Mathematics, 05.01.2021 17:00, brucewaynel1043

19. An aircraft maintaining a constant airspeed of 525 MPH is headed due south. The jet stream is 120 MPH in the northeasterly direction.

a. Sketch the scenario on the x-y axis.

b. Express the velocity of the plane relative to the air as a vector.

c. Express the velocity of the jet stream as a vector.

d. Find the velocity of the plane relative to the ground.

Find the actual speed and direction of the plane relative to the ground. Round to the

tenths.

e.

Answers: 1

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Answers

answered: donnafranks2003

a. Sketch is given in the attachment

b. $V_{p}$ = (0,-525)

c. $V_{w}$ = ( $\sqrt{2}$ , $\sqrt{2}$ )

d. $V_{g}$ = (60 $\sqrt{2}$ , 60 $\sqrt{2}$ - 525)

e. Actual Speed = | $V_{g}$ | = 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 = $V_{p}$

Velocity of the jet stream = $V_{w}$

Velocity of the plane relative to ground = $V_{g}$

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:

$V_{p}$ = | $V_{p}$ | cosФi + | $V_{p}$ | sinФj

where,

| $V_{p}$ | = magnitude of the vector.

| $V_{p}$ | = 525 MPH

Ф = 270° w.r.t to x - axis. (See the sketch in the attachment)

$V_{p}$ = | $V_{p}$ | cosФi + | $V_{p}$ | sinФj

Plug in the values into this equation to express it in the vector form as required.

$V_{p}$ = (525) cos270i + (525) sin270j

As, Cos270 = 0

and

Sin270 = -1

So,

$V_{p}$ = (525) (0)i + (525) (-1)j

$V_{p}$ = 0i -525j

$V_{p}$ = (0,-525)

c. Express the velocity of the jet stream as a vector.

Solution:

Velocity of the jet stream as a vector.

$V_{w}$ = | $V_{w}$ | cosФi + | $V_{w}$ | sinФj

where,

| $V_{w}$ | = magnitude of the vector.

| $V_{p}$ | = 120 MPH

Ф = 45° (See the sketch in the attachment)

$V_{w}$ = | $V_{w}$ | cosФi + | $V_{w}$ | sinФj

Plug in the values into this equation to express it in the vector form as required.

$V_{w}$ = (120) cos45i + (120) sin45j

As, Cos45 = $\frac{1}{\sqrt{2} }$

and

Sin45 = $\frac{1}{\sqrt{2} }$

So,

$V_{w}$ = (120) ( $\frac{1}{\sqrt{2} }$ )i + (120) ( $\frac{1}{\sqrt{2} }$ )j

$V_{w}$ = 60 $\sqrt{2}$ i +60 $\sqrt{2}$ j

$V_{w}$ = ( $\sqrt{2}$ , $\sqrt{2}$ )

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

$V_{g}$ = $V_{p}$ + $V_{w}$

$V_{g}$ = 0i -525j + 60 $\sqrt{2}$ i +60 $\sqrt{2}$ j

$V_{g}$ = 60 $\sqrt{2}$ i + 60 $\sqrt{2}$ - 525j

$V_{g}$ = (60 $\sqrt{2}$ , 60 $\sqrt{2}$ - 525)

e. Find the actual speed and direction of the plane relative to the ground. Round to the tenths.

Solution:

Actual Speed = | $V_{g}$ | = magnitude of the velocity of the plane relative to the ground.

Actual Speed = | $V_{g}$ | = $\sqrt{(60\sqrt{2}) ^{2} + (60\sqrt{2}-525) ^{2} }$

Actual Speed = | $V_{g}$ | = 448.3 MPH

Direction = Ф = $tan^{-1}$ ( $\frac{60\sqrt{2}-525 }{60\sqrt{2} }$ )

Direction = Ф = - 79.1°

To find out the direction with respect to x - axis.

Direction = Ф = 360 - 79.1

Direction = Ф = 280.9°

19. An aircraft maintaining a constant airspeed of 525 MPH is headed due south. The jet stream is

1

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