document.write( "Question 345253: Students are traveling in two cars to a football game 135 miles away. The first
\n" ); document.write( "car leaves on time and travels at an average speed of 45 miles per hour. The
\n" ); document.write( "second car starts 1/2 hour later and travels at ab average speed of 55 miles per
\n" ); document.write( "hour. How long will it take the second car to catch up to the first car? Will
\n" ); document.write( "the second car catch up to the first car before it arrives at the game?
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Algebra.Com's Answer #246898 by checkley77(12844)\"\" \"About 
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D=RT
\n" ); document.write( "135=45T FIRST CAR.
\n" ); document.write( "135=55(T-.5) SECOND CAR.
\n" ); document.write( "45T=55(T-.5
\n" ); document.write( "45T=55T-27.5
\n" ); document.write( "45T-55T=-27.5
\n" ); document.write( "10T=-27.5
\n" ); document.write( "T=27.5/10
\n" ); document.write( "T=2.75 HOURS FOR THE TRIP TIME
\n" ); document.write( "45*2.75=123.75 MILES IS THE POINT WHERE THEY MEET.
\n" ); document.write( "55(2.75-.5)=55*2.25=123.75 DITTO.
\n" ); document.write( "2.75-.5=2.25 HOURS THE SECOND CAR WILL CATH UP WITH THE FIRST CAR.
\n" ); document.write( "yes THEY WILL MEET BEFORE THEY ARRIVE @ THE FOOTBALL GAME.
\n" ); document.write( "123.75 MILES IS LESS THAN 135 MILES.
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