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

Step-by-step 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