Question #130958
A spelunker is surveying a cave. She follows a passage 180 m straight west, then 210 m in a direction 45° east of south, and then 280 m at 30° east of north. After a fourth displacement, she finds herself back where she started. (a) Use a scale drawing to determine the magnitude and direction of the fourth displacement. (b) Use component method to determine the magnitude and direction of the fourth displacement.
1
Expert's answer
2020-08-31T12:51:39-0400

a) See the figure:



Approximately, the magnitude is 150 m, the angle South of West is 45°.

b) Compute x- and y-components of each vector representing displacement:

1:x=180,     y=0.2:x=210 sin45°=148,     y=210 cos45°=148.3:x=280 sin30°=140,     y=280 cos30°=242.4:u=?     v=?1: x=-180,\\ \space \space \space \space \space y=0.\\ 2: x=210\text{ sin}45°=148,\\ \space \space \space \space \space y=-210\text{ cos}45°=-148.\\ 3: x=280\text{ sin}30°=140,\\ \space \space \space \space \space y=280\text{ cos}30°=242.\\ 4: u=?\\ \space \space \space \space \space v=?

sine she returned to the initial position, the displacement is 0, or x- and y-components of this vector are 0. These components, on the other hand, are the sum of components of vectors 1, 2, 3, and 4. This will help us find u and v:


Ox:0=180+148+140+u,u=108,Oy:0=0148+242+v,v=94.Ox:0=-180+148+140+u,\\u=-108,\\ Oy:0=0-148+242+v,\\ v=-94.

The magnitude of vector 4?


m=v2+u2=143 m.m=\sqrt{v^2+u^2}=143\text{ m}.

The angle (South of West) is


θ=atan94108=41°.\theta=\text{atan}\frac{94}{108}=41°.

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Comments

Gutema
29.03.23, 22:31

Interesting

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