Question

A spelunker is surveying a cave. She follows a passage 150m straight west, then 230m in a direction 45∘ east of south, and then 280 m at 30∘ east of north. After a fourth unmeasured displacement, she finds herself back where she started.

Answer 1

A spelunker is surveying a cave. She follows a passage 150m straight west, then 230m in a direction $45{}^{\circ}$ east of south, and then 280 m at $30{}^{\circ}$ east of north. After a fourth unmeasured displacement, she finds herself back where she started. Determine the magnitude and direction of the fourth displacement.

Solution:

Resultant vector $\vec{a}$ :

Second displacement:

$$\vec {a} = \overrightarrow {a _ {E}} + \overrightarrow {a _ {S}}$$

Along the horizontal axis:

$$a _ {E} = - 1 5 0 m + 2 3 0 m * \cos 4 5 ^ {o} = 1 2. 6 m$$

Along the vertical axis:

$$a _ {S} = 2 3 0 \mathrm {m} * \sin 4 5 ^ {o} = 1 6 2. 6 \mathrm {m}$$

Third displacement:

Resultant vector $\vec{b}$ :

$$\vec {b} = \overrightarrow {b _ {E}} + \overrightarrow {b _ {S}}$$

Along the horizontal axis:

$$b _ {E} = a _ {E} + 2 8 0 \mathrm {m} * \sin 3 0 ^ {o} = 1 5 2. 6 \mathrm {m}$$

Along the vertical axis:

$$b _ {S} = - a _ {S} + 2 8 0 \mathrm {m} * \cos 3 0 ^ {o} = 7 9. 8 \mathrm {m}$$

Vector to be found - a vector of the opposite vector $\vec{b}$ :

$$\vec {c} = - \vec {b}$$

Length of the vector $\vec{b}$ , Pythagorean Theorem:

$$\begin{array}{l} | b | = \sqrt {b _ {E} ^ {2} + s ^ {2}} = \sqrt {1 5 2 . 6 ^ {2} + 7 9 . 8 ^ {2}} \\ = 1 7 2. 7 m \\ \end{array}$$

Angle of the vector $\vec{c}$ :

$$\alpha = \arcsin \frac {b _ {E}}{b} = \arcsin \frac {1 5 2 . 6 m}{1 7 2 . 7 m} = 6 2 ^ {o}$$

So, the fourth displacement has magnitude $172.7m$ and direction at $62{}^{\circ}$ south of west

Answer: fourth displacement: 172.7m at $62{}^{\circ}$ south of west.