Question

1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant 80 mph from the west. Find the actual speed and direction of the aircraft.

Answer 1

Answer:

The magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction $(v_j)$ = 550 mph

Velocity of jet stream from west to east direction $(v_s)=80\ mph$

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:

$\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph$

Similarly, the velocity of the stream is, $\vec{v_s}=80\vec{i}$

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

$\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}$

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

$|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph$

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle $\theta$ with the x axis in the third quadrant.

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)$

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

Answer 2

The magnitude of the vectors is calculated by the Pythagorean theorem.  the actual speed of the jet stream is 496.67 mph.

Velocity:

Given here,  

Speed in southwest direction = 550 mph

Speed in west = 80 mph

Speed in southwest direction,

$\vec v_s_w = -550 \rm {\ cos (45)} \it i +\rm (-550 \rm sin (45 )\it j)\\\vec v_s_w = \rm -388 \it i - \rm 388 \it j$

Speed in west direction,

$\vec v_w = 80 i$

Add these values,

$\vec v = -5550 i - 550 j +80 i\\\\\vec v = -308.91 i - 388 j$

From the Pythagorean theorem,

$\vec v = \sqrt { -308.91 i - 388 j}\\\\\vec v = 496.67\rm \ mph$

Therefore, the actual speed of the jet stream is 496.67 mph.

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