document.write( "Question 671618: A Car A left City X, traveling at an average velocity of 50 miles per hour. Three hours later a Car B left City X traveling on the same road at 60 miles per hour. When will the Car B catch up with the Car A? How far will each car have traveled? \n" ); document.write( "
Algebra.Com's Answer #417490 by DrBeeee(684)\"\" \"About 
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Use d = rt
\n" ); document.write( "For car A we have
\n" ); document.write( "(1) dA = 50t
\n" ); document.write( "For car B we have
\n" ); document.write( "(2) dB = 60(t-3) because car B left 3 hrs after car A
\n" ); document.write( "When they meet they both travelled the same distance say dF, then we have
\n" ); document.write( "(3) dF = dA = dB
\n" ); document.write( "Setting (1) equal to (2) we get
\n" ); document.write( "(4) 50t = 60(t-3) or
\n" ); document.write( "(5) 50t = 60t - 180 or
\n" ); document.write( "(6) 10t = 180 or
\n" ); document.write( "(7) t = 18 hrs
\n" ); document.write( "The distance travelled by car A in 18 hrs is
\n" ); document.write( "(8) dF = 50*18 or
\n" ); document.write( "(9) dF = 900 mi
\n" ); document.write( "To check this, let's see how far car B travelled. Using t = 18 in (2) we get
\n" ); document.write( "(10) dB = 60(18-3) or
\n" ); document.write( "(11) dB = 60*15 or
\n" ); document.write( "(12) dB = 900 mi
\n" ); document.write( "Answer: Car B will catch up with car A 18 hours after car A left city X at a distance of 900 miles.
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