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- anonymous

An airplane takes 1 hour longer to go a distance of 816 miles flying against a headwind than on the return trip with a tailwind. If the speed of the wind is 50 miles per hour, find the speed of the plane in still air.

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- anonymous

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- amistre64

you can use distance equations to help solve this

- amistre64

distance = speed x time
the distance remains constant, the speeds remain constant, its the time and the interaction that vary
(a+w) t = d
(a-w) (t+1) = d

- anonymous

a=?
w=?

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- amistre64

w is wind speed, a is airplane speed, at least in my head

- amistre64

now using these 2 equations ... we can work up something that with any luck will give us a relationship that we can solve
(a+w) t = d
(a-w) (t+1) = d
-at -wt = -d
at +a -wt -w = d
a = w(2t+1)
but t = d/(a+w) so ..
a = w(2d/(a+w)+1)
a = 2wd/(a+w) +w
a(a+w) = 2wd +w(a+w)
a(a+w)-w(a+w) = 2wd
a^2 +aw -aw -w^2 = 2wd
a^2 -w^2 = 2wd
a^2 = 2wd + w^2

- amistre64

im sure theres a simpler process .... but i can never recall it off hand

- anonymous

i need to get the a?
isn't?

- amistre64

yes, our goal is to get the speed of the airplane (a)

- anonymous

it's not factorable using factoring.. can i use quadratic formula? but the answer may have square roots

- amistre64

i cant readily see a way to remove the time variable ... i keep coming up with solutions that are not a single point.

- amistre64

a = 50(2t + 1)

- amistre64

lets see how well this plays tho, lets assume it takes t=1 to get there
a = 50(3) = 150
150+50 = 200, not 816 so my thought is evidently not correct

- anonymous

aww :(

- amistre64

maybe this makes better sense?
|dw:1440940969684:dw|

- amistre64

the h+1 isnt setting right with me

- amistre64

do you see any flaws in my thinking?
speed = distance/time; example, miles per hour; or, m/h
distance = speed x time
distance = (total speed) x time
distance = (a/h + w/h) x time
distance = (a/h - w/h) x (time+1)

- anonymous

is h=1?

- amistre64

h is equal to 1 time span yes ....
(a+50) t = 816
(a-50) (t+1) = 816
my first run thru lets a = 290
t = 816/(290+50) = 2.4
and (290-50) (3.4) = 816, so it works out as long as the time stated was 2.4 hours
but this seems more complicated than what it usually is

- amistre64

a^2 = 2wd + w^2
a^2 = 2(50)(816) + (50)^2 = 84100
a = sqrt(84100) = 290 ...

- amistre64

i dont spose we have an answer key to check against

- IrishBoy123

i got that too

- amistre64

... how did you get it tho? :)

- amistre64

if we solve each setup for time, the time component can be eliminated ....
t = d/(a+w)
t = d/(a-w) - 1
d/(a+w) = d/(a-w) - 1
d(a-w) = d(a+w) - a^2+w^2
ad -wd = ad +wd - a^2+w^2
-wd = wd - a^2+w^2
a^2 = 2wd + w^2
is the same thing i got to start with ... so its bound to be good lol

- IrishBoy123

same as you pretty much
eg from your
(a+w) t = d
(a-w) (t+1) = d
you can go
\[\implies \frac{d}{a-w} = \frac{d}{a+w} + 1\]
then x (a-w)(a+w) on each side etc

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