document.write( "Question 757384: Vehicle A which is ahead of vehicle B by 2/3 mile is traveling at 53 mph. Vehicle B which is behind vehicle A is traveling 75 mph. How long and how far will it take vehicle B to catch vehicle A?\r
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document.write( "Same question but with A going 30 mph. \n" );
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Algebra.Com's Answer #460808 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! Vehicle A which is ahead of vehicle B by 2/3 mile is traveling at 53 mph. Vehicle B which is behind vehicle A is traveling 75 mph. How long and how far will it take vehicle B to catch vehicle A? \n" ); document.write( "Same question but with A going 30 mph. \n" ); document.write( "*** \n" ); document.write( "let x=distance A traveled \n" ); document.write( "x+2/3=distance B traveled to catch A \n" ); document.write( "travel time=distance/speed (same for A and B) \n" ); document.write( ".. \n" ); document.write( " \n" ); document.write( "75x=53x+(106/3 \n" ); document.write( "22x=106/3 \n" ); document.write( "x=106/66 \n" ); document.write( "x≈1.6 \n" ); document.write( "x+2/3≈2.27 \n" ); document.write( "travel time=1.6/53=.03 hr or≈1.81 min \n" ); document.write( "B will catch up to A in 1.81 min after traveling 2.27 mi \n" ); document.write( ".. \n" ); document.write( "For A going 30 mph \n" ); document.write( " \n" ); document.write( "75x=30x+20 \n" ); document.write( "45x=20 \n" ); document.write( "x=20/45 \n" ); document.write( "x≈0.44 \n" ); document.write( "x+2/3≈1.11 \n" ); document.write( "travel time=0.44/30=.015hr or≈0.89 min \n" ); document.write( "B will catch up to A in 0.89 min after traveling 1.11mi \n" ); document.write( " |