pre calculus, speed and bearing

Question

A fighter plane is flying with a bearing of 155 degrees at a speed of 110 mph towards enemy lines when it encounters a strong wind of 60 mph blowing FROM 250 degrees. 

 

A. Draw a vector diagram. Label the velocity vector of the plane and the velocity vector of the wind.

 

B. Find the plane's speed relative to the ground 

 

C. Find the plan's true bearing

Answer 1

A. Can't do graphical in this format.  Draw the vector for the plane, then draw the wind vector, putting the tail of the wind on the head of the plane's vector.  The line from the tail of the plane's vector to the head of the wind's vector is the resultant vector.

 

B. Multiple ways to do this, but since we're doing pre-calculus, let's break both into x and y components.  Remembering that headings start with zero at North (our y axis) and go around clockwise, so our equations are a bit different than what we conventionally use in physics.  For the plane, we have:

 

v_x = 110*sin(155) = 46.5

v_y = 110*cos(155) = -99.7

 

For the wind, subtract 180 from 250 to get 70 for the direction the wind is going, so we have:

 

v_x = 60*sin(70) = 56.4

v_y = 60*cos(70) = 20.5

 

Therefore, when we add these two, the resultant vector is:

 

<46.5,-99.7> + <56.4,20.5> = <102.9,-79.2>

 

Speed will be the magnitude of this vector, so:

 

v = √(102.9² + (-79.2)²) = 129.9 mph

 

C. To get bearing, we note that tan^-1(79.2/102.9) = 37.6.  Looking at what this means on your graph, note that the vector is 37.6^o South of East, or a bearing of 127.6^o.

 

The result should make intuitive sense, as the plane has a quartering tailwind.