IsTim
  • IsTim
A skater travels in a straight line 8.5*10^2m[25 degress N of E] and then 5.6810^2m in a straight line [21 degrees E of N]. The entire motion takes 4.2 min. What is the skater's displacement angle?
Physics
schrodinger
  • schrodinger
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IsTim
  • IsTim
Figured out the displacement. 1.4*10^3m. The angle's iffy.
JamesJ
  • JamesJ
The amount of time taken is irrelevant. For each leg of the journey figure out the angle of displacement. Call each vector, \( v_1 \) and \( v_2 \). Each vector has an x and a y component. Write \[ v_1 = (x_1,y_1), v_2 = (x_2,y_2) \] Then at the end the skater is at \( v_1 + v_2 = (x_1 + x_2, y_1 + y_2 ) \) and can calculate her angle of displacement by taking the tan of the ratio of the y displacement over the x displacement. It is identical in principle to the last problem you had.
IsTim
  • IsTim
It is the last problem I had. I just didn't ask for the angle that time. Reposted... Anyways, I usedA=sin^-1(asinB/b). Does that work?

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JamesJ
  • JamesJ
You should also draw a diagram so you can see at least directionally where the answer is. No, use tan(y/x)
IsTim
  • IsTim
|dw:1327905348977:dw|
IsTim
  • IsTim
Why do I use tan(y/x)? Doesn't that apply only for right-angle triangles?
TuringTest
  • TuringTest
It's nice to use if you get your vector in terms of components, which you haven't done yet -I got the right answer now btw,
IsTim
  • IsTim
So, Idraw that chart thing?
TuringTest
  • TuringTest
assuming I know what you mean by 'the chart thing' , yeah I do
TuringTest
  • TuringTest
or *I would
IsTim
  • IsTim
i hafta flip thru my notes now to remember chart thing.
TuringTest
  • TuringTest
V is the displacement vector:\[V_x=850\cos(25^{\circ})+568\cos(90^{\circ}-21^{\circ})\]\[V_y=850\sin(25^{\circ})+568\sin(90^{\circ}-21^{\circ})\]\[|V|=\sqrt{V_x^2+V_y^2}\]\[\theta=\tan^{-1}(\frac{V_x}{V_y})\]
IsTim
  • IsTim
So I'm allowed doing that? I never learned it like that. I never knew.
TuringTest
  • TuringTest
My way definitely gets the job done. Convert the vector into components, the add them. The new vectors components will be the sum of the old vectors components. You can find the new vectors magnitude from the pythagorean theorem, and its angle from an inverse trig function. I know people have other ways of doing it though.|dw:1327895183170:dw|
IsTim
  • IsTim
Ok. I'm converting them into components. I'll try doing what you say. But I'll be taking a break. thank you very much though.
TuringTest
  • TuringTest
welcome!
IsTim
  • IsTim
Turing? You here? I need help again!
IsTim
  • IsTim
I got 42.2 degrees, but the answer is 24!
anonymous
  • anonymous
what now?
IsTim
  • IsTim
I think I'm just gunna quit and move on.
anonymous
  • anonymous
y
IsTim
  • IsTim
I worked so long on this simple question. It's wasting my time by now.
anonymous
  • anonymous
i can try if you have the patience lol
IsTim
  • IsTim
I can wait. I'll do another question while I wait tho. K?
anonymous
  • anonymous
k
IsTim
  • IsTim
WAIT! I FIGURED IT OUT
IsTim
  • IsTim
Sorry for wasting your time! I'm Sorry!
IsTim
  • IsTim
Mth?
anonymous
  • anonymous
o
anonymous
  • anonymous
lol
anonymous
  • anonymous
np
IsTim
  • IsTim
Thank you everyone.

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