Question

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.

Answer 1

Answer:at an angle $36.77^{\circ}$ North of east

Explanation:

Given

Velocity of Wind $(v_w)=6.68 m/s$

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

In vector Form

$\vec{v_w}=-6.68\hat{i}$

$\vec{v_{bw}}=8.33(cos\theta \hat{i}+sin\theta \hat{j})$

To get the final velocity in North cos component of $v_{bw}$ will balance velocity of wind

$8.33\cos \theta =6.68$

$cos\theta =0.801$

$\theta =36.77^{\circ}$

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

at an angle $36.77^{\circ}$ North of east