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The nose of an ultralight plane is pointed south, and its airspeed indicator shows 44 m/s. The plane is in a 18 m/s wind blowing toward the southwest relative to earth.
a. letting x be east and y be north, find the components of \vec v_{\rm P/E} (the velocity of the plane relative to the earth.
b.Find the magnitude of \vec v_{\rm P/E}.
c.Find the direction of \vec v_{\rm P/E}.

Sagot :

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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)

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