document.write( "Question 483206: a fishing boat left dock 24 and traveled south at an average speed of 18 mph. A cruise ship left some time later traveling in the same direction at an average speed of 30 mph. After traveling for 6 hours the cruise ship caught up with the fishing boat. How long did the fishing boat travel before the cruise ship caught up?
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Algebra.Com's Answer #330713 by Maths68(1474)\"\" \"About 
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Ship’s Data
\n" ); document.write( "Speed = 30mph
\n" ); document.write( "Traveled time = 6 hours
\n" ); document.write( "Distance = Time * Speed
\n" ); document.write( "Distance = 6*30
\n" ); document.write( "Distance = 180m\r
\n" ); document.write( "\n" ); document.write( "Ship covers 180m to catch up with the boat.\r
\n" ); document.write( "\n" ); document.write( "Therefore boat has also traveled 180m.\r
\n" ); document.write( "\n" ); document.write( "Boat Data
\n" ); document.write( "Speed = 18mph
\n" ); document.write( "Time = ?
\n" ); document.write( "Distance =180m
\n" ); document.write( "Distance =Time traveled * Speed
\n" ); document.write( "180=Time * 18
\n" ); document.write( "Divide by 18 both sides
\n" ); document.write( "180/18=(Time *18)/18
\n" ); document.write( "10=Time
\n" ); document.write( "Time traveled= 10 hours\r
\n" ); document.write( "\n" ); document.write( "The fishing boat traveled for 10 hours.
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