Question

A student is flying west on a school trip from Winnipeg to Calgary in a jet that has an air velocity of 792 km/h.The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h [N] is _____° S of W. (give your answer with the correct number of significant digits and do not include units)

Answer 1

Answer:

The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h[N] is 4.5° S of W

Explanation:

The given parameters are;

Velocity of Jet = 792 km/h

Direction of jet velocity = West

Velocity of wind = 62.0 km/h

Direction of wind velocity = North

Therefore, the jet has to have a component of 62.0 km/h South of West to compensate for the wind velocity

The direction of the plane, θ° South of West (S of W) to compensate for the wind is given as follows;

$Tan \left (\theta \right )= \dfrac{62}{792} = \dfrac{31}{396}$

Therefore;

$\theta = tan^{-1}\left (\dfrac{31}{396} \right ) = 4.476^{\circ} \approx 4.5^{\circ}$

The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h[N] = 4.5° S of W.

Answer 2

The direction the plane would have to fly to is : 4.5° south of west

Given data :

Jet velocity = 792 km/h

wind velocity = 62 km/h

Given that the direction of the jet velocity is west and the direction of the wind velocity is north.

The direction of component velocity of the Jet would be 62 km/h south of west

Determine the direction ( ∅ ) of the plane south of west

Tan ( ∅ ) = 62 / 792

Therefore ( ∅ ) = Tan⁻¹ ( 62/792 )

                          ≈ 4.5° S of W

Hence we can conclude that The direction the plane would have to fly to is : 4.5° south of west

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