SOLUTION: A plane flies 720 mi against a steady 30 mi.h headwind and then returns to the same point with the wind. If the entire trip takes 10 hr what is the planes speed in still air?: A pl
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Question 83265: A plane flies 720 mi against a steady 30 mi.h headwind and then returns to the same point with the wind. If the entire trip takes 10 hr what is the planes speed in still air?: A plane flies 720 mi against a steady 30 mi.h headwind and then returns to the same point with the wind. If the entire trip takes 10 hr what is the planes speed in still air?
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Let x=speed (rate) of the plane in still air
speed against headwind =x-30
speed with headwind=x+30
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Time required travelling against the headwind=720/(x-30)
Time require travelling with the headwind=720/(x+30)
Now we are told that the total time was 10 hours, So:
720/(x-30)+720/(x+30)=10 multiply both sides by (x-30)(x+30)
720(x+30)+720(x-30)=10(x-30)(x+30) divide both sides by 10
72(x+30)+72(x-30)=(x-30)(x+30) get rid of parens
72x+2160+72x-2160=x^2-900 collect like terms
144x=x^2-900 subtract 144x from both sides
144x-144x=x^2-144x-900 simplify
x^2-144x-900=0 quadratic in standard form and this can be factored
(x-150)(x+6)=0
x=150mph-------------------------------------rate in still air
x=-6----------------discount negative value for speed
CK
720/(150-30)+720/(150+30)=10
6+4=10
10=10
