Question

an airplane with an airspeed of 160 km/h pointing due east encounters a crossing of 175 km/h directed from south. What is the resultant magnitude and direction of the airplane's velocity

Answer 1

The magnitude of the resultant velocity is 237 km/h

The direction of the resultant velocity is $47.6^{\circ}$ north of east

Explanation:

Since velocity is a vector, we can find the resultant velocity of the airplane by using rules of vector addition.

The two components of the plane's velocity are:

$v_x = 160 km/h$ due east

$v_y = 175 km/h$ due north (from south)

The two components are perpendicular to each other, so we can find the magnitude of the resultant velocity by using Pythagorean's theorem:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{160^2+175^2}=237 km/h$

The direction of the velocity instead can be found by using

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{175}{160})=47.6^{\circ}$

in a direction north of east.

Learn more about vector addition:

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