SOLUTION: Two trains leave their respective stations at the same time travelling towards each other. If train A is travelling at 100 kph and train B at 120 kph and the distance between the t

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Two trains leave their respective stations at the same time travelling towards each other. If train A is travelling at 100 kph and train B at 120 kph and the distance between the t      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 769637: Two trains leave their respective stations at the same time travelling towards each other. If train A is travelling at 100 kph and train B at 120 kph and the distance between the two stations is 110 km how long will it be before they meet up ? how far had each train travelled ?
sorry i know this is really easy we are working these out with graphs and i dont get it so i was wondering if any one had any alternatives and please show working ! Thanks :D

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +d+ = the distance in km that the slower train
will travel until they meet
+110+-+d+ = the distance the faster train will
travel until they meet
Let +t+ = the time elapsed until they meet
----------------------------------------
Slower train's equation:
(1) +d+=+100t+
Faster train's equation:
(2) +110+-+d+=+120t+
--------------------
(2) +d+=+-120t+%2B+110%0D%0AIf+%7B%7B%7B+d+ is the vertical axis and +t+ is the horizontal,
you can plot these.
The point (d,t) where the lines cross is when and where
they meet
Here's the plot:
+graph%28+400%2C+400%2C+-1%2C+2%2C+-20%2C+140%2C+100x%2C+-120x+%2B+110++%29+
Note that ( d,t ) = ( 0,0 ) is the starting point for one train
and ( 0,110) is where the other train starts
They meet at ( .5,50 )
Note that 50 km is measured from the slower train's starting point
--------------------
Substitute (1) into (2)
(2) +110+-+d+=+120t+
(2) +110+-+100t+=+120t+
(2) +220t+=+110+
(2) +t+=+.5+ hrs
and
(1) +d+=+100t+
(1) +d+=+100%2A.5+
(1) +d+=+50+ km
check:
(2) +110+-+d+=+120%2A.5+
(2) +d+=+110+-+60+
(2) +d+=+50+ km
----------------
So the slower train travels 50 km
and the faster train travels 110 - d = 60 km