vectors (coordinate geometry)

Question

A cessna plane is travelling at an air speed of 420 km/h with a heading of 80? . There is a wind blowing at a speed of 50 km/h from the south-east.Unit vectors i point east and j point north.

(a) Express the velocity P of the plane relative to the air, and the velocity W of the wind, in component form, giving the numerical values in km/h.

(b) Express the resultant velocity V of the plane, relative to the ground, in component form, giving numerical values in km/h.

(c) Find the magnitude and direction of the resultant velocity V of the plane, giving the magnitude in km/h t and the direction as a bearing.

Answer 1

Joe,

I wish this editor allowed for drawings, but we'll have to make do.

Remember that a heading of 80° is counted clockwise from 0° at true north. This is different from most math which considers angles counter-clockwise from 0 at due east. How you break vectors into components is a matter of personal preference, but the result is the same. Some people will use the angle given regardless of its reference. Others (including me) prefer to always reference the x-axis for consistency with what they do in other math areas. Based on referencing the x-axis, the i component is always Vcosθ while the j component is always Vsinθ. So:

 

the heading of 80º is equivalent to 90 - 80 = 10º above the x-axis.

 

P = 420cos10i + 420sin10j  or  P = (413.6i + 72.9j) kph

 

Wind from the southeast puts it on a heading of 315º or 45º above the -x axis.

 

W = -25√(2)i + 25√(2)j   or  W = (-35.4i + 35.4j) kph

 

For the resultant velocity, we just add the components of the plane and the wind:

 

V = (378i + 108j) kph

 

For the resultant magnitude, we apply Pythagorean theorem:

 

V = √(378² + 108²) = 393 kph

 

For the bearing, there is uncertainty about the format. Some math books use degrees CW from north as a bearing. Other books use degrees east or west of the north-south line as a bearing. I will assume the latter.

 

To find it, we use the components of the resultant with the tangent of the angle:

 

θ = arctan(108/378) = 15.9º above the +x axis   or   Bearing = N74º3'E

 

I rounded off to the minute for lack of sig figs.