An airplane is flying on a compass heading(bearing) of 340 degrees art 325 mph. A wind is blowing with the bearing 320 degrees at 40 mph, how do you find the component form of the velocity of the airplane?

Question

An airplane is flying on a compass heading(bearing) of 340 degrees art 325 mph. A wind is blowing with the bearing 320 degrees at 40 mph, how do you find the component form of the velocity of the airplane?

1 Answer

Impact of this question

34856 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-06T03:14:38

Below are the formulas for X- and Y-components of the resulting movement as well as the amplitude and angle of this movement (all angles are measured according to trigonometric standard).

Explanation:

In navigation the angle of the course (on a compass) is counted clockwise from the North (so, the direction to the North is $0^{o}$ $0^o$ , to the East is $90^{0}$ $90^0$ , to the South is $180^{o}$ $180^o$ and to the West is $270^{o}$ $270^o$ ).
The North on most maps is a vertically up direction.
In coordinate Geometry and Trigonometry, which we will use, angles are measured counterclockwise from the positive direction of the horizontal X-axis (the East on most maps).

Let's make a simple transformation into Trigonometric standard using the direction to the East as an X-axis.:
$340^{o}$ $340^o$ on a compass is $90^{o} + (360^{o} - 340^{o}) = 110^{o}$ $90^o +(360^o -340^o)=110^o$ counterclockwise from the X-axis.
$320^{o}$ $320^o$ on a compass is $90^{o} + (360^{o} - 320^{o}) = 130^{o}$ $90^o +(360^o -320^o)=130^o$ counterclockwise from the X-axis.

This is a problem on addition of two vectors. Each is defined by its amplitude and angle of direction:
airplane (vector $A$ $A$ ) has amplitude $325$ $325$ ( $m p h$ $mph$ ) and angle $110^{o}$ $110^o$ ;
wind (vector $W$ $W$ ) has amplitude $40$ $40$ ( $m p h$ $mph$ ) and angle $130^{o}$ $130^o$ .

To add these two vectors, we represent both as sums of X-component and Y-component:
$A_{X} = 345 \cdot cos (110^{o})$ $A_X = 345*cos(110^o)$
$A_{Y} = 345 \cdot sin (110^{o})$ $A_Y = 345*sin(110^o)$
$W_{X} = 40 \cdot cos (130^{o})$ $W_X = 40*cos(130^o)$
$W_{Y} = 40 \cdot sin (130^{0})$ $W_Y = 40*sin(130^0)$

Both X-components act along the same direction, both Y-components act along the same direction. So, we can add X-components to get an X-component of the resulting movement and add Y-components to get a Y-component of the resulting movement.

${(A + W)}_{X} = A_{X} + W_{X} = 345 \cdot cos (110^{o}) + 40 \cdot cos (130^{o})$ $(A+W)_X = A_X + W_X = 345*cos(110^o)+40*cos(130^o)$
${(A + W)}_{Y} = A_{Y} + W_{Y} = 345 \cdot sin (110^{o}) + 40 \cdot sin (130^{o})$ $(A+W)_Y = A_Y + W_Y = 345*sin(110^o)+40*sin(130^o)$

Knowing two components of the resulting vector of movement, we can easily determine the amplitude $| A + W |$ $|A+W|$ and direction $∠ (A + W)$ $/_(A+W)$ of the resulting vector:

$| A + W | = \sqrt{{(A + W)}_{X}^{2} + {(A + W)}_{Y}^{2}}$ $|A+W| = sqrt((A+W)_X^2+(A+W)_Y^2)$

$∠ (A + W) = arctan [\frac{{(A + W)}_{Y}}{{(A + W)}_{X}}]$ $/_(A+W) = arctan[(A+W)_Y/(A+W)_X]$

We leave the calculations to the person who suggested the problem.

If it's necessary to express the angle $∠ (A + W)$ $/_(A+W)$ in compass format ( $∠ α$ $/_alpha$ ) from its standard trigonometric format ( $∠ β$ $/_beta$ ), use the conversion formula
$∠ α = 360^{o} - (∠ β - 90^{o}) = 450^{o} - ∠ β$ $/_alpha = 360^o-(/_beta-90^o) = 450^o-/_beta$
(if the result is greater than $360^{o}$ $360^o$ , subtract $360^{o}$ $360^o$ )

Science

Math

Humanities

... and beyond

An airplane is flying on a compass heading(bearing) of 340 degrees art 325 mph. A wind is blowing with the bearing 320 degrees at 40 mph, how do you find the component form of the velocity of the airplane?

1 Answer

Explanation:

Related questions

Impact of this question