Question

a plane leaves the airport on a bearing of 45 degree travelling at 400 mph. the wind is blowing at bearing of 135 degree at the speed of 40mph. what is the actual velocity and direction of the plane?

Answer 1

$\vec{v_a}=\vec{v_p}+\vec{v_w}$

where $\vec{v_a}$ - vector of the actual velocity of the plane; $\vec{v_p}$ - vector of the own velocity of the plane; $\vec{v_w}$ - vector of the wind velocity.

$v_a=\sqrt{v_{ax}^2+v_{ay}^2}$

where $v_a$ - absolute value of the actual velocity of the plane; $v_{ax}$ - x-component of the $\vec{v_a}$; $v_{ay}$ - y-component of the $\vec{v_a}$.

$v_{ax}$ - the algebraic sum of the x-components of $\vec{v_p}$ and $\vec{v_w}$; $v_{ay}$ - the algebraic sum of the y-components of $\vec{v_p}$ and $\vec{v_w}$.

$v_{ax}=v_{px}+v_{wx}=v_p \cdot \cos{\alpha} + v_w \cdot \cos{\beta}$

$v_{ay}=v_{py}+v_{wy}=v_p \cdot \sin{\alpha} + v_w \cdot \sin{\beta}$

where $v_p$ and $v_w$ - absolute values of the own velocity of the plane and wind velocity, respectively; $v_{px}$ and $v_{wx}$ - x-components of the $\vec{v_p}$ and $\vec{v_w}$, respectively; $v_{py}$ and $v_{wy}$ - y-components of the $\vec{v_p}$ and $\vec{v_w}$​, respectively; $\alpha$ - angle of the plane rising; $\beta$ - angle of the wind direction.

$v_{ax}=400 \cdot \cos{45 \degree} + 40 \cdot \cos{135 \degree} = 400 \cdot \frac{\sqrt{2}}{2} - 40 \cdot \frac{\sqrt{2}}{2} \approx 254.56 \; mph$

$v_{ay}=400 \cdot \sin{45 \degree} + 40 \cdot \sin{135 \degree} = 400 \cdot \frac{\sqrt{2}}{2} + 40 \cdot \frac{\sqrt{2}}{2} \approx 311.13 \; mph$

$v_a=\sqrt{254.56^2+311.13^2} \approx 402.0 \; mph$

$\tan{\gamma}=\frac{v_{ay}}{v_{ax}}$

where $\gamma$ - angle of the actual direction of the plane.

$\tan{\gamma}=\frac{311.13}{254.56} \approx 1.222$

$\gamma \approx 50.7 \degree$

Answer: actual velocity of the plane is 402.0 mph, actual direction of the plane is 50.7$\degree$.