document.write( "Question 694208: Sara leaves home at 7am traveling at a rate of 45 mi/h. her son discovers that she has forgotten her briefcase and starts out to catch up with her. her son leaves at 7:30 am traveling at a rate of 55 mi/h. At what time will he over take his mother? \n" ); document.write( "
Algebra.Com's Answer #427790 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Sara 45 mph \n" ); document.write( "Son 55 mph \n" ); document.write( "Sara 07:00 \n" ); document.write( "Son 07:30 \n" ); document.write( "Difference in time= 00:30 => 0.50 hours \n" ); document.write( "Sara will have covered 22.50 miles Son starts \n" ); document.write( "catch up distance= 22.50 miles \n" ); document.write( "catch up speed = 55 -45 \n" ); document.write( "catch up speed = 10 mph \n" ); document.write( "Catchup time = catchup distance/catch up speed \n" ); document.write( "catch up time= 2.25 \n" ); document.write( "catch up time= 2.25 hours \n" ); document.write( "They will meet at 9:55 am mph \n" ); document.write( " \n" ); document.write( " |