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anonymous

  • one year ago

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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  1. amistre64
    • one year ago
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    you can use distance equations to help solve this

  2. amistre64
    • one year ago
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    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

  3. anonymous
    • one year ago
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    a=? w=?

  4. amistre64
    • one year ago
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    w is wind speed, a is airplane speed, at least in my head

  5. amistre64
    • one year ago
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    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

  6. amistre64
    • one year ago
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    im sure theres a simpler process .... but i can never recall it off hand

  7. anonymous
    • one year ago
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    i need to get the a? isn't?

  8. amistre64
    • one year ago
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    yes, our goal is to get the speed of the airplane (a)

  9. anonymous
    • one year ago
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    it's not factorable using factoring.. can i use quadratic formula? but the answer may have square roots

  10. amistre64
    • one year ago
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    i cant readily see a way to remove the time variable ... i keep coming up with solutions that are not a single point.

  11. amistre64
    • one year ago
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    a = 50(2t + 1)

  12. amistre64
    • one year ago
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    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

  13. anonymous
    • one year ago
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    aww :(

  14. amistre64
    • one year ago
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    maybe this makes better sense? |dw:1440940969684:dw|

  15. amistre64
    • one year ago
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    the h+1 isnt setting right with me

  16. amistre64
    • one year ago
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    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)

  17. anonymous
    • one year ago
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    is h=1?

  18. amistre64
    • one year ago
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    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

  19. amistre64
    • one year ago
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    a^2 = 2wd + w^2 a^2 = 2(50)(816) + (50)^2 = 84100 a = sqrt(84100) = 290 ...

  20. amistre64
    • one year ago
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    i dont spose we have an answer key to check against

  21. IrishBoy123
    • one year ago
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    i got that too

  22. amistre64
    • one year ago
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    ... how did you get it tho? :)

  23. amistre64
    • one year ago
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    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

  24. IrishBoy123
    • one year ago
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    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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