Question

1. 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 9.0 m/s east. to an observer on the ground, what is the magnitude of brendas velocity?

54.9 m/s
27.5 m/s
35.1 m/s
46.1 m/s

Explanation:

Step1: Analyze the velocity vectors

Brenda's velocity with respect to air \(v_{ba} = 45.0\ m/s\) (south) and air's velocity \(v_{ag}=9.0\ m/s\) (east). These two - velocity vectors are perpendicular to each other.

Step2: Use the Pythagorean theorem

The magnitude of Brenda's velocity with respect to the ground \(v_{bg}\) is given by \(v_{bg}=\sqrt{v_{ba}^{2}+v_{ag}^{2}}\).
Substitute \(v_{ba} = 45.0\ m/s\) and \(v_{ag}=9.0\ m/s\) into the formula:

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Answer:

46.1 m/s