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.

Expert's answer

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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