An airplane is traveling at a fixed altitude with an outside wind factor. The airplane is headed N 40° W at a speed of 600 miles per hour. As the airplane comes to a certain point, it comes across a wind in the direction N 45° E with a velocity of 80 miles per hour. What are the resultant speed and direction of the airplane? Round your answers to the nearest hundredth.

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nuuk nuuk · Answer 1 · 2020-05-06T06:22:07+02:00

Answer:[tex]\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18][/tex]

Explanation:

Given

Plane is initially flying with velocity of magnitude [tex]v=600\ mph[/tex]

at angle of [tex]40^{\circ}[/tex] with North towards west

Velocity of plane airplane can be written as

[tex]v_a=600(-\sin 40\hat{i}+\cos 40\hat{j})[/tex]

Now wind is encountered with speed of [tex]v=80\ mph[/tex] at angle of [tex]N45^{\circ}E[/tex]

[tex]v_w=80(\cos 45\hat{i}+\sin 45\hat{j})[/tex]

resultant velocity

[tex]\vec{v_R}=\vec{v_a} +\vec{v_w}[/tex]

[tex]\vec{v_R}=600(-\sin 40\hat{i}+\cos 40\hat{j})+ 80(\cos 45\hat{i}+\sin 45\hat{j})[/tex]

[tex]\vec{v_R}=\hat{i}[-385.67+56.56]+\hat{j}[459.62+56.56][/tex]

[tex]\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18][/tex]

for direction [tex]\tan \theta =\frac{516.18}{329.11}[/tex]

[tex]\tan \theta =1.568[/tex]

[tex]\theta =57.47^{\circ}[/tex] west of North