Answer:
a) vx = -12.7 m/s vy = -56.7m/s
b) v= 58.1 m/s
c) θ = 77.4º S of W
Explanation:
a)
- In order to get the components of the velocity of the plane relative to the earth, we need just to get the components of both velocities first:
- Since the nose of the plane is pointing south, if we take y to be north, and positive, this means that the velocity of the plane can be written as follows:
[tex]v_{ps} = -44m/s (1)[/tex]
- Since the wind is pointing SW, it's pointing exactly 45º regarding both directions, so we can find its components as follows (they are equal each other in magnitude)
[tex]v_{we} = - 18m/s * cos (45) = -12.7 m (2)[/tex]
[tex]v_{ws} = - 18m/s * cos (45) = -12.7 m (3)[/tex]
- The component of v along the x-axis is simply (2), as the plane has no component of velocity along this axis:
[tex]v_{e} = v_{x} = -12.7 m/s (4)[/tex]
- The component of v along the -y axis is just the sum of (1) and (3)
- [tex]v_{y} = -44 m/s + (-12.7m/s) = -56.7 m/s (5)[/tex]
b)
- We can find the magnitude of the velocity vector, just applying the Pythagorean Theorem to (4) and (5):
[tex]v = \sqrt{(-12.7m/s)^{2} + (-56.7m/s)^{2}} = 58.1 m/s (6)[/tex]
c)
- Taking the triangle defined by vx, vy and v, we can find the angle that v does with the negative x-axis, just using the definition of tangent, as follows:
[tex]tg_{\theta} =\frac{v_{y} }{v_{x} } = \frac{(-56.7m/s)}{(-12.7m/s} = 4.46 (7)[/tex]
- Taking tg⁻¹ from (7), we get:
tg⁻¹ θ = tg⁻¹ (4.46) = 77.4º S of W. (8)
