Question

qid 1026
an airplane has a velocity of 502 km/h relative to the moving air ($v_{pa}$). at the same time, a wind blows northward with a speed of 82 km/h relative to earth. the airplane is moving in an easterly direction relative to earth ($v_{pg}$). in what direction is the airplane heading (a)? (answer in terms of degrees south of east)
9.4° south of east
71.2° south of east
0.16° south of east

Explanation:

Step1: Analyze velocity vectors

We have velocity of plane relative to air $V_{pa} = 502$ km/h and velocity of wind relative to earth $V_{wa}=82$ km/h. The plane moves east - ward relative to earth $V_{pg}$. Using vector addition $V_{pg}=V_{pa}+V_{wa}$. Let the angle of the plane's heading (relative to the east - west direction) be $\theta$.

Step2: Use trigonometry

We know that $\sin\theta=\frac{V_{wa}}{V_{pa}}$. Substituting $V_{wa} = 82$ km/h and $V_{pa}=502$ km/h, we get $\sin\theta=\frac{82}{502}\approx0.163$.

Step3: Calculate the angle

Then $\theta=\sin^{- 1}(0.163)\approx9.4^{\circ}$.

Answer:

$9.4^{\circ}$ south of east