document.write( "Question 459862: anybody help me with this 2 questions...having headache....\r
\n" ); document.write( "\n" ); document.write( "a ahip traveling east at 25mph is 10mi from a harbor when another ship leaves the harbor traveling east at 35mph. How long does it take the second ship to catch up to the first ship?\r
\n" ); document.write( "\n" ); document.write( "Two cyclist start at the same time from opposite ends of a course that is 51mi long. one cyclist is riding at a rate of 16mph and the second cyclist is riding at a rate of 18mph. How long after they begin will they meet?
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Algebra.Com's Answer #315375 by mananth(16946)\"\" \"About 
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Ship I speed 25 mph
\n" ); document.write( "Ship II speed 35 mph\r
\n" ); document.write( "\n" ); document.write( "catch up distance = 10 miles
\n" ); document.write( "catchup speed = 35 -25 = 10 mph
\n" ); document.write( "catchup time = catchup distance / catch up speed
\n" ); document.write( "10/10
\n" ); document.write( "=1 hour to catch up\r
\n" ); document.write( "\n" ); document.write( "........
\n" ); document.write( "cyclist I speed - 16 mph
\n" ); document.write( "cyclist II speed =18 mph\r
\n" ); document.write( "\n" ); document.write( "riding towards each other.
\n" ); document.write( "combined speed = 34 mph
\n" ); document.write( "distance = 51 miles
\n" ); document.write( "t=d/r
\n" ); document.write( "51/34
\n" ); document.write( "=1.5 hours
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