document.write( "Question 51645: @ noon a train leaves bridgton heading east at 90 mi/h to cogsville, 450 mi away. at 12:15 p.m. a train leave cogsville heading west to bridgton at 100 mi/h. at what time will they pass each other? \n" ); document.write( "

Algebra.Com's Answer #34473 by josmiceli(19441) $\"\"$ $\"About$

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Where will the 1st train be at 12:15?
\n" ); document.write( "It will have travelled for 15 min at 90 mi/hr
\n" ); document.write( "r*t = d
\n" ); document.write( "90 mi/hr * 1/4 hr = 22.5 mi from Bridgton
\n" ); document.write( "Now it is (450 - 22.5) mi from Cogsville
\n" ); document.write( "Now the 2nd train leaves, and both will travel for the same
\n" ); document.write( "amount of time until they meet
\n" ); document.write( " $\"t%281%29+=+t%282%29\"$
\n" ); document.write( " $\"d%281%29%2Fr%281%29+=+d%282%29+%2F+r%282%29\"$
\n" ); document.write( " $\"d%281%29+%2F+90+=+d%282%29+%2F+100\"$
\n" ); document.write( " $\"d%281%29+=+.9%2Ad%282%29\"$
\n" ); document.write( " $\"d%281%29+%2B+d%282%29+=+450+-+22.5\"$
\n" ); document.write( " $\"d%281%29+%2B+d%282%29+=+427.5\"$
\n" ); document.write( " $\".9%2Ad%282%29+%2B+d%282%29+=+427.5\"$
\n" ); document.write( " $\"1.9%2Ad%282%29+=+427.5\"$
\n" ); document.write( " $\"d%282%29+=+225\"$
\n" ); document.write( " $\"d%281%29+=+427.5+-+225\"$
\n" ); document.write( " $\"d%281%29+=+202.5\"$
\n" ); document.write( " $\"t%281%29+=+t%282%29\"$
\n" ); document.write( " $\"202.5+%2F+90+=+225+%2F+100\"$
\n" ); document.write( " $\"2.25+=+2.25\"$
\n" ); document.write( "So, the trains travel for 2.25 hours until they meet. That is with
\n" ); document.write( "a starting time of 12:15 pm
\n" ); document.write( "So, they meet at 12:15 pm + 2 1/4 hrs = 2:30 pm
\n" ); document.write( " \n" ); document.write( "

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