A plane is heading due south with an airspeed of 350mph. A wind in the direction of 27 degrees is blowing at 45mph. Find the true course of airplane.

Question

dont forget bearings are measured from north, clockwise. Round result to hundredths place

Answer 1

For a picture, draw the airspeed vector on the negative y-axis (due south) and draw the wind speed vector in the first quadrant at 27 degrees east of north.

 

The vectors form two sides of a parallelogram.

 

Draw in the missing two sides with dotted lines.

 

A parallelogram has two diagonals.  The true speed vector is the diagonal that runs from the origin to the intersection of the two sides represented by dotted lines.

 

Find its magnitude by decomposing the wind speed vector into horizontal and vertical components, combining the vertical component with the plane's airspeed, then use the Pythagorean formula to form the resultant.

 

North and east are positive, south and west are negative.

 

The vertical component of the wind speed is  + 45cos27 = 40.095 mph North

 

The horizontal component of the wind speed is + 45sin27 = 20.430 mph East

The plane's speed is - 350 mph = 350 mph South

 

The vertical component of the true speed of the plane is - 350 + 40.095 = - 309.905 mph

 

The resultant is the square root of [(- 309.905)² + (20.430)²] = 310.58 mph

 

For the direction Θ, take the inverse tangent of the true speed's components.  Make any necessary adjustments due to the fact that the reference is the positive y-xis (North).

 

Θ = arctan[- 309.905/20.430] = - 86.22 degrees.  The true speed vector is 86.22 degrees below the x-axis.

 

Report the heading as 310.58 mph, 86.22 South of East.