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

- schrodinger

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- IsTim

Figured out the displacement. 1.4*10^3m. The angle's iffy.

- 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

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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## More answers

- JamesJ

You should also draw a diagram so you can see at least directionally where the answer is. No, use tan(y/x)

- IsTim

|dw:1327905348977:dw|

- IsTim

Why do I use tan(y/x)? Doesn't that apply only for right-angle triangles?

- 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

So, Idraw that chart thing?

- TuringTest

assuming I know what you mean by 'the chart thing' , yeah I do

- TuringTest

or *I would

- IsTim

i hafta flip thru my notes now to remember chart thing.

- 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

So I'm allowed doing that? I never learned it like that. I never knew.

- 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

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

welcome!

- IsTim

Turing? You here? I need help again!

- IsTim

I got 42.2 degrees, but the answer is 24!

- anonymous

what now?

- IsTim

I think I'm just gunna quit and move on.

- anonymous

y

- IsTim

I worked so long on this simple question. It's wasting my time by now.

- anonymous

i can try
if you have the patience lol

- IsTim

I can wait. I'll do another question while I wait tho. K?

- anonymous

k

- IsTim

WAIT! I FIGURED IT OUT

- IsTim

Sorry for wasting your time! I'm Sorry!

- IsTim

Mth?

- anonymous

o

- anonymous

lol

- anonymous

np

- IsTim

Thank you everyone.

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