document.write( "Question 1170177: A cruise ship going across a lake 15km wide travels a distance of 6km at a certain
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document.write( "speed and then finishes the trip at a speed 1 kph slower than its original speed. The
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document.write( "boat arrives 20 minutes later than if the original speed is maintained throughout the
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document.write( "trip. What is the original speed of the cruise ship?
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Algebra.Com's Answer #795038 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! A cruise ship going across a lake 15km wide travels a distance of 6km at a certain speed and then finishes the trip at a speed 1 kph slower than its original speed. \n" ); document.write( " The boat arrives 20 minutes later than if the original speed is maintained throughout the trip. \n" ); document.write( "What is the original speed of the cruise ship? \n" ); document.write( ": \n" ); document.write( "let s = original speed \n" ); document.write( "then \n" ); document.write( "(s-1) = slower speed \n" ); document.write( ": \n" ); document.write( "Change 20 min to \n" ); document.write( ": \n" ); document.write( "Total dist was 15 km therefore he traveled 6 km at s and 9 km at (s-1) \n" ); document.write( "Find the actual time take for the trip \n" ); document.write( " \n" ); document.write( "then \n" ); document.write( "actual time - normal time = 1/3 hr \n" ); document.write( " \n" ); document.write( "multiply by 3s(s-1), get rid of the denominators \n" ); document.write( "3(15s-6) - 15(3(s-1)) = s(s-1) \n" ); document.write( "45s - 18 - 45s + 45 = s^2 - s \n" ); document.write( "27 = s^2 - s \n" ); document.write( "A quadratic equation \n" ); document.write( "s^2 - s - 27 = 0 \n" ); document.write( "using the quadratic formula, the positive solution \n" ); document.write( "s = 5.72 km/h is the normal speed \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "seems like we deserve to get an integer here, check in time equation with calc \n" ); document.write( "( \n" ); document.write( "1.049 + 1.907 - 2.622 = .334 which is about 1/3 hr or 20 min \n" ); document.write( "seems to check out, if you find that I screwed this up, let me know. ankor@att.net \n" ); document.write( " |