document.write( "Question 520063: A car leaves a house in Atlanta at 8:00a.m. Traveling at an average speed of 45mph.
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document.write( "A van leaves the same house at 10:00a.m. Traveling at an average speed of 60mph.
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document.write( "In how many hours will the van catch up with the car? \n" );
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Algebra.Com's Answer #345920 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Car speed 45 mph \n" ); document.write( "Van speed 60 mph \n" ); document.write( "Car speed 08:00 \n" ); document.write( "Van speed 10:00 \n" ); document.write( "Difference in time= 02:00 => 2 hours \n" ); document.write( "Car speed will have covered 90 miles before van starts \n" ); document.write( "catch up distance= 90 miles \n" ); document.write( "catch up speed = 60 -45 mph \n" ); document.write( "catch up speed = 15 mph \n" ); document.write( "Catchup time = catchup distance/catch up speed \n" ); document.write( "catch up time= 90 / 15 \n" ); document.write( "catch up time= 6 hours \n" ); document.write( "They will meet at 10:00 \n" ); document.write( "Van speed speed = 60 mph \n" ); document.write( "Time to catch up = 6 hours\r \n" ); document.write( "\n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " |