SOLUTION: At noon, a jetliner left New York for Los Angeles 4000 kn away. At 1:00 P.M. New York time a jetliner, flying 10 km/hr faster than the first one, left Los Angeles for New York. T

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Question 674406: At noon, a jetliner left New York for Los Angeles 4000 kn away. At 1:00 P.M. New York time a jetliner, flying 10 km/hr faster than the first one, left Los Angeles for New York. The planes passed each other at 3:00 P.M.New York time. Find the speed of the planes.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the speed of the NY_LA plane = +s+
The NY-LA plane got a head start of +s%2A1+ km
The planes were then +4000+-+s+ km apart
Let +d+ = distance in km that the NY-LA plane
has to travel until it meets the LA-NY plane
----------------------
NY-LA plane's equation:
(1) +d+=+s%2A2+ ( note that time starts at 1 PM and ends at 3 PM )
LA-NY plane's equation:
(2) +4000+-+s+-+d+=+%28+s+%2B+10+%29%2A2+
------------------------------
Substitute (1) into (2)
(2) +4000+-+s+-+2s+=+%28+s+%2B+10+%29%2A2+
(2) +4000+-+3s+=+2s+%2B+20+
(2) +5s+=+4000+-+20+
(2) +5s+=+3980+
(2) +s+=+796+
+s+%2B+10+=+806+
The speeds of the planes are:
NY-LA - 796 km/hr
LA-NY - 806 km/hr
-----------------
check answer:
(1) +d+=+s%2A2+
(1) +d+=+796%2A2+
(1) +d+=+1592+ km
and
(2) +4000+-+s+-+d+=+%28+s+%2B+10+%29%2A2+
(2) +4000+-+796+-+d+=+806%2A2+
(2) +3204+-+d+=+1612+
(2) +d+=+3204+-+1612+
(2) +d+=+1592+ km
OK