Let the velocity of the airplane relative to the air be v_{p}_{a}, its velocity relative to the earth be v_{pg}, and the velocity of the jet stream relative to the earth be v_{ag}.

The speed of the airplane relative to the earth is expressed as:

$\overline{){{\mathbf{v}}}_{\mathbf{p}\mathbf{g}}{\mathbf{=}}{{\mathbf{v}}}_{\mathbf{p}\mathbf{a}}{\mathbf{+}}{{\mathbf{v}}}_{\mathbf{a}\mathbf{g}}}$

Let east be the +x direction and north be the +y direction

The x components of velocity, v_{pg} is given by:

v_{pgx} = v_{pg}cos(180°+45°) = v_{pg} cos (225°)

Similarly, v_{pax} = v_{pa} cos(180°+5.00°) = v_{pa} cos (185°)

and v_{agx} = v_{ag}cos (-20°)

Therefore, the expression for the x components of the airplane's speed relative to the earth becomes:

$\overline{){{\mathbf{v}}}_{\mathbf{p}\mathbf{g}}{\mathbf{c}}{\mathbf{o}}{\mathbf{s}}{\mathbf{(}}{\mathbf{225}}{\mathbf{\xb0}}{\mathbf{)}}{\mathbf{=}}{{\mathbf{v}}}_{\mathbf{p}\mathbf{a}}{\mathbf{c}}{\mathbf{o}}{\mathbf{s}}{\mathbf{(}}{\mathbf{185}}{\mathbf{\xb0}}{\mathbf{)}}{\mathbf{+}}{{\mathbf{v}}}_{\mathbf{a}\mathbf{g}}{\mathbf{c}}{\mathbf{o}}{\mathbf{s}}{\mathbf{(}}{\mathbf{-}}{\mathbf{20}}{\mathbf{\xb0}}{\mathbf{)}}}$

Likewise, the y component of the airplane's speed relative to the earth is then written as:

$\overline{){{\mathbf{v}}}_{{\mathbf{pg}}}{\mathbf{sin}}{\mathbf{(}}{\mathbf{225}}{\mathbf{\xb0}}{\mathbf{)}}{\mathbf{=}}{{\mathbf{v}}}_{{\mathbf{pa}}}{\mathbf{sin}}{\mathbf{(}}{\mathbf{185}}{\mathbf{\xb0}}{\mathbf{)}}{\mathbf{+}}{{\mathbf{v}}}_{{\mathbf{ag}}}{\mathbf{sin}}{\mathbf{(}}{\mathbf{-}}{\mathbf{20}}{\mathbf{\xb0}}{\mathbf{)}}}$

An airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 20° south of east. Its direction of motion relative to the Earth is 45.0° south of west, while its direction of travel relative to the air is 5.00° south of west.**(a)** What is the airplane’s speed relative to the air mass?**(b)** What is the airplane’s speed relative to the Earth?

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