Question

A jet moving at 500.0 km/ h due is in a region where the wind is moving at 120.0 km/h in a direction 30.00° north of east. What is the speed of the aircraft relative to the ground?

Answer 1

Answer:

The speed of the aircraft relative to the ground is 606.9 $m/s^2$

Explanation:

x-y coordinate system:  

x is Positive due East direction. Similarly  y is Positive due North Direction.

Now let us Decompose each vector into x-y

500 km/h due east = (500, 0)

120 km/h at 30 degree north of east

= $(120 cos(30), 120 \times sin(30))$

= $(120 \times \frac{\sqrt{(3)}}{2}, 120 \times \frac{1}{ 2})$

= $(60 \times \sqrt(3), 60)$

Adding  the vectors.

=$(500, 0) + (60 \times \sqrt{3}, 60)$

=$(500 + 60 \times \sqrt(3), 60)$

=$(500 + 60 \times 1.73, 60)$

=$(500 +103.8, 60)$

= (603.8, 60)

Returning back to polar form

Magnitude = $\sqrt{603.923^2 + 60^2} = 606.9$