Answer:
The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.
Explanation:
Given:
Speed of jumbo jet in southwesterly direction
= 550 mph
Velocity of jet stream from west to east direction ![(v_s)=80\ mph](https://tex.z-dn.net/?f=%28v_s%29%3D80%5C%20mph)
First let us draw a vectorial representation of the above velocity vectors.
Consider the south direction as negative y axis and west direction as negative x axis.
From the diagram,
The velocity of the jet can be represented as:
![\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph](https://tex.z-dn.net/?f=%5Cvec%7Bv_j%7D%3D-550%5Ccos%2845%29%5Cvec%7Bi%7D%2B%28-550%5Csin%2845%29%5Cvec%7Bj%7D%20%29%5C%5C%5C%5C%5Cvec%7Bv_j%7D%3D-388.91%5Cvec%7Bi%7D-388.91%5Cvec%7Bj%7D%5C%20mph)
Similarly, the velocity of the stream is, ![\vec{v_s}=80\vec{i}](https://tex.z-dn.net/?f=%5Cvec%7Bv_s%7D%3D80%5Cvec%7Bi%7D)
Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:
![\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D%3D%5Cvec%7Bv_j%7D%2B%5Cvec%7Bv_s%7D%5C%5C%5C%5C%5Cvec%7Bv%7D%3D-388.91%5Cvec%7Bi%7D-388.91%5Cvec%7Bj%7D%2B80%5Cvec%7Bi%7D%5C%5C%5C%5C%5Cvec%7Bv%7D%3D%28-388.91%2B80%29%5Cvec%7Bi%7D-388.91%5Cvec%7Bj%7D%5C%5C%5C%5C%5Cvec%7Bv%7D%3D%28-308.91%29%5Cvec%7Bi%7D-388.91%5Cvec%7Bj%7D)
Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,
![|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph](https://tex.z-dn.net/?f=%7C%5Cvec%7Bv%7D%7C%3D%5Csqrt%7B%28-308.91%29%5E2%2B%28-388.91%29%5E2%7D%5C%5C%5C%5C%7C%5Cvec%7Bv%7D%7C%3D496.67%5C%20mph)
As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle
with the x axis in the third quadrant.
The direction is given as:
![\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)](https://tex.z-dn.net/?f=%5Ctheta%3D%5Ctan%5E%7B-1%7D%28%5Cfrac%7Bv_y%7D%7Bv_x%7D%29%5C%5C%5C%5C%5Ctheta%3D%5Ctan%5E%7B-1%7D%28%5Cfrac%7B-388.91%7D%7B-308.91%7D%29%5C%5C%5C%5C%5Ctheta%3D51.54%5C%C2%B0%28Third%5C%20quadrant%29)
Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.