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after flying for 17 min in a wind blowing 38 km/h at an angle of $18^\circ$ south of east, an airplane pilot is over a town that is 55 km due north of the starting point. what is the speed of the airplane relative to the air?
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Explanation:

Step1: Convert time to hours

$t = \frac{17}{60} \approx 0.2833$ hours

Step2: Define displacement vectors

Let $\vec{v}_{pa}$ = plane's velocity relative to air, $\vec{v}_{ag}$ = wind velocity relative to ground, $\vec{v}_{pg}$ = plane's velocity relative to ground.
$\vec{v}_{ag} = 38\cos(18^\circ)\hat{i} - 38\sin(18^\circ)\hat{j}$
$\vec{v}_{pg} = \frac{55}{0.2833}\hat{j} \approx 194.14\hat{j}$ km/h

Step3: Relate velocities: $\vec{v}_{pg} = \vec{v}_{pa} + \vec{v}_{ag}$

Rearrange: $\vec{v}_{pa} = \vec{v}_{pg} - \vec{v}_{ag}$
$\vec{v}_{pa} = -38\cos(18^\circ)\hat{i} + (194.14 + 38\sin(18^\circ))\hat{j}$

Step4: Calculate components

$v_{pa,x} = -38\cos(18^\circ) \approx -36.17$ km/h
$v_{pa,y} = 194.14 + 38\sin(18^\circ) \approx 194.14 + 11.74 = 205.88$ km/h

Step5: Find magnitude of $\vec{v}_{pa}$

$v_{pa} = \sqrt{(-36.17)^2 + (205.88)^2}$
$v_{pa} = \sqrt{1308.27 + 42386.57} = \sqrt{43694.84} \approx 209.0$ km/h

Answer:

209.0 km/h