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

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:
${\stackrel{\to }{v}}_{gs}=\frac{800km}{2.00h}$
$=400k\frac{m}{h}\stackrel{^}{j}$
Where we set our coordinates so that the $+ve$ y-direction points north.
The ground velocity ${\stackrel{\to }{v}}_{as}$ is then equal to:
${\stackrel{\to }{v}}_{as}={v}_{as}{\mathrm{cos}70}^{\circ }\stackrel{^}{i}+{v}_{as}{\mathrm{sin}70}^{\circ }\stackrel{^}{j}$
$=\left(500k\frac{m}{h}\right){\mathrm{cos}70}^{\circ }\stackrel{^}{i}+\left(500k\frac{m}{h}\right){\mathrm{sin}70}^{\circ }\stackrel{^}{j}$
$=\left(171\stackrel{^}{i}+469\stackrel{^}{j}\right)k\frac{m}{h}$
Step 3
Therefore, the wind velocity can be obtained from:
${\stackrel{\to }{v}}_{gs}={\stackrel{\to }{v}}_{as}+{\stackrel{\to }{v}}_{w}$
$\left(400\stackrel{^}{j}\right)k\frac{m}{h}=\left(171\stackrel{^}{i}+469\stackrel{^}{j}\right)k\frac{m}{h}+{\stackrel{\to }{v}}_{w}$
$⇒{\stackrel{\to }{v}}_{w}=\left(-171\stackrel{^}{i}-69\stackrel{^}{j}\right)k\frac{m}{h}$
Step 4
Part a:
The magnitude of wind velocity is:
${v}_{w}=\sqrt{{\left(-171k\frac{m}{h}\right)}^{2}+{\left(-69k\frac{m}{h}\right)}^{2}}$
$=184k\frac{m}{h}$.
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.

esfloravaou

The destination is $\stackrel{\to }{D}=800km\stackrel{^}{j}$ 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 ${\stackrel{\to }{v}}_{pg}=\left(400k\frac{m}{h}\right)\stackrel{^}{j}$.
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 ${\stackrel{\to }{v}}_{ag}$. Thus, $\left(400k\frac{m}{h}\right)\stackrel{^}{j}=\left(500k\frac{m}{h}\right){\mathrm{cos}70}^{\circ }\stackrel{^}{i}+\left(500k\frac{m}{h}\right){\mathrm{sin}70}^{\circ }\stackrel{^}{j}+{\stackrel{\to }{v}}_{ag}$
which yields
${\stackrel{\to }{v}}_{ag}=\left(-171k\frac{m}{h}\right)\stackrel{^}{i}-\left(70.0k\frac{m}{h}\right)\stackrel{^}{j}$.
a) The magnitude of

b) The direction of ${\stackrel{\to }{v}}_{ag}$ is
$\theta ={\mathrm{tan}}^{-1}\left(\frac{-70.0k\frac{m}{h}}{-171k\frac{m}{h}}\right)={22.3}^{\circ }$ (south of west).

karton

Step 1
The speed of the plane relative to the ground is given by,
${v}_{gs}=\frac{D}{t}$
Here, D is the distance to the destination from starting point and t is the time of travel. Substituting the values,
${v}_{gs}=\frac{800km}{2.00h}$
=400km/h
Taking the north direction to be + y direction, the velocity of the plane relative to the ground is,
${\stackrel{\to }{v}}_{gs}=400km/h\stackrel{^}{j}$
The air velocity is given by,
${\stackrel{\to }{v}}_{as}=\left(500km/h\right)\mathrm{cos}{70}^{\circ }\stackrel{^}{i}+\left(500km/h\right)\mathrm{sin}{70}^{\circ }\stackrel{^}{j}$
$=\left(171\stackrel{^}{i}+469.8\stackrel{^}{j}\right)km/h$
The wind velocity is given by,
${\stackrel{\to }{v}}_{w}={\stackrel{\to }{v}}_{gs}-{\stackrel{\to }{v}}_{as}$
$=400km/h\stackrel{^}{j}-\left(171\stackrel{^}{i}+469.8\stackrel{^}{j}\right)km/h$
$=\left(171\stackrel{^}{i}-69.8\stackrel{^}{j}\right)km/h$
Step 2
a) The magnitude of the wind velocity is given by,
${v}_{w}=\sqrt{{v}_{w,i}^{2}+{v}_{w,j}^{2}}$
Here, ${v}_{w,i}$ is the i -component of wind velocity and ${v}_{w,j}$ is the j -component of wind velocity. Substituting the values,
${v}_{w}=\sqrt{\left(-171km/h{\right)}^{2}+\left(-69.8km/h{\right)}^{2}}$
$=184.7km/h$
b) The direction of the wind velocity is given by,
$\begin{array}{}\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{w,i}}{{v}_{w,j}}\right)\\ ={\mathrm{tan}}^{-1}\left(\frac{-69.8km/h}{-171km/h}\right)\\ ={22.2}^{\circ }\end{array}$
Therefore, the wind velocity will be directed ${22.2}^{\circ }$ south of west.

