document.write( "Question 302031: An express train leaves New York at 3pm and reaches Boston at 6pm. A slow train leaves Boston at 1.30pm and arrives at New York at 6pm. If both trains travel at constant speeds, at what time do they meet? \n" ); document.write( "
Algebra.Com's Answer #216651 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! An express train leaves New York at 3pm and reaches Boston at 6pm. \n" ); document.write( " A slow train leaves Boston at 1.30pm and arrives at New York at 6pm. \n" ); document.write( " If both trains travel at constant speeds, at what time do they meet? \n" ); document.write( ": \n" ); document.write( "From the given information: \n" ); document.write( "Express travel time = 3 hrs \n" ); document.write( "Slow train travel time = 4.5 hrs \n" ); document.write( ": \n" ); document.write( "Let d = dist from N.Y. to Boston \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "Let t = travel time of express when they meet \n" ); document.write( "then \n" ); document.write( "(t+1.5) = travel time of the slow train when they meet \n" ); document.write( ": \n" ); document.write( "Write a distance equation; dist = speed * time \n" ); document.write( ": \n" ); document.write( "Dist of each train = the total dist \n" ); document.write( " \n" ); document.write( "multiply thru by 9 to get rid of the denominators, results: \n" ); document.write( "3dt + 2d(t+1.5) = 9d \n" ); document.write( "3dt + 2dt + 3d = 9d \n" ); document.write( "5dt + 3d = 9d \n" ); document.write( "divide thru by d \n" ); document.write( "5t + 3 = 9 \n" ); document.write( "5t = 9 - 3 \n" ); document.write( "5t = 6 \n" ); document.write( "t = \n" ); document.write( "t = 1.2 hrs, which is 1 hr 12 min \n" ); document.write( ": \n" ); document.write( "3:00 + 1:12 = 4:12 pm the trains meet \n" ); document.write( " \n" ); document.write( " |