Question #5d46f

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Question #5d46f

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Answer 1 · 2017-07-18T05:28:13

$θ = {24.5}^{o}$ $theta = 24.5^"o"$

$v = 174$ $v = 174$ $km/h$ $"km/h"$

Explanation:

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

We're asked to find

(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 $210$ $210$ $km/h$ $"km/h"$ relative to the air (not affected by wind).

The velocity components of the wind relative to the earth are

$v_{x} = - 40$ $v_x = -40$ $km/h$ $"km/h"$

$v_{y} = 0$ $v_y = 0$

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

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

$v_{x} = (210 l km/h) cos θ$ $v_x = (210color(white)(l)"km/h")costheta$

$v_{y} = (210 l km/h) sin θ$ $v_y = (210color(white)(l)"km/h")sintheta$

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

$v_{x} = (210 l km/h) cos θ - 40 l km/h$ $v_x = (210color(white)(l)"km/h")costheta - 40color(white)(l)"km/h"$

$v_{y} = (210 l km/h) sin θ$ $v_y = (210color(white)(l)"km/h")sintheta$

We can use the trigonometric relationship

$tan α = \frac{v_{y}}{v_{x}}$ $tanalpha = (v_y)/(v_x)$

where $α$ $alpha$ is the desired angle of $30^{o}$ $30^"o"$ .

So, plugging in values, we have

$tan (30^{o}) = \frac{(210 l km/h) sin θ}{(210 l km/h) cos θ - 40 l km/h}$ $tan(30^"o") = ((210color(white)(l)"km/h")sintheta)/((210color(white)(l)"km/h")costheta - 40color(white)(l)"km/h")$

Neglecting units:

$tan (30^{o}) = \frac{210 sin θ}{210 cos θ - 40}$ $tan(30^"o") = (210sintheta)/(210costheta - 40)$

So,

$30^{o} = arctan (\frac{210 sin θ}{210 cos θ - 40})$ $30^"o" = arctan((210sintheta)/(210costheta - 40))$

What we can do is graph the two equations

$y = 30$ $y = 30$

and

$y = arctan (\frac{210 sin x}{210 cos x - 40})$ $y = arctan((210sinx)/(210cosx - 40))$

(make sure your calculator is in degree mode)

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

$θ = {24.5}^{o}$ $theta = color(red)(24.5^"o"$

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

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

$v_{y} = (210 l km/h) sin ({24.5}^{o}) = 87.2$ $v_y = (210color(white)(l)"km/h")sin(24.5^"o") = 87.2$ $km/h$ $"km/h"$

The speed $v$ $v$ is thus

$v = \sqrt{{(v_{x})}_{x}^{2} + {(v_{y})}_{y}^{2}} = \sqrt{{(151 l km/h)}^{2} + {(87.3 l km/h)}^{2}}$ $v = sqrt((v_x)^2 + (v_y)^2) = sqrt((151color(white)(l)"km/h")^2 + (87.3color(white)(l)"km/h")^2)$

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

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Question #5d46f

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