document.write( "Question 336482: a car A left city X traveling at an average velocity of 60 miles per hour. Two hours later a car B left city X traveling on the same road at an average velocity of 80 miles per hour. When will the car B catch up to the car A? How far will each car have traveled? \n" ); document.write( "
Algebra.Com's Answer #241196 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
a car A left city X traveling at an average velocity of 60 miles per hour. Two hours later a car B left city X traveling on the same road at an average velocity of 80 miles per hour. When will the car B catch up to the car A? How far will each car have traveled?
\n" ); document.write( "
\r\n" );
document.write( "Just for fun let's do it in our head the easy way, and then we'll do it by\r\n" );
document.write( "algebra.\r\n" );
document.write( "\r\n" );
document.write( "In 2 hours Car A, doing 60 mph is 120 miles ahead of Car B when Car B\r\n" );
document.write( "starts. Car B travels 80 mph which is 20 mph faster, so that's how fast\r\n" );
document.write( "he's approaching Car A at 80-60 or 20 mph.  So to catch up the 120 miles \r\n" );
document.write( "head start at 20 mph approach rate will take 6 hours since 120/20 = 6, \r\n" );
document.write( "and they will have traveled 80 mph*6 hours or 480 miles.\r\n" );
document.write( "\r\n" );
document.write( "Your teacher doesn't want you to do it that way, but by algebra. So\r\n" );
document.write( "here's how to do it by algebra, even though the \"head\" way is easier for\r\n" );
document.write( "this problem.\r\n" );
document.write( " \r\n" );
document.write( "Now by algebra:\r\n" );
document.write( "\r\n" );
document.write( "Make this chart:\r\n" );
document.write( "\r\n" );
document.write( "        distance    rate    time\r\n" );
document.write( "Car A    \r\n" );
document.write( "Car B    \r\n" );
document.write( "\r\n" );
document.write( "Let x be the time it takes B to catch up to A\r\n" );
document.write( "Then x+2 will be the time A has traveled since he \r\n" );
document.write( "travels for 2 hours longer.  So fill those in and\r\n" );
document.write( "also the rates at which they travel:\r\n" );
document.write( "    \r\n" );
document.write( "\r\n" );
document.write( "        distance    rate    time\r\n" );
document.write( "Car A                60      x+2 \r\n" );
document.write( "Car B                80       x\r\n" );
document.write( "\r\n" );
document.write( "Now use distance = rate × time to fill in the distances:\r\n" );
document.write( "\r\n" );
document.write( "        distance    rate    time\r\n" );
document.write( "Car A    60(x+2)     60      x+2 \r\n" );
document.write( "Car B      80x       80       x\r\n" );
document.write( "\r\n" );
document.write( "The two distances are equal, so make the equation:\r\n" );
document.write( "\r\n" );
document.write( "        60(x+2) = 80x\r\n" );
document.write( "\r\n" );
document.write( "Solve that and get x = 6 hours\r\n" );
document.write( "\r\n" );
document.write( "Then use either value for the distance and substitute 6 for x\r\n" );
document.write( "\r\n" );
document.write( "60(x+2) = 60(6+2) = 60(8) = 480 miles\r\n" );
document.write( "\r\n" );
document.write( "Or you could substitute in the other value for the distance:\r\n" );
document.write( "\r\n" );
document.write( "80x = 80(6) = 480 miles.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );