Applied Maths - Mechanics 18

還不明白的薑

2009-07-31 2:57 am

In a wind blowing from the South with constant speed w a helicopter flies horizontally with constant velocity in a direction @ East of North from a point A to a point B. The speed of the helicopter relative to the air is kw, where k > 1 . Find the speed of the helicopter along AB.

The helicopter returns from B to A with constant velocity and the same speed kw relative to the air, and in the same wind. Show that the total time for the two journeys is 2c( k^2 -sin^2 @ )^0.5 / w(k^2 -1) , where c = AB.

回答 (1)

用戶9112

2009-07-31 6:03 am

✔ 最佳答案

When flying from A to B, the result velocity V is making an angle of θ with the N direction.
The North direction resultant velocity is Vcosθ
This minus the speed of the South wind is the original North direction speed of the helicopter.
The East direction resultant velocity is Vsinθ which is the same as the East direction velocity of the helicopter.
Therefore,
(kw)2 = (Vcosθ-w)2 + (2Vsinθ)2
(kw)2 = V2 - 2wcosθV + w2
V2 - 2wcosθV + w2(1 – k2) = 0
V = {2wcosθ + √[(2wcosθ)2 – 4w2(1 – k2)]}/2, the other root is rejected since it is negative
V = wcosθ + √(k2w2 - w2sin2θ)
V = w[cosθ + √(k2 – sin2θ)]
Similarly when the helicopter flies from B to A, the velocity is
U = w[- cosθ + √(k2 – sin2θ)]
Total time taken = c/V + c/U
= c/w[cosθ + √(k2 – sin2θ)] + c/w[- cosθ + √(k2 – sin2θ)]
=(c/w)[- cosθ + √(k2 – sin2θ) + cosθ + √(k2 – sin2θ)]/[(k2 – sin2θ) - cos2θ]
=2c√(k2 – sin2θ)/[w(k2 - 1)]

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