Question

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.

Answer 1

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

Explanation:

Given

Plane is initially flying with velocity of magnitude $v=600\ mph$

at angle of $40^{\circ}$ with North towards west

Velocity of plane airplane can be written as

$v_a=600(-\sin 40\hat{i}+\cos 40\hat{j})$

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

$v_w=80(\cos 45\hat{i}+\sin 45\hat{j})$

resultant velocity

$\vec{v_R}=\vec{v_a} +\vec{v_w}$

$\vec{v_R}=600(-\sin 40\hat{i}+\cos 40\hat{j})+ 80(\cos 45\hat{i}+\sin 45\hat{j})$

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

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

for direction $\tan \theta =\frac{516.18}{329.11}$

$\tan \theta =1.568$

$\theta =57.47^{\circ}$ west of North