Answer
(a) The plane's velocity relative to the air is $160~km/h$ at an angle of $20.0^{\circ}$ north of east.
(b) The magnitude of the plane's velocity relative to the ground is $150.54~km/h$ at an angle of $21.3^{\circ}$ north of east.
(c) The wind speed is $10.0~km/h$ and it is blowing directly to the west.
Work Step by Step
(a) The plane flies at an airspeed of $160~km/h$ and heads at an angle of $20.0^{\circ}$ north of east. This is the plane's velocity relative to the air.
(b) We can find the east component of the plane's velocity relative to the ground:
$v_x = \frac{(320~km)~cos~20.0^{\circ}-20~km}{2.0~h} = 140.35~km/h$
We can find the north component of the plane's velocity relative to the ground:
$v_y = \frac{(320~km)~sin~20.0^{\circ}}{2.0~h} = 54.72~km/h$
We can find the magnitude of the plane's velocity relative to the ground:
$\sqrt{(140.35~km/h)^2+(54.72~km/h)^2} = 150.54~km/h$
We can find the angle $\theta$ north of east:
$tan~\theta = \frac{54.72}{140.35}$
$\theta = tan^{-1}(\frac{54.72}{140.35})$
$\theta = 21.3^{\circ}$
The magnitude of the plane's velocity relative to the ground is $150.54~km/h$ at an angle of $21.3^{\circ}$ north of east.
(c) We can find the x-component of the wind's velocity $v_{wx}$:
$140.35~km/h = (160~km/h)~cos~20.0^{\circ}+v_{wx}$
$v_{wx} = 140.35~km/h - (160~km/h)~cos~20.0^{\circ}$
$v_{wx} = -10.0~km/h$
$v_{wx} = 10.0~km/h$ (west)
We can find the y-component of the wind's velocity $v_{wy}$:
$54.72~km/h = (160~km/h)~sin~20.0^{\circ}+v_{wy}$
$v_{wx} = 54.72~km/h - (160~km/h)~sin~20.0^{\circ}$
$v_{wx} = 0$
The wind speed is $10.0~km/h$ and it is blowing directly to the west.
