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A jet plane is traveling with a velocity of 720 kilometers/hour due north. There is a crosswind of 40 kilometers/hour from the east. Find the resultant velocity and the direction of the jet plane.

Verified Answer :

v = 721 km/h
θ = 3.2°

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skyluke89

Answer: 721.1 km/h, at 86.6 degrees north of west

Explanation:

The magnitude of the velocity of the plane can be found by calculating the resultant of the two velocities:

\(R=\sqrt{(720 km/h)^2+(40 km/h)^2}=721.1 km/h\)

The direction can be found as follows:

\(\theta = tan^{-1} (\frac{720 km/h}{40 km/h})=86.8^{\circ}\)

where 720 km/h represents the velocity towards north, while 40 km/h represents the velocity towards west, so the angle is measured with respect to east. Therefore:

the direction is \(86.8^{\circ}\) north of west

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