Mary Keefe

2022-01-07

A light plane attains an airspeed of 500 km/h. The pilot sets out for a destination 800 km due north but discovers that the plane must be headed $20.0}^{\circ$ east of due north to fly there directly. The plane arrives in 2.00 h. What were the (a) magnitude and (b) direction of the wind velocity?

Joseph Lewis

Beginner2022-01-08Added 43 answers

Step 1

Given: Air speed$=500k\frac{m}{h},D=800km$ , due north,

$\theta ={20}^{\circ}$ east of due north, $t=2.00h$ .

Step 2

The air speed is the speed of the plane relative to the air while ground speed is its speed relative to the ground.

The plane arrives in 2.00 hours after moving a distance of 800km, this means that its ground speed is:

$\overrightarrow{v}}_{gs}=\frac{800km}{2.00h$

$=400k\frac{m}{h}\hat{j}$

Where we set our coordinates so that the$+ve$ y-direction points north.

The ground velocity$\overrightarrow{v}}_{as$ is then equal to:

$\overrightarrow{v}}_{as}={v}_{as}{\mathrm{cos}70}^{\circ}\hat{i}+{v}_{as}{\mathrm{sin}70}^{\circ}\hat{j$

$=\left(500k\frac{m}{h}\right){\mathrm{cos}70}^{\circ}\hat{i}+\left(500k\frac{m}{h}\right){\mathrm{sin}70}^{\circ}\hat{j}$

$=(171\hat{i}+469\hat{j})k\frac{m}{h}$

Step 3

Therefore, the wind velocity can be obtained from:

$\overrightarrow{v}}_{gs}={\overrightarrow{v}}_{as}+{\overrightarrow{v}}_{w$

$\left(400\hat{j}\right)k\frac{m}{h}=(171\hat{i}+469\hat{j})k\frac{m}{h}+{\overrightarrow{v}}_{w}$

$\Rightarrow {\overrightarrow{v}}_{w}=(-171\hat{i}-69\hat{j})k\frac{m}{h}$

Step 4

Part a:

The magnitude of wind velocity is:

$v}_{w}=\sqrt{{(-171k\frac{m}{h})}^{2}+{(-69k\frac{m}{h})}^{2}$

$=184k\frac{m}{h}$ .

Step 5

Part b:

Its direction is:

$\mathrm{tan}\theta =\frac{{v}_{wy}}{{v}_{wx}}$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{wy}}{{v}_{wx}}\right)$

$={\mathrm{tan}}^{-1}\left(\frac{-69}{-171}\right)$

$={22}^{\circ}$ , south of west.

Given: Air speed

Step 2

The air speed is the speed of the plane relative to the air while ground speed is its speed relative to the ground.

The plane arrives in 2.00 hours after moving a distance of 800km, this means that its ground speed is:

Where we set our coordinates so that the

The ground velocity

Step 3

Therefore, the wind velocity can be obtained from:

Step 4

Part a:

The magnitude of wind velocity is:

Step 5

Part b:

Its direction is:

esfloravaou

Beginner2022-01-09Added 43 answers

The destination is $\overrightarrow{D}=800km\hat{j}$ where we orient axes so that + y points north and + x points east.

This takes two hours, so the (constant) velocity of the plane (relative to the ground) is$\overrightarrow{v}}_{pg}=\left(400k\frac{m}{h}\right)\hat{j$ .

This must be the vector sum of the plane’s velocity with respect to the air which has (x,y) components$\left(500{\mathrm{cos}70}^{\circ},500{\mathrm{sin}70}^{\circ}\right)$ , and the velocity of the air (wind) relative to the ground $\overrightarrow{v}}_{ag$ . Thus, $\left(400k\frac{m}{h}\right)\hat{j}=\left(500k\frac{m}{h}\right){\mathrm{cos}70}^{\circ}\hat{i}+\left(500k\frac{m}{h}\right){\mathrm{sin}70}^{\circ}\hat{j}+{\overrightarrow{v}}_{ag}$

which yields

$\overrightarrow{v}}_{ag}=(-171k\frac{m}{h})\hat{i}-\left(70.0k\frac{m}{h}\right)\hat{j$ .

a) The magnitude of

$\overrightarrow{v}}_{ag}\text{}is\text{}\left|{\overrightarrow{v}}_{ag}\right|=\sqrt{{(-171k\frac{m}{h})}^{2}+{(-70.0k\frac{m}{h})}^{2}}\}=185k\frac{m}{h$

b) The direction of$\overrightarrow{v}}_{ag$ is

$\theta ={\mathrm{tan}}^{-1}\left(\frac{-70.0k\frac{m}{h}}{-171k\frac{m}{h}}\right)={22.3}^{\circ}$ (south of west).

This takes two hours, so the (constant) velocity of the plane (relative to the ground) is

This must be the vector sum of the plane’s velocity with respect to the air which has (x,y) components

which yields

a) The magnitude of

b) The direction of

karton

Expert2022-01-11Added 613 answers

Step 1

The speed of the plane relative to the ground is given by,

Here, D is the distance to the destination from starting point and t is the time of travel. Substituting the values,

=400km/h

Taking the north direction to be + y direction, the velocity of the plane relative to the ground is,

The air velocity is given by,

The wind velocity is given by,

Step 2

a) The magnitude of the wind velocity is given by,

Here,

b) The direction of the wind velocity is given by,

Therefore, the wind velocity will be directed

nick1337

Expert2023-05-22Added 777 answers

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