Plane Flying with Wind Present

[Masrat_A]

Homework Statement

A pilot flies a plane from A to B, 400 miles due North of A. If a Westward wind of $20 mph$ is present, determine the direction the pilot needs to aim the plane relative to the air and the speed at which he needs to fly the plane in order to arrive at B in two hours.

Homework Equations

See below.

The Attempt at a Solution

Could anyone check my work, please, and point out any possible mistakes?

Diagram: http://imgur.com/a/3F3Jr

a) Speed

$400 (AB) + 40 (BC) - 40 = 400 m (AB)$
$AB = 400 m$

$| BC | = 2*40 = 40 m$
$BC = 40 m$

$d = | AC | = \sqrt {400^2+40^2}$
$d = 402 h$
$402 m/2 h = 201mi/h$

b) Degree

$sin \theta = BC/AC$
$\theta = sin^{-1}(BC/AC)$
$sin^{-1}(BC/AC) = sin^{-1}(40/402)$
$sin^{-1}(40/402) = 6^\circ$

 

[MarkFL]

I would work the problem using vectors. Let $0<v$ be the speed of the plane in mph. Orient our coordinate axes such that the plane is initially at the origin.

The velocity vectors are:

Wind: $v_W=20\left\langle -1,0\right\rangle$

Plane: $v_P=v\left\langle \cos(\theta),\sin(\theta)\right\rangle$

And so, after 2 hours we want the sum of the displacement vectors to be at point B:

$2\left(20\left\langle -1,0\right\rangle+v\left\langle \cos(\theta),\sin(\theta)\right\rangle\right)=400\left\langle 0,1\right\rangle$

Equating corresponding components, there results:

$v\cos(\theta)=20$

$v\sin(\theta)=200$

And so the plane's bearing, found by dividing the latter by the former, and complimentary to $\theta$ should be:

$90^{\circ}-\arctan\left(10\right)\approx6^{\circ}$

And the planes speed (in mph), found by squaring and adding, should be:

$v=\sqrt{200^2+20^2}\approx201$

These results agree with yours.

 

[CWatters]

Just a quick intervention...

I haven't checked your working but I note the problem asks for

the direction the pilot needs to aim the plane relative to the air

 

[jbriggs444]

Just a quick intervention...

I haven't checked your working but I note the problem asks for

As long as we are working strictly within the realm of classical mechanics, the direction that a plane is "aimed" will be the same regardless of the state of motion of the coordinate system against which one measures the aim angle.

The plane's track will certainly vary. Its path traced out over the ground will be due north while its track (e.g. as left as a smoke-trail in the sky) will be diagonal. But the plane's aim will be invariant.

The discrepancy between track and aim can be visually seen because the plane will be "crabbing" from the point of view of a ground observer -- moving partially sideways due to the wind.

 

[CWatters]

Yes I was confusing "relative to the air" with "relative to the wind direction".