Question

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

`\frac{a}{ {v}^(2) - {w}^(2) } ( \sqrt{ {v}^(2) - {w}^(2) \ { \sin \ }^(2) \alpha } + w \cos( \alpha )`
[using relative velocity /\ ]​

Answer 1

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)