document.write( "Question 19517: A Car Starts on a trip and travels at a speed of 55 mph. Two hours later a second car starts on the same trip and travels at a speed of 65 mph. When the second car has been on the road for x hours, the first car has traveled(x+2)55 and the second card has traveled 65x miles. At time x the distance between the first car and the second car is how many miles? \n" ); document.write( "
Algebra.Com's Answer #10034 by kapilsinghi(68)\"\" \"About 
You can put this solution on YOUR website!
Assuming the cars start from the same starting point\r
\n" ); document.write( "\n" ); document.write( "Assuming the the cars are travling in the same direction \r
\n" ); document.write( "\n" ); document.write( " distance between the to cars
\n" ); document.write( "= 65x - 55(x+2)
\n" ); document.write( "= 65x - 55x - 110
\n" ); document.write( "= 10x - 110\r
\n" ); document.write( "\n" ); document.write( "if the distance come out to be negative then the care with speed 65MPH is still behind the 55MPH care if it is positive then the case is other way round\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Assuming the the cars are travling in the same direction \r
\n" ); document.write( "\n" ); document.write( " distance between the to cars (will be positive)
\n" ); document.write( "= 65x + 55(x+2)
\n" ); document.write( "= 65x + 55x + 110
\n" ); document.write( "= 110x + 110\r
\n" ); document.write( "\n" ); document.write( "Assuming the the cars are travling at 90degrees to each \r
\n" ); document.write( "\n" ); document.write( "then the distance between them
\n" ); document.write( "= (65x^2 + (55(x+2))^2) ^ 1/2\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "kapilsinghi123@gmail.com
\n" ); document.write( "
\n" ); document.write( "
\n" );