document.write( "Question 908533: car A travels at 20 mph and car B travels at 15 mph; car A is 200 m behind car B; how long does car A have to travel before it catches up with car B? \n" ); document.write( "
Algebra.Com's Answer #551105 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! Algebraic solution \n" ); document.write( "t*(20-15)=200 \n" ); document.write( "t*5=200 \n" ); document.write( "t=200/5 \n" ); document.write( "t=40 hours \n" ); document.write( "check \n" ); document.write( "40(20-15)=200 \n" ); document.write( "40(5)=200 \n" ); document.write( "200=200 \n" ); document.write( "ok \n" ); document.write( "200 15 20 0 0 \n" ); document.write( "Gaining speed 20-15= 5 mph \n" ); document.write( "Headstart 200 =200 miles \n" ); document.write( "Time to catch up =40 hours \n" ); document.write( " \n" ); document.write( " |