## Related questions with answers

An airplane flying due north has an air speed of 500 m/h. There is a northwesterly wind of 60 km/h. Find the true speed and direction of the plane relative to the ground.

Solution

VerifiedSince the airplane is flying due north it has a velocity vector of

$\mathbf v _{ {\color{#c34632}{\text{airplane}}}} = \big < 0 , 500 \big >$

Since wind velocity vector is northwesterly

$\mathbf v _{ {\color{#c34632}{\text{wind }}}} = \big <60 \cos 135 , 60 \sin 135 \big >$

whereas the true speed of the airplane is affected by the wind velocity, thus the contribution of the wind speed to the airplane speed is the projection of wind velocity vector onto airplane velocity vector.

The contribution of the wind velocity to the airplane velocity is

$\begin{align*} \mathbf v_{ {\color{#c34632}{\text{cont }}}}&=\mathbf {proj_{v_{{\color{#c34632}{\text{airplane}}}}} \,\,{v_{ {\color{#c34632}{\text{wind }}}}}}\\\\ &= \Big (\dfrac{ \mathbf v_{{\color{#c34632}{\text{airplane}}}} \, \cdot \, \mathbf v_{{\color{#c34632}{\text{wind}}}} }{ || \mathbf v_{{\color{#c34632}{\text{airplane}}}} ||^2 }\Big) \, \, \mathbf v_{{\color{#c34632}{\text{airplane}}}}\\\\ &= \Big[ \dfrac{ \big < 0 , 500 \big > \, \cdot \, \big <-30\sqrt 2 , 30\sqrt 2 \big > }{(500)^2 }\,\Big ]\,\, \big < 0 , 500 \big >\\\\ &= \big < 0 , 30\sqrt 2 \big > \end{align*}$

Therefore, the true velocity of the airplane is

$\begin{align*} \mathbf v _{ {\color{#c34632}{\text{true}}}} &= \mathbf v _{ {\color{#c34632}{\text{airplane}}}} +\mathbf v _{ {\color{#c34632}{\text{cont}}}} \\\\ &= \big < 0 , 500 \big > + \big < 0 , 30\sqrt 2 \big >\\\\ &= \big < 0 , 500+ 30 \sqrt 2 \big > \\\\ &= \big < 0 , 542.42 \big > \\\\ \end{align*}$

consequently airplane speed relative to the ground is

$|| \mathbf v _{ {\color{#c34632}{\text{true}}}} ||=542.42$

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