Relative Velocity of Aircraft: Mag & Dir Before & After 16:30

[markosheehan]

an aircraft flies horizontally at a speed 200 km hr^-1 relative to the air. its position at 15:00 is p. a wind with speed 100 km hr^-1 blows from the east from 15:00 to 16:30 . the wind then changes to a speed of 50 km hr^-1 from the south. the aircraft adjusts its heading so it maintains a south westerly course all the time.

find the magnitude and direction of the of the aircrafts velocity relative to the air both before and after 16:30. also if airport q is 600 km south west of p find the time when the aircraft rrives at q.i could kind of get the answer for before 16:30 but i can not the the right answer of 34.8 degrees west of south for after 16:30. could anyone explain this to me? i am also confused on the wind and the direction in the aircraft should fly at for both the times. i am trying to label Va=aircraft vector , Vw=wind vector and Vaw =vector of the aircraft relative to the wind.

 

[MarkFL]

I would begin by stating the vectors we know...let $\vec{A}$ represent the aircraft, $\vec{W}$ the wind, and $\vec{R}$ the resultant.

So, from 15:00 to 16:30, we have:

\(\displaystyle \vec{A}=200\left\langle \cos(\theta),\sin(\theta) \right\rangle\)

\(\displaystyle \vec{W}=100\left\langle -1,0 \right\rangle\)

And what we want is:

\(\displaystyle \vec{R}=\vec{A}+\vec{W}=|R|\left\langle \cos\left(\frac{5\pi}{4}\right),\sin\left(\frac{5\pi}{4}\right) \right\rangle=-\frac{|R|}{\sqrt{2}}\left\langle 1,1 \right\rangle\)

This gives us the following system:

\(\displaystyle 200\cos(\theta)-100=-\frac{|R|}{\sqrt{2}}\)

\(\displaystyle 200\sin(\theta)=-\frac{|R|}{\sqrt{2}}\)

This implies:

\(\displaystyle 200\cos(\theta)-100=200\sin(\theta)\)

\(\displaystyle 2\cos(\theta)-1=2\sin(\theta)\)

Solving this, and using the solution in the 3rd quadrant, we obtain:

\(\displaystyle \theta=2\left(\pi-\arctan\left(\frac{2+\sqrt{7}}{3}\right)\right)\approx245.70481105463543^{\circ}\)

And so:

\(\displaystyle |R|\approx257.793547457352\,\frac{\text{km}}{\text{hr}}\)

Now, can you do the same thing, except with the wind vector that applies after 16:30?

 

[markosheehan]

thanks for your reply. i am sorry but i think the answer acceptable on my course is using the cosine and sine rule.
do you know how to do it using that method?

could you show me a diagram of the vectors of the wind,plane and the resultant. i can not visualize the directions.

thanks

 

[MarkFL]

thanks for your reply. i am sorry but i think the answer acceptable on my course is using the cosine and sine rule.

That would have been nice to know up front. :p

do you know how to do it using that method?

Yes, I just find vector addition easier.

could you show me a diagram of the vectors of the wind,plane and the resultant. i can not visualize the directions.

thanks

The wind vector points to the west and the resultant points to the southwest. So the aircraft's vector will complete the triangle.

Using the Law of Sines, we may write:

\(\displaystyle \frac{\sin(\theta)}{|R|}=\frac{\sin\left(\frac{\pi}{4}\right)}{200}=\frac{1}{200\sqrt{2}}\implies |R|=200\sqrt{2}\sin(\theta)\)

Using the Law of Cosines, we may write:

\(\displaystyle R^2=100^2+200^2-2\cdot100\cdot200\cos(\theta)=100^2\left(5-4\cos(\theta)\right)\)

These two equation imply:

\(\displaystyle 8\sin^2(\theta)=5-4\cos(\theta)\)

Using a Pythagorean identity, we may write:

\(\displaystyle 8\left(1-\cos^2(\theta)\right)=5-4\cos(\theta)\)

In standard quadratic form, we have:

\(\displaystyle 8\cos^2(\theta)-4\cos(\theta)-3=0\)

Using the quadratic formula and discarding the negative root, we obtain:

\(\displaystyle \cos(\theta)=\frac{1+\sqrt{7}}{4}\)

And thus:

\(\displaystyle \theta=\arccos\left(\frac{1+\sqrt{7}}{4}\right)\approx24.29518894536457^{\circ}\)

And so the direction $\alpha$ of the aircraft is:

\(\displaystyle \alpha=270^{\circ}-\theta\approx245.70481105463543^{\circ}\)

And we have the resultant in kph:

\(\displaystyle |R|=200\sqrt{2}\left(\frac{1+\sqrt{7}}{4}\right)=50\sqrt{2}\left(1+\sqrt{7}\right)\approx257.793547457352\)

Now for the new wind direction, we have the wind vector pointing north, and the resultant still points to the southwest. Can you find the new direction and speed of the aircraft relative to the ground (resultant)?

 

[markosheehan]

Sorry I should of mentioned it.

Thanks I can do it from here!

 

[MarkFL]

Sorry I should of mentioned it.

Thanks I can do it from here!

Just so you can check you work, I will complete the problem. :D

Spoiler

Using the Law of Sines, we may state:

\(\displaystyle \frac{200}{\sin\left(\frac{3\pi}{4}\right)}=\frac{|R|}{\sin(\theta)}\implies |R|=200\sqrt{2}\sin(\theta)\)

And using the Law of Cosines:

\(\displaystyle R^2=50^2+200^2-2\cdot50\cdot200\cos(\theta)=100^2\left(\frac{17}{4}-2\cos(\theta)\right)\)

These two equations imply:

\(\displaystyle 32\cos^2(\theta)-8\cos(\theta)-15=0\)

Discard the negative root, and we have

\(\displaystyle \cos(\theta)=\frac{1+\sqrt{31}}{8}\implies \theta=\arccos\left(\frac{1+\sqrt{31}}{8}\right)\approx34.81793259684111^{\circ}\)

This is west of south. And so the resultant in kph is:

\(\displaystyle |R|\approx161.49485779096793\)

Now, as of 16:30 the aircraft had traveled in km:

\(\displaystyle d_1\approx\frac{3}{2}(257.793547457352)=386.69032118602803\)

This leaves:

\(\displaystyle 600-386.69032118602803=213.30967881397197\)

km to go, which will take a time (in hr) of:

\(\displaystyle t\approx\frac{213.30967881397197}{161.49485779096793}\approx1.32084502089888\)

So, the aircraft arrives at the airport at (to the nearest minute) at 17:49. :D