document.write( "Question 746316: A ship leaves a port at 12:00 Noon and sails East at speed of 10 miles/hour. Another ship leaves
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document.write( "the same port at 1:00 PM and sails North at a speed of 20 miles/hour. At what time are the two
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document.write( "ships going to be 50 miles apart from each other? (Hint: Distance = Speed * Time) \n" );
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Algebra.Com's Answer #454404 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! A ship leaves a port at 12:00 Noon and sails East at speed of 10 miles/hour. \n" ); document.write( " Another ship leaves the same port at 1:00 PM and sails North at a speed of 20 miles/hour. \n" ); document.write( "At what time are the two ships going to be 50 miles apart from each other? \n" ); document.write( ": \n" ); document.write( "Let t = sailing time of the Northbound ship (From 1 PM) \n" ); document.write( "then \n" ); document.write( "(t+1) = sailing time of the Eastbound \n" ); document.write( ": \n" ); document.write( "This is a pythag problem a^2 + b^2 = c^2; where: \n" ); document.write( "a = 20t \n" ); document.write( "b = 10(t+1) \n" ); document.write( "c = 50 mi \n" ); document.write( "(20t)^2 + (10t+10)^2 = 50^2 \n" ); document.write( "400t^2 + 100t^2 + 200t + 100 = 2500 \n" ); document.write( "A quadratic equation \n" ); document.write( "500t^2 + 200t + 100 - 2500 = 0 \n" ); document.write( "500t^2 + 200t - 2400 = 0 \n" ); document.write( "simplify, divide by 100 \n" ); document.write( "5t^2 + 2t - 24 = 0 \n" ); document.write( "Factors to \n" ); document.write( "(5t+12)(t-2) = 0 \n" ); document.write( "the positive solution \n" ); document.write( "t = 2 hrs, therefore at 3 PM, they will be 50 mi apart \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this out \n" ); document.write( "2(20) = 40 mi traveled by northbound ship \n" ); document.write( "3(10) = 30 mi traveled by the eastbound \n" ); document.write( "Enter into your calc; enter: \n" ); document.write( " \n" ); document.write( " |