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.
\n" ); document.write( "Car B left Melbourne for aDELAIDE at 1.00 pm same day and traveled at an average speed of 90 km/h.
\n" ); document.write( "At what time will Car B catch Car A?
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Algebra.Com's Answer #524999 by mananth(16946)\"\" \"About 
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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
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