Anovice pilot sets a plane's controls, thinking the plane will

The plane's velocity is intended to be $2.50\times {10}^2\text{ km/h}$ towards the north. This can be represented as a vector pointing upwards with a magnitude of $2.50\times {10}^2\text{ km/h}$. We can express this vector as $v_p=250\text{ km/h}·\hat{j}$, where $\hat{j}$ is the unit vector in the y-direction.
The wind's velocity is given as $75\text{ km/h}$ towards the southeast. We can break down this velocity into its x and y components. The angle between the wind's velocity and the positive x-axis is ${45}^{\circ }$ (since southeast is halfway between east and south). Using trigonometry, we can find the x and y components of the wind's velocity as $v_{wx}=75\text{ km/h}·\cos ({45}^{\circ })=\frac{75\text{ km/h}}{\sqrt{2}}$ and $v_{wy}=-75\text{ km/h}·\sin ({45}^{\circ })=-\frac{75\text{ km/h}}{\sqrt{2}}$.
Now, we can add the plane's velocity vector and the wind's velocity vector to obtain the resultant velocity vector. Adding the x and y components separately, we get:
$v_{rx}=v_{px}+v_{wx}=0+\frac{75\text{ km/h}}{\sqrt{2}}=\frac{75\text{ km/h}}{\sqrt{2}}$
$v_{ry}=v_{py}+v_{wy}=250\text{ km/h}+(-\frac{75\text{ km/h}}{\sqrt{2}})=250\text{ km/h}-\frac{75\text{ km/h}}{\sqrt{2}}$
Therefore, the resultant velocity vector can be expressed as $v_r=v_{rx}·\hat{i}+v_{ry}·\hat{j}$. Plugging in the values, we have:
$v_r=\frac{75\text{ km/h}}{\sqrt{2}}·\hat{i}+(250\text{ km/h}-\frac{75\text{ km/h}}{\sqrt{2}})·\hat{j}$
Simplifying further, we get:
$v_r\approx \frac{75}{\sqrt{2}}\text{ km/h}·\hat{i}+(250-\frac{75}{\sqrt{2}})\text{ km/h}·\hat{j}$
Therefore, the plane's resultant velocity is approximately $\frac{75}{\sqrt{2}}\text{ km/h}$ towards the east and $(250-\frac{75}{\sqrt{2}})\text{ km/h}$ towards the north.

Result:
${\mathbf{v}}_{\mathbf{r}}=2.50\times {10}^2\mathrm{k}\mathrm{m}/\mathrm{h}·\hat{\mathbf{j}}+{\mathbf{v}}_{\mathbf{w}\mathbf{e}}+{\mathbf{v}}_{\mathbf{w}\mathbf{s}}$
Solution:
Given:
${\mathbf{v}}_{\mathbf{p}}=2.50\times {10}^2\mathrm{k}\mathrm{m}/\mathrm{h}$ (to the north) and
${\mathbf{v}}_{\mathbf{w}}=75\mathrm{k}\mathrm{m}/\mathrm{h}$ (toward the southeast).
First, we represent the velocities as vectors. Since ${\mathbf{v}}_{\mathbf{p}}$ is to the north, its vector representation is ${\mathbf{v}}_{\mathbf{p}}=2.50\times {10}^2\mathrm{k}\mathrm{m}/\mathrm{h}·\hat{\mathbf{j}}$, where $\hat{\mathbf{j}}$ represents the unit vector in the north direction.
Similarly, since ${\mathbf{v}}_{\mathbf{w}}$ is toward the southeast, we can break it down into two components: one component in the east direction and another in the south direction. Let's denote the east component as ${\mathbf{v}}_{\mathbf{w}\mathbf{e}}$ and the south component as ${\mathbf{v}}_{\mathbf{w}\mathbf{s}}$.
Using trigonometry, we can determine the magnitudes of ${\mathbf{v}}_{\mathbf{w}\mathbf{e}}$ and ${\mathbf{v}}_{\mathbf{w}\mathbf{s}}$:
${\mathbf{v}}_{\mathbf{w}\mathbf{e}}=v_w·\cos ({45}^{\circ })=75\mathrm{k}\mathrm{m}/\mathrm{h}·\cos ({45}^{\circ })$,
${\mathbf{v}}_{\mathbf{w}\mathbf{s}}=v_w·\sin ({45}^{\circ })=75\mathrm{k}\mathrm{m}/\mathrm{h}·\sin ({45}^{\circ })$.
The vector representation of ${\mathbf{v}}_{\mathbf{w}}$ is the sum of ${\mathbf{v}}_{\mathbf{w}\mathbf{e}}$ and ${\mathbf{v}}_{\mathbf{w}\mathbf{s}}$, i.e.,
${\mathbf{v}}_{\mathbf{w}}={\mathbf{v}}_{\mathbf{w}\mathbf{e}}+{\mathbf{v}}_{\mathbf{w}\mathbf{s}}$.
Finally, we can calculate the resultant velocity ${\mathbf{v}}_{\mathbf{r}}$ by adding ${\mathbf{v}}_{\mathbf{p}}$ and ${\mathbf{v}}_{\mathbf{w}}$:
${\mathbf{v}}_{\mathbf{r}}={\mathbf{v}}_{\mathbf{p}}+{\mathbf{v}}_{\mathbf{w}}$.
Therefore, the plane's resultant velocity is ${\mathbf{v}}_{\mathbf{r}}=2.50\times {10}^2\mathrm{k}\mathrm{m}/\mathrm{h}·\hat{\mathbf{j}}+{\mathbf{v}}_{\mathbf{w}\mathbf{e}}+{\mathbf{v}}_{\mathbf{w}\mathbf{s}}$.

