Honors.smartdude992 wrote:
physics 1 right??Bert10099 wrote:
Problem to solve is:
An airplane trip involves three legs, with two stopovers. The first leg is due east for 620km, the second leg is south-east (45 degrees) for 440km, and the third leg is a 53 degrees south of west for 550km. What is the planes total displacement?
Karma for solution.
Meh. They're both easy.nukchebi0 wrote:
Ugh, I remember this stuff from trigonometry last year.
Calculus>trigonometry.
I have verified the validity of the 960.84km as well.
I know...but Calculus is still better.SenorToenails wrote:
Meh. They're both easy.nukchebi0 wrote:
Ugh, I remember this stuff from trigonometry last year.
Calculus>trigonometry.
I agree. It has far more uses.nukchebi0 wrote:
I know...but Calculus is still better.
What happens if the plane started from the north or south pole?
What do you mean, what would happen?Veridic wrote:
What happens if the plane started from the north or south pole?
The direction system is completely arbitrary. You could impose whatever coordinate system you wanted, and as long as you stay consistent, the answer will be equivlient.
How do you go east when at the north pole?
You guys are just starting this? Or is this your first semester of physics?Bert10099 wrote:
Honors.smartdude992 wrote:
physics 1 right??Bert10099 wrote:
Problem to solve is:
An airplane trip involves three legs, with two stopovers. The first leg is due east for 620km, the second leg is south-east (45 degrees) for 440km, and the third leg is a 53 degrees south of west for 550km. What is the planes total displacement?
Karma for solution.
Azimuthal projections provide a better method of picking a direction. If you select an arbitrary reference line, you can better determine which direction to travel. Cardinal directions are not equipped to deal with north/south poles.Veridic wrote:
How do you go east when at the north pole?
Because the problem references the point of origin and the end point in relation to the point of origin, it doesn't matter about locationSenorToenails wrote:
What do you mean, what would happen?Veridic wrote:
What happens if the plane started from the north or south pole?
The direction system is completely arbitrary. You could impose whatever coordinate system you wanted, and as long as you stay consistent, the answer will be equivlient.
Now if it said you start at north pole, and it wants to know how far the end point is from a point 274 miles at 131 degrees fromt he point of origin (point of reference/important/etc), it just means once you get the second triangle to find distance from point of origin, you must find a third triangle the distance from ending point to point of origin and the distance from point of origin to the point of reference
Like I said -- arbitrary reference system.Blehm98 wrote:
Because the problem references the point of origin and the end point in relation to the point of origin, it doesn't matter about locationSenorToenails wrote:
What do you mean, what would happen?Veridic wrote:
What happens if the plane started from the north or south pole?
The direction system is completely arbitrary. You could impose whatever coordinate system you wanted, and as long as you stay consistent, the answer will be equivlient.
Now if it said you start at north pole, and it wants to know how far the end point is from a point 274 miles at 131 degrees fromt he point of origin (point of reference/important/etc), it just means once you get the second triangle to find distance from point of origin, you must find a third triangle the distance from ending point to point of origin and the distance from point of origin to the point of reference
hmm, this is kinda wrong. You have to take the earth´s curvature into consideration.
Not on a basic level. You're only looking at vectors in x and y terms, no zstef10 wrote:
hmm, this is kinda wrong. You have to take the earth´s curvature into consideration.
ye, of course.bennisboy wrote:
Not on a basic level. You're only looking at vectors in x and y terms, no zstef10 wrote:
hmm, this is kinda wrong. You have to take the earth´s curvature into consideration.
but its still wrong.
gay go back and edit you thread after i gave the answer..SenorToenails wrote:
Break it into components.
x-direction: 620 + cos(45)*440 - sin(37)*550
y-direction: sin(45)*440 + cos(37)*550
displacement = sqrt((x-direction)^2 + (y-direction)^2)
Break it into parts, and find the legs of the triangle. Then solve for the hypotenuse.
http://img135.imageshack.us/img135/9739/solutionqx4.jpg
jokes on you. took me longer to pulll up autocad2008 than it did to find the answer.
Pretty awesome. I added the image in my edit. My solution, however, was posted before yours.DankmanHightimes wrote:
gay go back and edit you thread after i gave the answer..SenorToenails wrote:
Break it into components.
x-direction: 620 + cos(45)*440 - sin(37)*550
y-direction: sin(45)*440 + cos(37)*550
displacement = sqrt((x-direction)^2 + (y-direction)^2)
Break it into parts, and find the legs of the triangle. Then solve for the hypotenuse.
http://img135.imageshack.us/img135/9739/solutionqx4.jpg
jokes on you. took me longer to pulll up autocad2008 than it did to find the answer.
Soh Cah Toa!
The pi of the plane will force it to go southward, which then it increases it's velocity while its moving downwards, but once the plane is near 100ft, it will then shift northward and then the pilots enjoy a nice apple pie.
The displacement = the volume, if it crashes into the ocean.
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