a plane decides to travel along AB towards the North with a velocity of V.
Prove that the time taken to reach the destination due to the strong winds blowing from the north to east at an angle of alpha is
[tex] \frac{a}{ {v}^{2} - {w}^{2} } ( \sqrt{ {v}^{2} - {w}^{2} \ { \sin \ }^{2} \alpha } + w \cos( \alpha ) [/tex]
[using relative velocity /\ ]​

Respuesta :

Step-by-step explanation:

The wind has a speed of w and a direction α with the vertical.  The x component of that speed is w sin α.  The y component is -w cos α.

In order to stay on the north trajectory AB, the plane must have a horizontal speed of -w sin α.  The plane's speed is v, so using Pythagorean theorem, the y component of the plane's speed is:

v² = (-w sin α)² + vᵧ²

v² = w² sin²α + vᵧ²

vᵧ = √(v² − w² sin²α)

The total vertical speed is therefore √(v² − w² sin²α) − w cos α.

If a is the length of AB, then the time is:

t = a / [√(v² − w² sin²α) − w cos α]

To rationalize the denominator, we multiply by the conjugate.

t = a / [√(v² − w² sin²α) − w cos α] × [√(v² − w² sin²α) + w cos α] / [√(v² − w² sin²α) + w cos α]

t = a [√(v² − w² sin²α) + w cos α] / (v² − w² sin²α − w² cos²α)

t = a [√(v² − w² sin²α) + w cos α] / (v² − w²)

Ver imagen MathPhys