To solve this problem using graphical techniques, we can represent the velocities as vectors and use vector addition to find the resultant velocity. Let's denote the plane's velocity as ${\overrightarrow{v}}_p$ and the wind's velocity as ${\overrightarrow{v}}_w$. We are given:
${\overrightarrow{v}}_p=2.50\times {10}^2\text{km/h}$ (to the north) and
${\overrightarrow{v}}_w=75\text{km/h}$ (toward the southeast).
We need to determine the resultant velocity ${\overrightarrow{v}}_r$ of the plane.
First, let's represent the plane's velocity vector ${\overrightarrow{v}}_p$ on a coordinate system. Since it is going north, its velocity vector will be vertical and pointing upward. Let's denote this vector as ${\overrightarrow{v}}_{p_y}$.
${\overrightarrow{v}}_{p_y}=2.50\times {10}^2\text{km/h}\hat{j}$, where $\hat{j}$ represents the unit vector in the y-direction (north).
Next, let's represent the wind's velocity vector ${\overrightarrow{v}}_w$ on the same coordinate system. Since it is going toward the southeast, its velocity vector will point in the southeast direction. Let's resolve this vector into its x and y components. The magnitude of the wind's velocity vector can be found using the Pythagorean theorem:
$|{\overrightarrow{v}}_w|=\sqrt{v_{w_x}^2+v_{w_y}^2}=\sqrt{{75}^2+{75}^2}\text{km/h}$.
The x-component of ${\overrightarrow{v}}_w$ can be found using trigonometry:
$v_{w_x}=|{\overrightarrow{v}}_w|\cos (\theta )=75\cos ({45}^{\circ })\text{km/h}$.
Similarly, the y-component of ${\overrightarrow{v}}_w$ can be found as:
$v_{w_y}=|{\overrightarrow{v}}_w|\sin (\theta )=75\sin ({45}^{\circ })\text{km/h}$.
Now, we can represent the wind's velocity vector ${\overrightarrow{v}}_w$ in terms of its components:
${\overrightarrow{v}}_w=v_{w_x}\hat{i}+v_{w_y}\hat{j}$, where $\hat{i}$ represents the unit vector in the x-direction (east).
To find the resultant velocity ${\overrightarrow{v}}_r$, we need to add the plane's velocity vector ${\overrightarrow{v}}_p$ and the wind's velocity vector ${\overrightarrow{v}}_w$:
${\overrightarrow{v}}_r={\overrightarrow{v}}_p+{\overrightarrow{v}}_w$.
Now, we substitute the values and perform the vector addition:
${\overrightarrow{v}}_r=(0\hat{i}+2.50\times {10}^2\hat{j})+(75\cos ({45}^{\circ })\hat{i}+75\sin ({45}^{\circ })\hat{j})$.
Simplifying the expression, we get:
${\overrightarrow{v}}_r=75\cos ({45}^{\circ })\hat{i}+(2.50\times {10}^2+75\sin ({45}^{\circ }))\hat{j}$.
Finally, we can calculate the magnitude and direction of ${\overrightarrow{v}}_r$:
$|{\overrightarrow{v}}_r|=\sqrt{(75<br>\cos ({45}^{\circ }){)}^2+(2.50\times {10}^2+75\sin ({45}^{\circ }){)}^2}\text{km/h}$.
${\theta }_r={\tan }^{-1}\left(\frac{2.50\times {10}^2+75\sin ({45}^{\circ })}{75\cos ({45}^{\circ })}\right)$.
Hence, the resultant velocity of the plane is $|{\overrightarrow{v}}_r|$ km/h, directed at an angle of ${\theta }_r$ with respect to the x-axis.