 2022-01-06

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south.
(a) If the airspeed of the plane (its speed in still air) is 320.0 km/h (about 200 mi/h), in which direction should the pilot head?
(b) What is the speed of the plane over the ground? Draw a vector diagram. alexandrebaud43

Step 1
We have
$\frac{{\stackrel{\to }{v}}_{P}}{G}$, the velocity of the plane relative to the ground, (due west)
$\frac{{\stackrel{\to }{v}}_{P}}{A}$, the velocity of the plane relative to the air (320km/h), and
$\frac{{\stackrel{\to }{v}}_{A}}{G}$, the velocity of the air relative to the ground (80km/h, due south).
$\frac{{\stackrel{\to }{v}}_{P}}{G}=\frac{{\stackrel{\to }{v}}_{P}}{A}+\frac{{\stackrel{\to }{v}}_{A}}{G}$.
Step 2
a) $\mathrm{sin}\theta =\frac{\frac{{v}_{A}}{G}}{\frac{{v}_{P}}{A}}=\frac{80\frac{km}{h}}{320\frac{km}{h}}$
$\theta ={14}^{\circ }$, north of west.
b) Pythagorean theorem (right triangle):
$\frac{{v}_{P}}{G}=\sqrt{{v}_{\frac{P}{A}}^{2}-{v}_{\frac{A}{G}}^{2}}=\sqrt{{\left(320\frac{km}{h}\right)}^{2}-{\left(80\frac{km}{h}\right)}^{2}}=310\frac{km}{h}$. Serita Dewitt

An airplane pilot wishes to fly due west. A wind of 80.0 km/h is blowing toward the south.
(a) If the airspeed of the plane (its speed in still air) is 320.0 km/h, in which direction should the pilot head?
(b) What is the speed of the plane over the ground? Draw a vector diagram.​
Explanation:
$\mathrm{sin}\varphi =80\frac{km}{h}/320\frac{km}{h}$
$\varphi ={14}^{\circ }$ north of west
Using Pythagorean theorem
${\sqrt{320km/h}}^{2}-{\sqrt{80km/h}}^{2}$
$=310\frac{km}{h}$ karton

To solve this problem, we can break down the velocities into their respective components. Let's denote the following:
- ${V}_{\text{plane}}$: Velocity of the plane in still air (320.0 km/h).
- ${V}_{\text{wind}}$: Velocity of the wind (80.0 km/h).
- ${\theta }_{\text{plane}}$: Direction in which the plane is heading (unknown).
- ${\theta }_{\text{wind}}$: Direction of the wind (south).
(a) To determine the direction in which the pilot should head (${\theta }_{\text{plane}}$), we need to find the resultant velocity by adding the velocity of the plane to the velocity of the wind. This can be calculated using the components:

The resultant velocity in the x-direction is given by . Since the pilot wishes to fly due west, the x-component of the resultant velocity should be zero:

Using the given values, we can substitute in the expressions for the components and solve for ${\theta }_{\text{plane}}$:
${V}_{\text{plane}}\mathrm{cos}\left({\theta }_{\text{plane}}\right)+{V}_{\text{wind}}\mathrm{cos}\left({\theta }_{\text{wind}}\right)=0$
Substituting the values, we have:
$320.0\mathrm{cos}\left({\theta }_{\text{plane}}\right)+80.0\mathrm{cos}\left(90°\right)=0$
Simplifying further, we get:
$320.0\mathrm{cos}\left({\theta }_{\text{plane}}\right)+0=0$
$320.0\mathrm{cos}\left({\theta }_{\text{plane}}\right)=0$
Since the cosine function is zero when the angle is 90° or 270°, we can conclude that the pilot should head either due north or due south.
(b) To find the speed of the plane over the ground, we need to calculate the resultant velocity magnitude using the components:

The magnitude of the resultant velocity is given by:

