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.