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A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s. The airspeed of the bee (i.e., its speed relative to the air) is 8.33 m/s. In which direction should the bee head in order to fly directly to the flower, due North relative to the ground? Answer in units of â—¦ East of North.

Respuesta :

Answer:at an angle [tex]36.77^{\circ}[/tex] North of east

Explanation:

Given

Velocity of Wind [tex](v_w)=6.68 m/s[/tex]

Velocity of bee relative to wind [tex](v_{bw})=8.33 m/s[/tex]

In vector Form

[tex]\vec{v_w}=-6.68\hat{i}[/tex]

[tex]\vec{v_{bw}}=8.33(cos\theta \hat{i}+sin\theta \hat{j})[/tex]

To get the final velocity in North cos component of [tex]v_{bw}[/tex] will balance velocity of wind

[tex]8.33\cos \theta =6.68[/tex]

[tex]cos\theta =0.801[/tex]

[tex]\theta =36.77^{\circ}[/tex]

Therefore Final velocity is 8.33sin(36.77)=4.98 m/s due to North

at an angle [tex]36.77^{\circ}[/tex] North of east

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