document.write( "Question 1208940: A motorboat can maintain a constant speed of 16 miles per hour relative to the water. The boat makes a trip upstream to a certain point 20 minutes; the return trip take 15 minutes. What is the speed of the current? \r
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #847482 by math_tutor2020(3816)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "c = speed of the current in mph\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Going against the current (i.e. upstream) the boat's speed goes from 16 mph to 16-c mph. \n" ); document.write( "distance = rate*time \n" ); document.write( "d = r*t \n" ); document.write( "d = (16-c)*(20/60) \n" ); document.write( "d = (1/3)(16-c)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Going with the current (downstream) means the boat goes from 16 mph to 16+c mph. \n" ); document.write( "d = r*t \n" ); document.write( "d = (16+c)*(15/60) \n" ); document.write( "d = (1/4)(16+c)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The two equations involve the same distance. \n" ); document.write( "This allows us to equate the right hand sides to solve for c. \n" ); document.write( "(1/3)(16-c) = (1/4)(16+c) \n" ); document.write( "12*(1/3)(16-c) = 12*(1/4)(16+c) \n" ); document.write( "4(16-c) = 3(16+c) \n" ); document.write( "64-4c = 48+3c \n" ); document.write( "64-48 = 3c+4c \n" ); document.write( "16 = 7c \n" ); document.write( "c = 16/7 \n" ); document.write( "c = 2.2857 mph approximately\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Side notes:
|