Question

A plane is flying with an airspeed of 190 miles per hour and heading 155°. The wind currents are running at 40 miles per hour at 160° clockwise from due north. Use vectors to find the true course and ground speed of the plane. (Round your answers to the nearest ten for the speed and to the nearest whole number for the angle.)

Answer 1

Answer:

Ground speed: 230 miles per hour.

Course: 156° clockwise from due north.

Explanation:

The plane speed is 190 heading 155° clockwise due north, is east is the horizontal of the coordinate plane (positive x-axis), the angle from the x-axis is:

$\alpha =360^o-(155^o-90^o)=295^o$

The 360° is to avoid using negative angles, and the -90° is because the angle given in the data is measured from the north. Now do the same with the speed of the wind current (40 miles per hour at 160° clockwise from due north):

$\beta =360^o-(160^o-90^o)=290^o$

To find the course and ground speed of the plane find the components in x and y for both speeds and add them up.

$v_{px}=190cos(295^o)=80.30$

$v_{py}=190sin(295^o)=-172.20$

$v_{cx}=40cos(290^o)=13.68$

$]v_{cx}=40sin(290^o)=-37.59$

$v_x=v_{px}+v_{cx}\\v_x=80.30+13.68=93.98$

$v_y=v_{py}+v_{cy}\\v_y=-172.20-37.59=-209.79$

The ground speed is the magnitude of this new vector:

$| {{\to} \atop {V}}| =\sqrt{v_x^2+v_y^2} \\| {{\to} \atop {V}}| =\sqrt{(93.98)^2+(-209.79)^2} =229.87\approx230$

The course is the angle:

$\gamma=arctan(\frac{v_y}{v_x} )=arctan(-0.4480)=-24.13^o\approx-24^o$

This value of $\gamma=-24^o$ is the angle between the negative y-axis and the speed vector (see the graph). For the angle clockwise due north add 180°.

$180^o-24^o=156^o$