The plane can fly at 800km/h and there is a wind of 100km/h at 40.0 degrees South of West.

(b) Find the resultant velocity if the plane steers due East at 800km/h:

Treating the plane and wind as vectors we place the planes vector starting at the origin and moving along the positive x-axis 800 units; then we draw a vector from this point below the axis 40 degreesĀ and aiming back towards the origin with length 100 units. The vector drawn from the origin to the end of the placed vector is the resultant.

Label the origin A, the end of the plane's vector B, and the end of the wind's vector C. Then the resultant is the vector from A to C.

AB=800,BC=100, `m/_ABC=40^@` . We can use trigonometry to find AC:

`AC^2=800^2+100^2-2(800)(100)cos40^@`

`AC^2~~527432.8891`

`AC~~726.25`

**So the resultant speed is 726.3km/h**

We can use trigonometry to find the angle:

`(sinA)/100=(sin40^@)/726.25`

`sinA~~.0885`

`A=sin^(-1)(.0885)~~5.078`

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The resultant is 726.3km/h at 5.1 degrees South of East

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(a) If the resultant is to be a velocity of 400km/h heading due East:

Let the resultant be labelled AB with A at the origin and B 400 units along the positive x-axis. Let C be placed so that BC=100 and `m/_ABC=140^@` , Then the steering velocity will be AC:

`AC^2=400^2+100^2-2(400)(100)cos140^@`

`AC^2~~231283.5554`

`AC~~480.9195`

Thus the steering speed should be 480.9km/h

To determine the direction we find `m/_A` :

`(sinA)/100=(sin140)/480.92`

`sinA~~.13366`

`A~~sin^(-1).1337~~7.68`

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To achieve a heading of due East at 400km/h the plane should fly 480.9km/h at a heading of 7.7 degrees north of East.

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