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IsTim
 4 years ago
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?
IsTim
 4 years ago
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?

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IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Figured out the displacement. 1.4*10^3m. The angle's iffy.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2The 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
 4 years ago
Best ResponseYou've already chosen the best response.0It 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?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2You should also draw a diagram so you can see at least directionally where the answer is. No, use tan(y/x)

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Why do I use tan(y/x)? Doesn't that apply only for rightangle triangles?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0It'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
 4 years ago
Best ResponseYou've already chosen the best response.0So, Idraw that chart thing?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0assuming I know what you mean by 'the chart thing' , yeah I do

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0i hafta flip thru my notes now to remember chart thing.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0V 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
 4 years ago
Best ResponseYou've already chosen the best response.0So I'm allowed doing that? I never learned it like that. I never knew.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0My 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
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. 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.

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Turing? You here? I need help again!

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I got 42.2 degrees, but the answer is 24!

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I think I'm just gunna quit and move on.

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I worked so long on this simple question. It's wasting my time by now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i can try if you have the patience lol

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I can wait. I'll do another question while I wait tho. K?

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry for wasting your time! I'm Sorry!
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