anonymous
  • 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.
Algebra
jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
you can use distance equations to help solve this
amistre64
  • 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
  • anonymous
a=? w=?

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amistre64
  • amistre64
w is wind speed, a is airplane speed, at least in my head
amistre64
  • 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
  • amistre64
im sure theres a simpler process .... but i can never recall it off hand
anonymous
  • anonymous
i need to get the a? isn't?
amistre64
  • amistre64
yes, our goal is to get the speed of the airplane (a)
anonymous
  • anonymous
it's not factorable using factoring.. can i use quadratic formula? but the answer may have square roots
amistre64
  • 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
  • amistre64
a = 50(2t + 1)
amistre64
  • 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
  • anonymous
aww :(
amistre64
  • amistre64
maybe this makes better sense? |dw:1440940969684:dw|
amistre64
  • amistre64
the h+1 isnt setting right with me
amistre64
  • 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
  • anonymous
is h=1?
amistre64
  • 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
  • amistre64
a^2 = 2wd + w^2 a^2 = 2(50)(816) + (50)^2 = 84100 a = sqrt(84100) = 290 ...
amistre64
  • amistre64
i dont spose we have an answer key to check against
IrishBoy123
  • IrishBoy123
i got that too
amistre64
  • amistre64
... how did you get it tho? :)
amistre64
  • 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
  • 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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