91Ó°ÊÓ

The given data can be listed below.

Relative velocity is defined as the velocity of an object relative to another observer. It is the time rate of change of relative position of one object with respect to another object.

Relative speed is the rate at which one thing moves or is at rest in comparison to another item that moves or is at rest.

The diagram of velocity using vector addition is given as below.

The velocity of windspeed relative to earth is given by,

$\begin{array}{rcl}{v}_{M/E}& =& -10\cos \left({45}^o\right)\hat{i}-10\sin \left({45}^o\right)\hat{j}\\ & =& \left(-7.07\hat{i}-7.07\hat{j}\right)m/s\end{array}$

The component of velocity of plane relative to earth is given by,

$v_{P/E}=v_{P/W}+v_{W/E}$

Substitute all the values in the above equation.

role="math" localid="1659676885475" $\begin{array}{rcl}{v}_{P/E}& =& \left[-7.07\hat{i}-7.07\hat{j}-35\hat{j}\right]m/s\\ & =& \left[-7.07\hat{i}-42\hat{j}\right]m/s\end{array}$

Thus, the components of v_P/E are $\begin{array}{rcl}& & \left[-7.07\hat{i}-42\hat{j}\right]m/s\end{array}$.

The magnitude of velocity of plane relative to earth is given by,

$v_{PE}=\sqrt{v_x^2+v_y^2}$

Here, v_x is the x-component of velocity of plane and v_y is the y-component.

Substitute all the values in the above,

$\begin{array}{rcl}{v}_{PE}& =& \sqrt{{\left(-7.07m/s\right)}^2+{\left(-42m/s\right)}^2}\\ & =& \sqrt{49.9849{\left(m/s\right)}^2+1764{\left(m/s\right)}^2}\\ & =& 42.6m/s\end{array}$

The direction of the plane is given by,

${\mathrm{θ}}_2={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$

Substitute all the values in the above,

$\begin{array}{rcl}{\mathrm{θ}}_2& =& {\tan }^{-1}\left(\frac{-7.07}{42}\right)\\ & =& 0.168\\ & =& 9.6^o\end{array}$

Hence, the magnitude and direction of velocity are 42.6m/s and 9.6^o.