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.

Question

amistre64 · Answer

you can use distance equations to help solve this

amistre64 · Answer

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 · Answer

a=?
w=?

amistre64 · Answer

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

amistre64 · Answer

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 · Answer

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

anonymous · Answer

i need to get the a?
 isn't?

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

a = 50(2t + 1)

amistre64 · Answer

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 · Answer

aww :(

amistre64 · Answer

maybe this makes better sense?

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