document.write( "Question 362228: Two trucks leave a warehouse at the same time, traveling in opposite directions. The rate of the faster truck exceeds that of the slower truck by 5 miles per hour. After 5 hours they are 600 miles apart. What are the rates of the trucks? \n" ); document.write( "
Algebra.Com's Answer #258142 by HasanSahin(52)![]() ![]() You can put this solution on YOUR website! If they have constant velocities in the initial condition: \n" ); document.write( "Let's say v1 = speed of faster truck and v2 = speed of slower truck.. \n" ); document.write( "The distance they travel in a certain time is X = V*t where t is the time. \n" ); document.write( "Let's say x1 = distance of faster truck from warehouse and x2 = distance of slower truck from warehouse..\r \n" ); document.write( "\n" ); document.write( "So that >> x1 = v1*t and x2 = v2*t\r \n" ); document.write( "\n" ); document.write( "Total distance after 5 hours is \r \n" ); document.write( "\n" ); document.write( "x1 + x2 = v1*5 + v2*5 = 600 miles (1)\r \n" ); document.write( "\n" ); document.write( "We know that v1 = v2 + 5 (2)\r \n" ); document.write( "\n" ); document.write( "And therefore, put v1 in the (2)nd equation into the (1)st equation such that;\r \n" ); document.write( "\n" ); document.write( "(v2 + 5)*5 + v2*5 = 600 >>\r \n" ); document.write( "\n" ); document.write( "5*v2 + 25 + 5*v2 = 600 >>\r \n" ); document.write( "\n" ); document.write( "10*v2 = 575 miles >>\r \n" ); document.write( "\n" ); document.write( "v2 = 57.5 miles per hour and if you put v2 into any equation you'll get >>\r \n" ); document.write( "\n" ); document.write( "v1 = 62.5 miles per hour\r \n" ); document.write( "\n" ); document.write( "RF.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |