91Ó°ÊÓ

1. The time of plane for flying is,

$$\begin{gathered} \mathbf{∆}\mathit{t}\mathbf{=}\mathbf{15}\mathbf{}\mathit{m}\mathit{i}\mathit{n} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathit{h} \end{gathered}$$

2. The velocity of airlocalid="1660895904400" ${\stackrel{→}{\mathbf{V}}}_{\mathbf{A}\mathbf{G}}\mathbf{=}\mathbf{42}\mathbf{}\mathit{k}\mathit{m}\mathbf{/}\mathit{h}$

3. The angle made is$\mathit{θ}\mathbf{=}{\mathbf{20}}^{\mathbf{°}}\mathbf{}\mathit{s}\mathit{o}\mathit{u}\mathit{t}\mathit{h}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{e}\mathit{a}\mathit{s}\mathit{t}$

4. The distance covered by plane is $s=55km$

Use the concept of the relative velocity and draw the figure according to the directions of plane and wind. We can use expression of velocity to find the velocity of pilot relative to ground. By using vector addition law, we can find speed of the air plane relative to the air.

Formula:

${\stackrel{→}{\mathbf{V}}}_{\mathbf{PG}}={\stackrel{→}{\mathbf{V}}}_{PA}+{\stackrel{→}{\mathbf{V}}}_{\mathbf{AG}}$

$velocity=\frac{displacement}{timeinterval}$

The velocity of the plane relative to the ground ${\stackrel{→}{\mathbf{V}}}_{\mathbf{AG}}$is,

$$\begin{gathered} velocity=\frac{displacement}{timeinterval} \\ {\stackrel{→}{\mathbf{V}}}_{\mathbf{PG}}=\frac{55km}{0.25h} \\ =220\frac{km}{h}\hat{j} \end{gathered}$$

The velocity of air with respect to ground ${\stackrel{→}{\mathbf{V}}}_{\mathbf{AG}}$is,

$$\begin{gathered} {\stackrel{→}{\mathbf{V}}}_{\mathbf{AG}}=\left(42kmkm/h\right)\left[\cos 20\hat{i}-\sin 20\hat{j}\right] \\ =\left[\left(39.47km/hr\right)\hat{i}-\left(14.36km/h\right)\hat{j}\right] \end{gathered}$$

From the figure, using vector addition,

$$\begin{gathered} {\stackrel{→}{\mathbf{V}}}_{\mathbf{PG}}={\stackrel{→}{\mathbf{V}}}_{PA}+{\stackrel{→}{\mathbf{V}}}_{\mathbf{AG}} \\ {\stackrel{→}{\mathbf{V}}}_{PA}={\stackrel{→}{\mathbf{V}}}_{\mathbf{PG}}-{\stackrel{→}{\mathbf{V}}}_{\mathbf{AG}} \\ =\left(220km/h\right)\hat{j}-\left(39.47km/h\right)\hat{i}+\left(14.36km/h\right)\hat{j} \\ {\stackrel{→}{\mathbf{V}}}_{PA}=-\left(39.47km/h\right)\hat{i}+\left(234.36km/h\right)\hat{j} \end{gathered}$$

In magnitude,

$$\begin{gathered} V_{PA}=\sqrt{{\left(39.47km/h\right)}^2+{\left(234.36km/h\right)}^2} \\ =237.66km/h \\ ≈2.4×{10}^{2}km/h \end{gathered}$$

Therefore, the speed of the airplane relative to the air is $2.4×{10}^{2}km/h$.