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Sagot :
Answer:
a. Sketch is given in the attachment
b. [tex]V_{p}[/tex] = (0,-525)
c. [tex]V_{w}[/tex] = ([tex]\sqrt{2}[/tex],[tex]\sqrt{2}[/tex])
d. [tex]V_{g}[/tex] = (60[tex]\sqrt{2}[/tex], 60[tex]\sqrt{2}[/tex] - 525)
e. Actual Speed = |[tex]V_{g}[/tex]| = 448.3 MPH
Direction = Ф = - 79.1°
Direction = Ф = 280.9° with respect to x - axis.
Step-by-step explanation:
a. Sketch the scenario on the x-y axis.
Where,
Velocity of the plane relative to air = [tex]V_{p}[/tex]
Velocity of the jet stream = [tex]V_{w}[/tex]
Velocity of the plane relative to ground = [tex]V_{g}[/tex]
Solution:
Note: Solution is given in the attachment. Please refer to the attachment for the sketch.
b. Express the velocity of the plane relative to the air as a vector.
Solution:
[tex]V_{p}[/tex] = |[tex]V_{p}[/tex]| cosФi + |[tex]V_{p}[/tex]| sinФj
where,
|[tex]V_{p}[/tex]| = magnitude of the vector.
|[tex]V_{p}[/tex]| = 525 MPH
Ф = 270° w.r.t to x - axis. (See the sketch in the attachment)
[tex]V_{p}[/tex] = |[tex]V_{p}[/tex]| cosФi + |[tex]V_{p}[/tex]| sinФj
Plug in the values into this equation to express it in the vector form as required.
[tex]V_{p}[/tex] = (525) cos270i + (525) sin270j
As, Cos270 = 0
and
Sin270 = -1
So,
[tex]V_{p}[/tex] = (525) (0)i + (525) (-1)j
[tex]V_{p}[/tex] = 0i -525j
[tex]V_{p}[/tex] = (0,-525)
c. Express the velocity of the jet stream as a vector.
Solution:
Velocity of the jet stream as a vector.
[tex]V_{w}[/tex] = |[tex]V_{w}[/tex]| cosФi + |[tex]V_{w}[/tex]| sinФj
where,
|[tex]V_{w}[/tex]| = magnitude of the vector.
|[tex]V_{p}[/tex]| = 120 MPH
Ф = 45° (See the sketch in the attachment)
[tex]V_{w}[/tex] = |[tex]V_{w}[/tex]| cosФi + |[tex]V_{w}[/tex]| sinФj
Plug in the values into this equation to express it in the vector form as required.
[tex]V_{w}[/tex] = (120) cos45i + (120) sin45j
As, Cos45 = [tex]\frac{1}{\sqrt{2} }[/tex]
and
Sin45 = [tex]\frac{1}{\sqrt{2} }[/tex]
So,
[tex]V_{w}[/tex] = (120) ([tex]\frac{1}{\sqrt{2} }[/tex])i + (120) ([tex]\frac{1}{\sqrt{2} }[/tex])j
[tex]V_{w}[/tex] = 60[tex]\sqrt{2}[/tex]i +60[tex]\sqrt{2}[/tex]j
[tex]V_{w}[/tex] = ([tex]\sqrt{2}[/tex],[tex]\sqrt{2}[/tex])
d. Find the velocity of the plane relative to the ground.
Solution:
Velocity of the plane relative to the ground = Sum of velocity of plane relative to the air and velocity of the jet stream
[tex]V_{g}[/tex] = [tex]V_{p}[/tex] + [tex]V_{w}[/tex]
[tex]V_{g}[/tex] = 0i -525j + 60[tex]\sqrt{2}[/tex]i +60[tex]\sqrt{2}[/tex]j
[tex]V_{g}[/tex] = 60[tex]\sqrt{2}[/tex]i + 60[tex]\sqrt{2}[/tex] - 525j
[tex]V_{g}[/tex] = (60[tex]\sqrt{2}[/tex], 60[tex]\sqrt{2}[/tex] - 525)
e. Find the actual speed and direction of the plane relative to the ground. Round to the tenths.
Solution:
Actual Speed = |[tex]V_{g}[/tex]| = magnitude of the velocity of the plane relative to the ground.
Actual Speed = |[tex]V_{g}[/tex]| = [tex]\sqrt{(60\sqrt{2}) ^{2} + (60\sqrt{2}-525) ^{2} }[/tex]
Actual Speed = |[tex]V_{g}[/tex]| = 448.3 MPH
Direction = Ф = [tex]tan^{-1}[/tex] ([tex]\frac{60\sqrt{2}-525 }{60\sqrt{2} }[/tex])
Direction = Ф = - 79.1°
To find out the direction with respect to x - axis.
Direction = Ф = 360 - 79.1
Direction = Ф = 280.9°
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