Substituting the given values and simplifying, we get:
${V}_{\text{resultant}}=\sqrt{{320.0\mathrm{cos}\left({\theta }_{\text{plane}}\right)+80.0\mathrm{cos}\left(90°\right)}^{2}+{\left(320.0\mathrm{sin}\left({\theta }_{\text{plane}}\right)+80.0\mathrm{sin}\left(90°\right)\right)}^{2}}$
Now, since $\mathrm{cos}\left(90°\right)=0$ and $\mathrm{sin}\left(90°\right)=1$, we can simplify further:
${V}_{\text{resultant}}=\sqrt{{320.0\mathrm{cos}\left({\theta }_{\text{plane}}\right)}^{2}+{\left(320.0\mathrm{sin}\left({\theta }_{\text{plane}}\right)+80.0\right)}^{2}}$
Squaring both sides of the equation, we get:
${{V}_{\text{resultant}}}^{2}={320.0\mathrm{cos}\left({\theta }_{\text{plane}}\right)}^{2}+{\left(320.0\mathrm{sin}\left({\theta }_{\text{plane}}\right)+80.0\right)}^{2}$
Expanding and simplifying the equation, we have:
${{V}_{\text{resultant}}}^{2}=102400.0{\mathrm{cos}}^{2}\left({\theta }_{\text{plane}}\right)+102400.0{\mathrm{sin}}^{2}\left({\theta }_{\text{plane}}\right)+51200.0\mathrm{sin}\left({\theta }_{\text{plane}}\right)+6400.0$
Since ${\mathrm{cos}}^{2}\left({\theta }_{\text{plane}}\right)+{\mathrm{sin}}^{2}\left({\theta }_{\text{plane}}\right)=1$, we can simplify further:
${{V}_{\text{resultant}}}^{2}=102400.0+51200.0\mathrm{sin}\left({\theta }_{\text{plane}}\right)+6400.0$
Now, we can substitute $\mathrm{sin}\left({\theta }_{\text{plane}}\right)=1$ (since the pilot should head either due north or due south) and calculate:
${{V}_{\text{resultant}}}^{2}=102400.0+51200.0+6400.0$
${{V}_{\text{resultant}}}^{2}=160000.0$
Taking the square root of both sides, we find:
${V}_{\text{resultant}}=400.0$
Therefore, the speed of the plane over the ground is 400.0 km/h. user_27qwe

Step 1:
(a) The wind is blowing towards the south with a velocity of 80.0 km/h. Let's denote the velocity of the wind as $\stackrel{\to }{{V}_{\text{wind}}}\right\}$. Since the plane wants to fly due west, its resulting velocity, $\stackrel{\to }{{V}_{\text{resultant}}}$, should be in the west direction.
To find the direction in which the pilot should head, we need to subtract the velocity of the wind from the velocity of the plane in still air. Mathematically, we can write:
$\stackrel{\to }{{V}_{\text{resultant}}}=\stackrel{\to }{{V}_{\text{plane}}}-\stackrel{\to }{{V}_{\text{wind}}}$
Using vector subtraction, we get:
$\stackrel{\to }{{V}_{\text{resultant}}}=\left(320.0\phantom{\rule{0.167em}{0ex}}\text{km/h}\right)\stackrel{^}{i}-\left(80.0\phantom{\rule{0.167em}{0ex}}\text{km/h}\right)\stackrel{^}{j}$
where $\stackrel{^}{i}$ represents the unit vector in the west direction and $\stackrel{^}{j}$ represents the unit vector in the south direction.
Simplifying this expression, we get:
$\stackrel{\to }{{V}_{\text{resultant}}}=320.0\phantom{\rule{0.167em}{0ex}}\text{km/h}\phantom{\rule{0.167em}{0ex}}\stackrel{^}{i}-80.0\phantom{\rule{0.167em}{0ex}}\text{km/h}\phantom{\rule{0.167em}{0ex}}\stackrel{^}{j}$
Therefore, the pilot should head in the direction $320.0\phantom{\rule{0.167em}{0ex}}\text{km/h}$ due west.
Step 2:
(b) To find the speed of the plane over the ground, we need to calculate the magnitude of the resultant velocity $\stackrel{\to }{{V}_{\text{resultant}}}$. Mathematically, we can write:

Substituting the values, we have:

Evaluating this expression, we get:

Therefore, the speed of the plane over the ground is $\sqrt{108800.0\phantom{\rule{0.167em}{0ex}}{\text{km/h}}^{2}}\approx 329.86\phantom{\rule{0.167em}{0ex}}\text{km/h}$.

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