# Question 5d46f

Jul 18, 2017

$\theta = {24.5}^{\text{o}}$

$v = 174$ $\text{km/h}$

#### Explanation:

For this problem, I'll take the positive $x$-direction as east, and the positive $y$-direction to be *north*.

(a) the direction the plane must face so that it travels to the island in a straight line

(b) its speed (which I'll assume is the speed relative to the earth)

We now know the plane has a velocity of $210$ $\text{km/h}$ relative to the air (not affected by wind).

The velocity components of the wind relative to the earth are

${v}_{x} = - 40$ $\text{km/h}$

${v}_{y} = 0$

(it's traveling west, and thus has no vertical component)

The components of the plane's velocity relative to the air are

${v}_{x} = \left(210 \textcolor{w h i t e}{l} \text{km/h}\right) \cos \theta$

${v}_{y} = \left(210 \textcolor{w h i t e}{l} \text{km/h}\right) \sin \theta$

If we factor in the wind, the plane's velocity components relative to the earth are

v_x = (210color(white)(l)"km/h")costheta - 40color(white)(l)"km/h"

${v}_{y} = \left(210 \textcolor{w h i t e}{l} \text{km/h}\right) \sin \theta$

We can use the trigonometric relationship

$\tan \alpha = \frac{{v}_{y}}{{v}_{x}}$

where $\alpha$ is the desired angle of ${30}^{\text{o}}$.

So, plugging in values, we have

$\tan \left({30}^{\text{o") = ((210color(white)(l)"km/h")sintheta)/((210color(white)(l)"km/h")costheta - 40color(white)(l)"km/h}}\right)$

Neglecting units:

$\tan \left({30}^{\text{o}}\right) = \frac{210 \sin \theta}{210 \cos \theta - 40}$

So,

${30}^{\text{o}} = \arctan \left(\frac{210 \sin \theta}{210 \cos \theta - 40}\right)$

What we can do is graph the two equations

$y = 30$

and

$y = \arctan \left(\frac{210 \sin x}{210 \cos x - 40}\right)$

(make sure your calculator is in degree mode)

and find where they intersect; the $x$-value will be the angle at which the plane must fly, and it is found to be

theta = color(red)(24.5^"o"

Now, using this angle and the velocity components, we can find the speed of the airplane relative to the earth:

v_x = (210color(white)(l)"km/h")cos(24.5^"o") - 40color(white)(l)"km/h" $= 151$ $\text{km/h}$

${v}_{y} = \left(210 \textcolor{w h i t e}{l} \text{km/h")sin(24.5^"o}\right) = 87.2$ $\text{km/h}$

The speed $v$ is thus

$v = \sqrt{{\left({v}_{x}\right)}^{2} + {\left({v}_{y}\right)}^{2}} = \sqrt{{\left(151 \textcolor{w h i t e}{l} \text{km/h")^2 + (87.3color(white)(l)"km/h}\right)}^{2}}$

= color(blue)(174 color(blue)("km/h"#