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

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.

Answer 2

does anyone know why the answer with x and y components for this question are (-,+)? I thought it would be switched since the vector is in quadrant 4 meaning x is positive and y is negative? And why the degree wasn't found with the actual numbers but with both positive?