Question

3 brenda is flying a small airplane. her velocity, with respect to the air around her is 45.0 m/s south. the air is also moving with a velocity of 9.90 m/s east. to an observer on the ground, what is the angle of brendas velocity? 17.8° east of south 12.4° east of south 12.4° south of east 17.8° east of south

Explanation:

Step1: Analyze the velocity components

Let \(v_{plane - air}=45.0\ m/s\) (south - y - component) and \(v_{air}=9.90\ m/s\) (east - x - component).

Step2: Use the tangent function

The tangent of the angle \(\theta\) (east of south) is given by \(\tan\theta=\frac{v_{air}}{v_{plane - air}}\).
Substitute \(v_{air} = 9.90\ m/s\) and \(v_{plane - air}=45.0\ m/s\) into the formula: \(\tan\theta=\frac{9.90}{45.0}\).

Step3: Calculate the angle

\(\theta=\arctan(\frac{9.90}{45.0})\).
Using a calculator, \(\theta=\arctan(0.22)\approx12.4^{\circ}\).

Answer:

12.4° east of south