An airplane flying at 82km/hr at 30 degrees east of north encounters a 40km/hr wind due west. What is the speed of the plane relative to the ground?
Apparently the first velocity you mentioned is the air speed, since you are asking for the ground speed. The problem should have made that more lear.
The velocity of the plane with respect to the ground is the vector sum of the wind speed and the velocity of the plane with respect to the air.
Vgy = 82 cos30 = 71.0 km/h (north component)
Vgx = 82 sin30 - 40 = 1.0 km/h (east component)
resultant = 71.0 km/h, 0.8 degrees E of N
To find the speed of the plane relative to the ground, we need to combine the velocity of the plane and the velocity of the wind.
First, let's break down the velocity of the plane into its north and east components.
The north component of the plane's velocity (V_north) can be found using trigonometry. We can use the given angle of 30 degrees east of north and the total velocity of the plane (82 km/hr) to find the north component.
V_north = 82 km/hr * sin(30°)
= 82 km/hr * 0.5
= 41 km/hr
The east component of the plane's velocity (V_east) can also be found using trigonometry. With the same angle, we can use the total velocity of the plane (82 km/hr) to find the east component.
V_east = 82 km/hr * cos(30°)
= 82 km/hr * √3/2
≈ 71.15 km/hr
Now, let's consider the velocity of the wind. Since the wind is blowing due west, its velocity is solely in the west direction, and its magnitude is 40 km/hr.
Now we can combine the eastward component of the plane's velocity with the westward component of the wind's velocity to find the plane's resultant velocity.
Resultant velocity = V_east - V_wind
= 71.15 km/hr - 40 km/hr
= 31.15 km/hr
Therefore, the speed of the plane relative to the ground is approximately 31.15 km/hr.