SOLUTION: An airliner traveling from Toronto to Vancouver took 5 h to cover the 3900km trip against a headwind. The return trip, traveling with a tailwind that was twice the speed of the he

Algebra ->  Linear-equations -> SOLUTION: An airliner traveling from Toronto to Vancouver took 5 h to cover the 3900km trip against a headwind. The return trip, traveling with a tailwind that was twice the speed of the he      Log On


   

Question 1077331: An airliner traveling from Toronto to Vancouver took 5 h to cover the 3900km trip against a headwind. The return trip, traveling with a tailwind that was twice the speed of the headwind, took 4h and 20min.
a) How fast were the headwind and the tailwind on the two trips?
b) How fast would the airliner have flown without a wind?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed of the plane with no wind in km/hr
Let +w+ = the speed of the headwind in km/hr
+2w+ = the speed of the tailwind
+20+ min = +1%2F3+ hrs
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For Toronto to Vancouver:
(1) +3900+=+%28+s+-+w+%29%2A5+
For Vancouver to Toronto:
(2) +3900+=+%28+s+%2B+2w+%29%2A%2813%2F3%29+
--------------------------------
(1) +780+=+s+-+w+
(1) +s+=+780+%2B+w+
Plug this result into (2)
(2) +3900+=+%28+780+%2B+w+%2B+2w+%29%2A%2813%2F3%29+
(2) +3900+=+%28+780+%2B+3w+%29%2A%2813%2F3%29+
(2) +11700+=+10140+%2B+39w+
(2) +39w+=+1560+
(2) +w+=+40+
and
+2w+=+80+
-----------------------
(a)
the headwind is 40 km/hr
The tailwind is 80 km/hr
------------------------
(b)
First I have to find +s+
(1) +3900+=+%28+s+-+w+%29%2A5+
(1) +3900+=+%28+s+-+40+%29%2A5+
(1) +3900+=+5s+-+200+
(1) +5s+=+4100+
(1) +s+=+820+
Let +t+ = time between the cities without wind
For Toronto to Vancouver:
(1) +3900+=+s%2At+
(1) +3900+=+820t+
(1) +t+=+4.756+ hrs
and
For Vancouver to Toronto:
(2) +3900+=+s%2At+
(2) +3900+=+820%2At+
I get the same answer
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check my math & get another opinion