An airplane airspeed indicator reads 100 , m/ | vedclass

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(B) Let $\vec{V}_p$ be the velocity of the airplane relative to the air and $\vec{V}_w$ be the velocity of the wind relative to the ground.Given: $|\v

Vedclass · Accepted Answer

$113$

Vedclass · Answer

(B) Let $\vec{V}_p$ be the velocity of the airplane relative to the air and $\vec{V}_w$ be the velocity of the wind relative to the ground.Given: $|\vec{V}_p| = 100 \, m/s$ at $37^{\circ}$ east of north,and $\vec{V}_w = 20 \, m/s$ towards east.The velocity of the airplane relative to the ground is $\vec{V}_g = \vec{V}_p + \vec{V}_w$.In vector components:$\vec{V}_p = 100 \sin 37^{\circ} \hat{i} + 100 \cos 37^{\circ} \hat{j} = 100(0.6) \hat{i} + 100(0.8) \hat{j} = 60 \hat{i} + 80 \hat{j} \, m/s$.$\vec{V}_w = 20 \hat{i} \, m/s$.$\vec{V}_g = (60 + 20) \hat{i} + 80 \hat{j} = 80 \hat{i} + 80 \hat{j} \, m/s$.The speed relative to the ground is $|\vec{V}_g| = \sqrt{80^2 + 80^2} = 80\sqrt{2} \approx 80 	imes 1.414 = 113.12 \, m/s$.Thus,the speed is closest to $113 \, m/s$.

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