Answer :
Answer:
The speed of the aircraft relative to the ground is 606.9 [tex]m/s^2[/tex]
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
= [tex](120 cos(30), 120 \times sin(30))[/tex]
= [tex](120 \times \frac{\sqrt{(3)}}{2}, 120 \times \frac{1}{ 2})[/tex]
= [tex](60 \times \sqrt(3), 60)[/tex]
Adding the vectors.
=[tex](500, 0) + (60 \times \sqrt{3}, 60)[/tex]
=[tex](500 + 60 \times \sqrt(3), 60)[/tex]
=[tex](500 + 60 \times 1.73, 60)[/tex]
=[tex](500 +103.8, 60)[/tex]
= (603.8, 60)
Returning back to polar form
Magnitude = [tex]\sqrt{603.923^2 + 60^2} = 606.9[/tex]