Question

an airplane flies 33 m/s due east while experiencing a cross - wind. the resultant velocity is determined to be 35.11 m/s east - northeast. what is the velocity of the wind? a 22 m/s north b 27 m/s north c 18 m/s north d 12 m/s north

Explanation:

Step1: Consider vector addition

Let the velocity of the airplane due - east be $\vec{v}_{a}=33\ m/s$ (east - direction) and the velocity of the wind be $\vec{v}_{w}$ (north - direction) and the resultant velocity be $\vec{v}_{r}$. The magnitude of the resultant velocity $v_{r} = 35.11\ m/s$ and is in the east - north - east direction.
We know that, using the Pythagorean theorem for vector addition since the airplane's velocity and the wind's velocity are perpendicular (east and north directions are perpendicular), $v_{r}^{2}=v_{a}^{2}+v_{w}^{2}$.

Step2: Solve for the wind velocity

We can re - arrange the formula $v_{r}^{2}=v_{a}^{2}+v_{w}^{2}$ to solve for $v_{w}$. So, $v_{w}=\sqrt{v_{r}^{2}-v_{a}^{2}}$.
Substitute $v_{a} = 33\ m/s$ and $v_{r}=35.11\ m/s$ into the formula:
\[

$$\begin{align*} v_{w}&=\sqrt{(35.11)^{2}-(33)^{2}}\\ &=\sqrt{(35.11 + 33)(35.11-33)}\\ &=\sqrt{(68.11)\times(2.11)}\\ &=\sqrt{143.7121}\\ &\approx12\ m/s \end{align*}$$

\]

Answer:

d. $12\ m/s$ north