Need help with two physics questions!


1) The north component of the airplane velocity is 260 km/h.
2) The direction of the plane is [tex]24^{\circ}[/tex] north of east.
Explanation:
1)
In this problem, we have to resolve the velocity vector into its components.
Taking east as positive x-direction and north as positive y-direction, the components of the velocity along the two directions are given by:
[tex]v_x = v cos \theta[/tex]
[tex]v_y = v sin \theta[/tex]
where
v is the magnitude of the velocity
[tex]\theta[/tex] is the angle between the direction of the velocity and the positive x-axis (the east direction)
For the airplane in this problem,
v = 750 km/h
[tex]\theta=20^{\circ}[/tex]
So, the two components are
[tex]v_x = (750)(cos 20)=704.8 km/h[/tex]
[tex]v_y = (750)(sin 20)=256.5 km/h[/tex]
So, the component in the north direction is 256.5 km/h, so approximately 260 km/h.
2)
In this problem, we have to use vector addition.
In fact, the motion of the plane consists of two displacements:
- A first displacement of 220 km in the east direction
- A second displacement of 100 km in the north direction
Using the same convention of the same problem (x = east and y = north), we can write
[tex]d_x = 220 km[/tex]
[tex]d_y = 100 km[/tex]
Since the two vectors are perpendicular to each other, we can find their magnitude using Pythagorean's theorem:
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(220)^2+(100)^2}=241.7 km[/tex]
And the direction is given by
[tex]\theta=tan^{-1}(\frac{d_y}{d_x})=tan^{-1}(\frac{100}{220})=24^{\circ}[/tex] north of east.
Learn more about vectors here:
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