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
Jun 6, 2016

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#, to the East is #90^0#, to the South is #180^o# and to the West is #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# on a compass is #90^o +(360^o -340^o)=110^o# counterclockwise from the X-axis.
#320^o# on a compass is #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#) has amplitude #325# (#mph#) and angle #110^o#;
wind (vector #W#) has amplitude #40# (#mph#) and angle #130^o#.

To add these two vectors, we represent both as sums of X-component and Y-component:
#A_X = 345*cos(110^o)#
#A_Y = 345*sin(110^o)#
#W_X = 40*cos(130^o)#
#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*cos(110^o)+40*cos(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|# and direction #/_(A+W)# of the resulting vector:

#|A+W| = sqrt((A+W)_X^2+(A+W)_Y^2)#

#/_(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)# in compass format (#/_alpha#) from its standard trigonometric format (#/_beta#), use the conversion formula
#/_alpha = 360^o-(/_beta-90^o) = 450^o-/_beta#
(if the result is greater than #360^o#, subtract #360^o#)