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An airplane is pushed 22 degrees eastward off its northward course by a jet stream traveling at a speed of 136.73 km/hr. The new speed is 365 km/hr. Find the plane's original velocity.

Respuesta :

As the speed of airplane is change due to jet stream

So the net speed is given as

[tex]v_{net} = v_{plane} + v_{stream}[/tex]

now we can rearrange it as

[tex]v_{plane} = v_{net} - v_{stream}[/tex]

now by the formula of vector difference we have

[tex]v_{plane} = \sqrt{v_{net}^2 + v_{stream}^2 - 2v_{net}v_{stream}cos\theta}[/tex]

now plug in all values

[tex]v_{plane} = \sqrt{365^2 + 136.73^2 - 2* 365* 136.73*cos22}

[tex]v_{plane} = 243.7 km/hr[/tex]

so above is the speed of the plane

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