document.write( "Question 575458: Car A passes point O heading east at a constant rate of 40 mph; Car B passes the same point one hour later heading north at a constant rate of 60 mph. Express the distance between the cars as a function of time t, where t is measured starting when car B passes point O. \n" ); document.write( "
Algebra.Com's Answer #369481 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "At , A is 40 miles from O and B is at O. As time increases, A moves away from O at 40 mph, so at any positive time hours, A is miles from O. At the same time, B is miles from O, but in a direction at right angle to A's direction of travel.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hence, the distance A has traveled at time is one leg of a right triangle and the distance B has traveled at the same time is the other leg of a right triangle.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "With a tip of the hat to Mr. Pythagoras:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "You can simplify if you like.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "John
\n" ); document.write( "
\n" ); document.write( "My calculator said it, I believe it, that settles it
\n" ); document.write( "
\"The

\n" ); document.write( " \n" ); document.write( "
\n" );