# A pilot wishes to fly a plane due north relative to the ground. The airspeed of the plane is 200 km/h and the wind is blowing from the west to east at 90 km/h. In what direction should the plane head? What is the ground speed of the plane?

Since the wind is blowing to the east, a plane headed due north will drift off coarse toward the east.Hence, to compensate for this crosswind, the plane must head west of due north. We must use addition of velocity vectors to relate the plane relative to the ground `v_(pG)^(->)` ,...

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Since the wind is blowing to the east, a plane headed due north will drift off coarse toward the east.Hence, to compensate for this crosswind, the plane must head west of due north. We must use addition of velocity vectors to relate the plane relative to the ground `v_(pG)^(->)` , the plane relative to the air `v_(pA)^(->)` , and the air relative to the ground `v_(AG)^(->)` .

The three velocities are related by the velocity vector addition formula.

`eq. (1) :->` `v_(ac)^(->)=v_(ab)^(->)+v_(bc)^(->)`

We want to know the ground speed of the plane which is `v_(pG)^(->)` . Therefore let `a=p` , `c=G` , and `b=A` in `eq. (1)` .

`v_(pG)^(->)=v_(pA)^(->)+v_(AG)^(->)`

This makes the velocity addition diagram that is given below. From the diagram you can tell that the sine of the angle theta equals the ratio of two of the vectors.

`sin(theta)=v_(AG)^(->)/v_(pA)^(->)=(90 (km)/h)/(200 (km)/h)=9/20`

Therefore the plane must travel relative to the air,

`theta=sin^-1(9/20)~~27^@` west of north.

We can now use the Pythagorean theorem to find the magnitude of `v_(pG)^(->)` .

`|v_(pG)^(->)|=sqrt(|v_(pA)^(->)|^2+|v_(AG)^(->)|^2)`

`v_(pG)^(->)=sqrt((200 (km)/h)^2+(90 (km)/h))^2=180 (km)/h`

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