Question

A jet heads in the direction N30 degreesW at a speed of 600mi/h. If the jet experiences a 50mi/h crosswind flowing due east, find (a) the true velocity of the jet. b) the true speed of the jet c) the direction of the jet.

Answer 1

First question is to figure out what N30 degreesW means as a compass heading!

Answer 2

In the first quadrant?

Answer 3

Well, North is usually straight up. Perhaps it means 30 degrees west of north? The equivalent of 120 degrees measured on the unit circle? That would be second quadrant.

Answer 4

Yeah I think that was what the W was for

Answer 5

So you find the x and y components of the jet's vector, and of the wind vector (this one is easy, it's all x component), and you add them to get the resultant vector. The magnitude of the resultant vector will be the true speed of the jet, and the direction of the resultant vector will the be the direction of the jet's travel.

Answer 6

Want to guide me through all that? haha

Answer 7

Well, same formulas you just used.

\[J_x = ||v||\cos\theta_{jet}\]\[J_y = ||v||\sin\theta_{jet}\]\[W_x = ||w||\cos\theta_{wind}\]\[W_y=||w||\sin\theta_{wind}\]
\[v = 600\]\[\theta_{jet} = 120^\circ\]\[w=50\]\[\theta_{wind} = 0^\circ\]
\[R_x = J_x+W_x\]\[R_y= J_y+W_y\]\[||R|| = \sqrt{R_x^2 + R_y^2}\]\[\theta_{R} = \tan^{-1}(\frac{R_y}{R_x})\]

Answer 8

Wait so what would |V| equal?

Answer 9

that's just the magnitude of v, so 600

Answer 10

Okay this might take me a little time so hold on haha :P

Answer 11

So far I have \[J _{x}=600\cos 120\]\[J _{y}=600\sin120\] \[W _{x}=50\cos0\] \[W _{y}=50 \sin0\]

Answer 12

Okay, what are the values of all of those things?

Answer 13

You can make a \(^\circ\) with ^\circ by the way

Answer 14

-1/2
square root 3/2
1
0
But what am I supposed to do with the 600's and 50's?

Answer 15

multiply!

Answer 16

Ohhh okay I just wasn't sure

Answer 17

So -300
150 square root 3
50
0

Answer 18

Want to check your math for the value of \(J_y\)?

Answer 19

Oh oops 300 square root 3

Answer 20

yes

now add the corresponding components together to get the components of the resultant vector

Answer 21

Wait what?

Answer 22

You have to add the x component of the jet vector plus the x component of the wind vector to get the x component of the resultant vector

You have to add the y component of the jet vector plus the y component of the wind vector to get the y component of the resultant vector

Answer 23

Ohhhh okay... So x=-250 and y=150 square root 3

Answer 24

Yes. So what will the magnitude of the vector that has those two components be? That will be the effective speed (over the ground) of the jet

Answer 25

\[\sqrt{(-250)^{2}}+(150\sqrt{3})^2\]

Answer 26

And I got that it equals 514.781

Answer 27

I got a slightly different number. Probably because I didn't go back to using the wrong value of \(J_y\) as you did :-)

Answer 28

I'm sorry I didn't notice it earlier when you said "x=-250 and y = 150 square root 3"

\[J_y = 600\sin 120^\circ = 600*\frac{\sqrt{3}}{2} = 300\sqrt{3} \approx 519.6\]

Answer 29

Oops haha yeah that was my problem

Answer 30

So that gives us a magnitude of \[\sqrt{(-250)^2 + (150\sqrt{3})^2} = \sqrt{62500+3*90000} = \sqrt{62500+270000} \]\[\qquad = \sqrt{332500} \approx 576.6\]

Answer 31

So, having the 50 mph crosswind knocks about half of 50 mph off the jet's forward speed (and changes the direction of travel as well).

Answer 32

Wait why did you use 150 square root of 3?

Answer 33

Sorry, I typed your expression there, but actually used the correct value to the right :-)

\((300\sqrt{3})^2 = 300*300*3 = 3*90000\)

Answer 34

Ohhh haha no problem (:

Answer 35

So now we have the tricky part: figuring out the direction the dang airplane is flying.

Answer 36

Great haha I was hoping the tricky part was over :P

Answer 37

Tricky for me, at least!

Answer 38

What we're going to do is change that to be just from the y-axis to the vector, so that the angle is less than 90 degrees (we'll add the 90 degrees back on to get the final heading).

That gives us a triangle with a hypotenuse of