Given,
The initial eastward velocity of the airplane. v₁=90 km/hr
The speed of the wind from the north, v₂=40 km/hr
In vector notation, the velocity of the airplane is
[tex]\vec{v_1}=90\hat{i}[/tex]And the velocity of the wind in vector notation is
[tex]\vec{v_2}=-40\hat{j}[/tex]Thus the resultant velocity is,
[tex]\begin{gathered} \vec{V}=\vec{v_1}+\vec{v_2} \\ =90\hat{i}-40\hat{j} \end{gathered}[/tex]The magnitude of this velocity, i.e., resulting speed of the airplane is
[tex]\begin{gathered} V=\sqrt[]{90^2+40^2} \\ =98.5\text{ km/hr} \end{gathered}[/tex]The direction of the resulting velocity of the airplane is,
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{-40}{90}) \\ =-24^{\circ} \end{gathered}[/tex]The negative sign in the angle is given in the clockwise direction from the positive x-axis.
Thus the speed of the plane is 98.5 km/hr and the direction is -24°
