going 69 miles per hour. A second truck leaves two hours later going 82 mph. At what time will the two trucks meet up \n" ); document.write( "

Algebra.Com's Answer #410721 by josmiceli(19441) $\"\"$ $\"About$

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The 1st truck has a head start of
\n" ); document.write( " $\"+d%5B1%5D+=+69%2A2+\"$
\n" ); document.write( " $\"+d%5B1%5D+=+138+\"$ mi
\n" ); document.write( "----------------
\n" ); document.write( "Now start a stopwatch when the 2nd truck leaves
\n" ); document.write( "Both trucks will travel for the same amount
\n" ); document.write( "of time until they meet
\n" ); document.write( "Let $\"+d+\"$ = distance 2nd truck travels
\n" ); document.write( "Let $\"+t+\"$ = time on stopwatch
\n" ); document.write( "------------------
\n" ); document.write( "1st truck's equation:
\n" ); document.write( " $\"+d+-+138+=+69t+\"$
\n" ); document.write( "2nd truck's equation:
\n" ); document.write( " $\"+d+=+82t+\"$
\n" ); document.write( "-----------
\n" ); document.write( " $\"+d+=+69t+%2B+138+\"$
\n" ); document.write( "By substitution:
\n" ); document.write( " $\"+82t+=+69t+%2B+138+\"$
\n" ); document.write( " $\"+13t+=+138+\"$
\n" ); document.write( " $\"+t+=+10.6154+\"$ hrs
\n" ); document.write( " $\"+.6154%2A60+=+36.9231+\"$
\n" ); document.write( " $\"+.9231%2A60+=+55.4+\"$
\n" ); document.write( "---------------------
\n" ); document.write( "10 hrs 36 min and 55 sec is the time elapsed
\n" ); document.write( "after 10 AM + 2 hrs or 12 PM
\n" ); document.write( "They will meet at 10:36:55 PM
\n" ); document.write( " \n" ); document.write( "

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