nick1337

To solve this problem, we can break down the velocity of the plane into two components: the northward component and the eastward component. Let's denote the magnitude of the northward component as ${v}_{n}$ and the magnitude of the eastward component as ${v}_{e}$. We also need to find the magnitude and direction of the wind velocity, denoted as ${v}_{w}$ and ${\theta }_{w}$ respectively.
The northward component of the plane's velocity can be calculated using the equation:
${v}_{n}={v}_{p}\mathrm{cos}\left({\theta }_{p}\right)$
where ${v}_{p}$ is the airspeed of the plane (500 km/h) and ${\theta }_{p}$ is the angle the plane is heading with respect to due north (20.0 degrees east of due north). Converting the angle to radians, we have ${\theta }_{p}=\frac{20\pi }{180}$.
Substituting the given values, we get:
${v}_{n}=500\mathrm{cos}\left(\frac{20\pi }{180}\right)$
The eastward component of the plane's velocity can be calculated using the equation:
${v}_{e}={v}_{p}\mathrm{sin}\left({\theta }_{p}\right)$
Substituting the values, we have:
${v}_{e}=500\mathrm{sin}\left(\frac{20\pi }{180}\right)$
Since the wind is affecting the plane's velocity, we can express the overall velocity of the plane in terms of the wind velocity:
${v}_{\text{total}}={v}_{p}+{v}_{w}$
where ${v}_{\text{total}}$ is the magnitude of the total velocity of the plane and ${v}_{w}$ is the magnitude of the wind velocity.
Given that the plane traveled a distance of 800 km due north in 2.00 hours, we can calculate the magnitude of the total velocity of the plane:
${v}_{\text{total}}=\frac{800}{2.00}$
Now, we can equate the components of the total velocity and the individual components of the plane's velocity and wind velocity:
${v}_{n}=\frac{800}{2.00}$
${v}_{e}=0$
${v}_{w}=\frac{800}{2.00}-{v}_{p}$
Finally, we can calculate the direction of the wind velocity using the tangent function:
$\mathrm{tan}\left({\theta }_{w}\right)=\frac{{v}_{e}}{{v}_{n}}$
Now let's substitute the values and calculate the solution:
The magnitude of the northward component of the plane's velocity:
${v}_{n}=500\mathrm{cos}\left(\frac{20\pi }{180}\right)\approx 472.92\phantom{\rule{0.167em}{0ex}}\text{km/h}$
The magnitude of the eastward component of the plane's velocity:
${v}_{e}=500\mathrm{sin}\left(\frac{20\pi }{180}\right)\approx 170.92\phantom{\rule{0.167em}{0ex}}\text{km/h}$
The magnitude of the total velocity of the plane:
${v}_{\text{total}}=\frac{800}{2.00}=400\phantom{\rule{0.167em}{0ex}}\text{km/h}$
The magnitude of the wind velocity:
${v}_{w}=\frac{800}{2.00}-500=-100\phantom{\rule{0.167em}{0ex}}\text{km/h}$
The direction of the wind velocity:
$\mathrm{tan}\left({\theta }_{w}\right)=\frac{{v}_{e}}{{v}_{n}}$
${\theta }_{w}=\mathrm{arctan}\left(\frac{{v}_{e}}{{v}_{n}}\right)$
Substituting the values:
${\theta }_{w}=\mathrm{arctan}\left(\frac{170.92}{472.92}\right)\approx {20.42}^{\circ }$
Therefore, the (a) magnitude of the wind velocity is $100$ km/h (opposite direction of the plane's motion) and (b) the direction of the wind velocity is ${20.42}^{\circ }$ east of due south.

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