A plane flying from Townsville, Australia, has an air speed of 264 m/s in a direction 5.0° south of west. It is in the jet stream, which is blowing at 37 m/s in a direction 15° south of east. In this problem you are going to be asked about the velocity of the airplane relative to the Earth.

Question

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Respuesta :

aristocles aristocles · Answer 1 · 2019-09-14T03:57:09+02:00

Answer:

Speed of plane is 300.5 m/s at angle of 6.22 degree South of West

Explanation:

Air speed of the plane is given as

v = 264 m/s in direction 5 degree South of West

So we have

[tex]v_1 = 264 cos5 \hat i + 264 sin5 \hat j[/tex]

[tex]v_1 = 263 \hat i + 23 \hat j[/tex]

Also we have speed of air is given as

v = 37 m/s at 15 degree South of West

so it is

[tex]v_2 = 37 cos15\hat i + 37 sin15 \hat j[/tex]

[tex]v_2 = 35.74 \hat i + 9.58 \hat j[/tex]

So the net speed of plane with respect to ground is given as

[tex]v_p = v_1 + v_2[/tex]

[tex]v_p = (263 \hat i + 23 \hat j) + (35.74 \hat i + 9.58 \hat j)[/tex]

[tex]v_p = 298.74 \hat i + 32.58\hat j[/tex]

so it is

[tex]v_p = \sqrt{298.74^2 + 32.58^2}[/tex]

[tex]v_p = 300.5 m/s[/tex]

direction is given as

[tex]\theta =tan^{-1} \frac{v_y}{v_x}[/tex]

[tex]\theta = tan^{-1} \frac{32.58}{298.74}[/tex]

[tex]\theta = 6.22 degree[/tex]