document.write( "Question 870627: Car A left Melbourne for Adelaide at 11.00 am and traveled at at an average speed of 70 km/h.
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document.write( "Car B left Melbourne for aDELAIDE at 1.00 pm same day and traveled at an average speed of 90 km/h.
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document.write( "At what time will Car B catch Car A? \n" );
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Algebra.Com's Answer #524999 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Car A 70 mph \n" ); document.write( "Car B 90 mph \n" ); document.write( "Car A 11:00 \n" ); document.write( "Car B 01:00 \n" ); document.write( "Difference in time= 14:00 => 2.00 hours \n" ); document.write( "Car A will have covered 140.00 miles when Car B starts \n" ); document.write( "catch up distance= 140.00 miles \n" ); document.write( "catch up speed = 90 -70 \n" ); document.write( "catch up speed = 20 mph \n" ); document.write( "Catchup time = catchup distance/catch up speed \n" ); document.write( "catch up time= 140/20= 7 \n" ); document.write( "catch up time= 7.00 hours \n" ); document.write( "They will meet at 08:00 pm \n" ); document.write( " \n" ); document.write( " |