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Emergency Landing

  • Thread starter COCoNuT
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  • #1
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Emergency Landing. A plane leaves the airport in Galisto and flies a distance 155 km at an angle 68.0 Degrees east of north and then changes direction to fly a distance 270 km at an angle 48.0 degrees south of east, after which it makes an immediate emergency landing in a pasture.


1.) When the airport sends out a rescue crew, how far should this crew fly to go directly to this plane? <--- know how to do this already and got the answer

2.) In which direction? (Express your answer as an angle measured south of east). <---- need help for this

ok i need help on #2, here's my work....

155cos(22)i +155sin(22)j
270sin(42)i +270cos(42)J

add those....

324.37i + 258.7j

arctan(y/x) = arctan(258.7/329.37) = 38 Degrees <--- my answer... What am i doing incorrectly? (please help within the next 5 hours if you can, that's when the hw is due). thanks
 

Answers and Replies

  • #2
chroot
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How did you arrive at your second vector (270 sin 42, 270 cos 42)?

It seems to me it should be (270 cos -48, 270 sin -48).

- Warren
 
  • #3
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155*cos(22) + 155sin(22)
270*cos(-48) + 270sin(-48)
-----------------------------
324.37 + (-142.58)

arctan (-142.58/324.37) = -24 Degrees south of east (answer must be in south of east)

is it ok to have a negative answer?
 
  • #4
chroot
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Angles increase counterclockwise from the positive x-axis. Negative angles increase clockwise. -24 degrees is in the southeast quadrant. It's just 24 degrees south of east. Draw a picture if you don't see it.

- Warren
 
  • #5
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Helpppp

Hey... I am soo confused. What answer do you get so that I can see what the heck I am doing wrong?
 
  • #6
chroot
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Sharayah,

It is our policy not to show explicit answers, but only provide a "path" the student might follow to solve the problem.

What exactly are you confused about? Have you been assigned the same question?

- Warren
 
  • #7
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Blah

I need help on the first question.
 
  • #8
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How do I approach it?
 
  • #9
chroot
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Sharayah,

As we have explained already in this thread, break each vector into components, then add the components. The length of a vector (x1, x2) is

[tex]L = \sqrt{x_1^2 + x_2^2}[/tex]

which is nothing more than the Pythagorean theorem.

- Warren
 
  • #10
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That is odd? I tryed that and ended up with the answer 307 km. Is that right? Ohh and sorry I barely came into the thread.
 
  • #11
chroot
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Sharayah,

It doesn't seem to be correct. Please show me the steps you used to come up with 307 km.

- Warren
 
  • #12
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I took 155^2 and added 265^2.... then took the sqrt of that

In numerical terms sqrt(155^2+265^2)
 
  • #13
chroot
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And where did you come up with the 155 and 265?

- Warren
 
  • #14
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Emergency Landing. A plane leaves the airport in Galisto and flies a distance 155 at an angle 68.0 east of north and then changes direction to fly a distance 265 at an angle 48.0 south of east, after which it makes an immediate emergency landing in a pasture.
 
  • #15
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Aren't those my two x distances?
 
  • #16
chroot
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The problem posted originally had 270 km as the magnitude of the second vector, not 265 km.

What makes you think you can just add the lengths of the vectors in the way you suggested? As I explained, you need to break the vectors into components and then add the components to find the sum of the vectors. Do you know how to do this?

- Warren
 
  • #17
COCoNuT said:
Emergency Landing. A plane leaves the airport in Galisto and flies a distance 155 km at an angle 68.0 Degrees east of north and then changes direction to fly a distance 270 km at an angle 48.0 degrees south of east, after which it makes an immediate emergency landing in a pasture.


1.) When the airport sends out a rescue crew, how far should this crew fly to go directly to this plane? <--- know how to do this already and got the answer

2.) In which direction? (Express your answer as an angle measured south of east). <---- need help for this

ok i need help on #2, here's my work....

155cos(22)i +155sin(22)j
270sin(42)i +270cos(42)J

add those....

324.37i + 258.7j

arctan(y/x) = arctan(258.7/329.37) = 38 Degrees <--- my answer... What am i doing incorrectly? (please help within the next 5 hours if you can, that's when the hw is due). thanks
I added vectors and got:

x(crew) = sin68*155+cos48*270
y(crew = cos68*155-sin48*270

took arctan(x/y) and got 23.7.. degrees south of east
 
  • #18
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I am sorry... I do not know how... I am new in physics and have not had a math class in over 2 years... so I am a little lost... but I should get it.
 
  • #19
chroot
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Do you know what it means to represent a vector in terms of its components?

- Warren
 
  • #20
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The components are the 'getting the sin and cos' right? If so I need a little guidence on how to determine wether or not it is cos or sin?!
 
  • #21
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Oh wait. Would it be like A is 155km @68 degree
 
  • #22
chroot
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Here's all the guidance you'll ever need:

The x-component, listed first in notation like (2, 5), is always the cosine part. The y-component, listed second in that notation, is always the sine part.

When you use the sine and cosine functions, the supplied angles must always be measured counterclockwise from the positive x-axis. I will call this "standard representation." This question involves phrases like "east of north" and so on, and is a complication. You have to convert "68 degrees east of north" into a number of degrees counterclockwise from the positive x-axis. If you draw "68 degrees east of north" on a piece of paper, you'll see that it's the same as 22 degrees in the standard representation. Thus, the first vector is 150 km at 22 degrees in standard representation.

This is one common method of specifying a vector -- giving its length and direction. Another common specification is by "components." The components of a vector represent how far long the x-axis, then how far along the y-axis you must go to get to the end of the vector.

In other words, a vector that is 100 units long and points at 45 degrees is the same as going (100/sqrt 2) units in the x direction and then (100/sqrt 2) units in the y-direction. It's just the pythagorean theorem. Try drawing this vector on a piece of paper and make sure you understand this before going on. Let me know if you don't understand this part.

So components just mean "how far along x, then how far along y?" To convert a vector into its components, use the following rule:

[tex]\mathbf u = ( |u| cos \theta, |u| sin \theta )[/tex]

Where [itex]|u|[/itex] is the length of u, and [itex]\theta[/itex] is the angle of the vector in standard representation.

The two vectors given are thus

[tex](150 \cos 22^o, 150 \sin 22^o)[/tex]

and

[tex](270 \cos -48^o, 270 \sin -48^o)[/tex]

To add two vectors with components (a, b) and (c, d), just add their components:

(a, b) + (c, d) = (a + c, b + d)

Once you've added the components, use the length formula I gave you above to find the total distance.

If all of this is still entirely baffling to you, you really need to speak up now and see your teacher about it. Much of the material you are going to cover later in your class depends upon a mastery of this.

- Warren
 
  • #23
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Now I am getting 269 ..... 155sin(68)+265cos(48) <---- doing this then taking the sqrt?! Am I even getting it?
 
  • #24
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ok thank you
 
  • #25
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Thank You Warren for your help.
 

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