document.write( "Question 919886: Towns A and B are 540km apart. A train leaves A and travels to B at a constant speed. A second train leaves A 1h later and, after 2h, reaches a point passed by the first train 40min earlier. The second train then increases speed by 5km/h and arrives at B at the same time as the first train. What were the speeds of the trains when they left A?
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Algebra.Com's Answer #558278 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! Towns A and B are 540km apart. \n" ); document.write( " A train leaves A and travels to B at a constant speed. \n" ); document.write( " A second train leaves A 1h later and, after 2h, reaches a point passed by the first train 40min earlier. \n" ); document.write( " The second train then increases speed by 5km/h and arrives at B at the same time as the first train. \n" ); document.write( " What were the speeds of the trains when they left A? \n" ); document.write( ": \n" ); document.write( "A-----------------p-----------540 km-----------------------------B \n" ); document.write( ": \n" ); document.write( "1st train got to p in 3 hr - 40 min or 2 \n" ); document.write( "2nd train got to p in 2 hrs, (after leaving 1 hr later than the 1st train) \n" ); document.write( "We can say that 2nd train travels the same dist in 2 hrs, as the 1st train in 2 \n" ); document.write( " time ratio; 1st train to 2nd train 2.33:2 or in integers 7:6 ratio \n" ); document.write( ": \n" ); document.write( "let s = speed of the 1st train \n" ); document.write( "then \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "speed varies inversely with time therefore \n" ); document.write( " \n" ); document.write( "and \n" ); document.write( "( \n" ); document.write( "The distance to p (dist = speed * time) = 2* \n" ); document.write( ": \n" ); document.write( "2nd trains time to p = 3 hrs after the 1st train started \n" ); document.write( "Write a time equation \n" ); document.write( "3 hrs to p + time traveling from p to B = 1st train travel time \n" ); document.write( "3 + \n" ); document.write( "It's such a pain in the you know what, to deal with these fractions, try to use decimals and see what we get \n" ); document.write( "3 + \n" ); document.write( "multiply by s(1.167s+5) \n" ); document.write( "3s(1.167s+5) + s(540-2.33s) = 540(1.167s+5) \n" ); document.write( "3.5s^2 + 15s + 540s - 2.33s^2 = 630s + 2700 \n" ); document.write( "Combine like terms \n" ); document.write( "3.5s^2 - 2.33s^2 - 630s + 15s + 540s - 2700 = 0 \n" ); document.write( "1.167s^2 - 75s - 2700 = 0 \n" ); document.write( "using the quadratic formula, I got a postive solution of \n" ); document.write( "s = 90 km/hr is the speed of the 1st train \n" ); document.write( "then \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Let's see if we lucked out and got this right, find the times. \n" ); document.write( "1st train travel time = 540/90 = 6 hrs \n" ); document.write( "2nd train travel time to p \n" ); document.write( "2 * 105 = 210 km to p \n" ); document.write( "2nd train speed to B = 110 km/hr (5 km/hr more) \n" ); document.write( " \n" ); document.write( "2nd train total time \n" ); document.write( "1 hr before starting \n" ); document.write( "2 hrs to p \n" ); document.write( "3 hrs p to B \n" ); document.write( "------------ \n" ); document.write( "6 hrs, same as 1st train!!!\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |