document.write( "Question 67706: A train leaves a station at 9:00 A.M. traveling at a constant rate of speed. Two hours later a second train, traveling at an average rate of 25 miles per hour more than the first train, leaves the same station, going in the same direction, on a parallel track. If the faster train overtakes the first train at 3:00 P.M. on the same day, what is the average rate of speed of the first train. \n" ); document.write( "
Algebra.Com's Answer #48168 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A train leaves a station at 9:00 A.M. traveling at a constant rate of speed. Two hours later a second train, traveling at an average rate of 25 miles per hour more than the first train, leaves the same station, going in the same direction, on a parallel track. If the faster train overtakes the first train at 3:00 P.M. on the same day, what is the average rate of speed of the first train. \n" ); document.write( "----------- \n" ); document.write( "If train two overtakes train one at 3:00 train two must have traveled \n" ); document.write( "from 11:00 to 3:00 or 4 hours. Train one left two hours earlier so \n" ); document.write( "it must have traveled for 6 hours. \n" ); document.write( "--------------- \n" ); document.write( "Train one DATA: \n" ); document.write( "Time = 6 hours ; Rate = x mph; Distance = r*t = 6x miles \n" ); document.write( "----------- \n" ); document.write( "Train two DATA: \n" ); document.write( "Time = 4 hurs ; Rate = x+25 mph ; Distance = r*t = 4(x+25) \n" ); document.write( "------------ \n" ); document.write( "EQUATION: \n" ); document.write( "distance = distance \n" ); document.write( "6x=4(x+25) \n" ); document.write( "6x=4x+100 \n" ); document.write( "2x=100 \n" ); document.write( "x=50 mph (speed of 1st train) \n" ); document.write( "-------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |