Question

A spelunker is surveying a cave. She follows a passage that takes her a distance 184 m straight west, then a distance 220 m in a direction 30.0 degrees south of east, then a distance 104 m at 80.0 degrees east of north. After a fourth unmeasured displacement she find herself back where she started. Use the method of components to find the magnitude and direction of the fourth displacement.

Answer 1

To determine the magnitude and direction of the fourth displacement, vector addition and the method of components are used by breaking down each movement into its east/west and north/south components. The fourth displacement equals the negative sum of the first three displacements' components, and its magnitude and direction can be calculated using the Pythagorean theorem and the inverse tangent respectively.

Step-by-step explanation:

The subject of this question is Physics, specifically vector displacement and the method of components. To find the magnitude and direction of the spelunker's fourth displacement, we need to calculate the resultant of the first three displacements using vector addition. Let's break each displacement into its components:

• First displacement: 184 m West (we can consider West as negative East)
• Second displacement: 220 m at 30° South of East (East component = 220 m × cos(30°), South component = 220 m × sin(30°))
• Third displacement: 104 m at 80° East of North (North component = 104 m × cos(80°), East component = 104 m × sin(80°))

Adding up the components:

• East/West component = -184 m + (220 m × cos(30°)) + (104 m × sin(80°))
• North/South component = (220 m × sin(30°)) + (104 m × cos(80°)) (South and down are taken as negative)

Since the spelunker ends up back where she started, the sum of these components must equal zero. Solving these equations, we find the components of the fourth displacement. Then, we can find the magnitude of the fourth displacement using the Pythagorean theorem and the direction using the inverse tangent function.

Answer 2

Refer to the diagram shown below.

Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.

The first distance is 184 m west. It is represented by
d₁ = -184 i

The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j

The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j

Let the fourth distance be
d₄ = a i + b j

Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.

The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95

The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94

The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m

The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°

Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.