A light plane attains an airspeed of $\mathbf{500}\mathbf{}\mathit{k}\mathit{m}\mathbf{/}\mathit{h}$. The pilot sets out for a destination $\mathbf{800}\mathbf{}\mathit{k}\mathit{m}$due north but discovers that the plane must be headed role="math" localid="1655442378207" $\mathbf{20}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$east of due north to fly there directly. The plane arrives in$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{h}\mathit{r}$. What were the (a) magnitude and (b) direction of the wind velocity?

1. The velocity of a plane with respect to air is ${\mathit{V}}_{\mathbf{P}}\mathbf{=}\mathbf{500}\mathbf{}\mathit{k}\mathit{m}\mathbf{/}\mathit{h}$

2. The distance that should be covered to reach the destination point is $\mathit{x}\mathbf{=}\mathbf{800}\mathbf{}\mathit{k}\mathit{m}$.

3. The time taken to reach the destination is $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{h}$

The operation of combining two or more vectors into a vector sum is known as vector addition. Using the vector diagram and resolving vectors into its components, we can find the magnitude and the velocity of the resultant vector.

Formulae:

Magnitude of the vector$\stackrel{\xcancel{}}{V}$is,

$V=\sqrt{V_x^2+V_y^2}$

The direction of vector is given by,

$\tan \theta =\frac{V_y}{V_x}$

The plane covers 800 km distance in$\mathbf{2}\mathbf{}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathit{s}\mathbf{}$. So, it’s velocity with respect to ground is,

$$\begin{gathered} V=\frac{x}{t} \\ =\frac{800km}{2h} \\ =400km/h \\ \stackrel{\xcancel{}}{V}=\left(400km/h\right)\hat{j} \end{gathered}$$

The velocity of the plane with respect to air can be written in the vector notation as,

localid="1660896163587" $$\begin{gathered} {\overrightarrow{V}}_p=V_p\sin \left({20}^{°}\right)\hat{i}+V_p\cos \left({20}^{°}\right)\hat{j} \\ =\left(500km/h\right)\sin \left({20}^{°}\right)\hat{i}+\left(500km/h\right)\cos \left({20}^{°}\right)\hat{j} \end{gathered}$$

From the vector diagram, the resultant velocity of the plane with respect to ground can be written as,

localid="1660896182707" $$\begin{gathered} \overrightarrow{V}={\overrightarrow{V}}_p+{\overrightarrow{V}}_w \\ \left(400km/h\right)\hat{j}=\left(500km/h\right)\sin {20}^{°}\hat{i}+\left(500km/h\right){\cos }^{°}\hat{j}+V_w \\ {\overrightarrow{V}}_w=-171\hat{i}-70\hat{j} \end{gathered}$$

Magnitude of the vectorlocalid="1660896193515" ${\overrightarrow{V}}_w$is,

$$\begin{gathered} V_w=\sqrt{{C_{wx}}^2+{V_{wy}}^2} \\ =\sqrt{{\left(-171km/s\right)}^2+{\left(-70km/h\right)}^2} \\ =184.7km/h \\ \approx 185km/h \end{gathered}$$

Therefore, the speed of wind is $\mathbf{185}\mathbf{}\mathit{k}\mathit{m}\mathbf{/}\mathit{h}$.

The direction of the wind velocity$\left(V_w\right)$is given by

$$\begin{gathered} \tan \theta =\frac{V_{wy}}{V_{wx}} \\ =\frac{-70km/h}{-171km/h} \\ \theta ={\tan }^{-1}\left(\frac{-70km/h}{-171km/h}\right) \\ =22.{26}^{°} \\ \approx 22.3^{°} \end{gathered}$$

From the figure, we can say that the direction of wind is $\mathbf{22}\mathbf{.}{\mathbf{3}}^{\mathbf{°}}\mathbf{}\mathit{s}\mathit{o}\mathit{u}\mathit{t}\mathit{h}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{w}\mathit{e}\mathit{s}\mathit{t}$.