Question

While surveying a cave, a spelunker follows a passage 190 m straight west, then 220 m in a direction 45.0 ∘ east of south, and then 270 m at 30.0 ∘ east of north. After a fourth unmeasured displacement, she finds herself back where she started. Use vector components to find the magnitude and direction of the fourth displacement.

∣r⃗4∣∣ =

θ4 =

Answer 1

let +x = east. –x = west
+y = north, –y = south

initially a spelunkerat the origin, (0, 0)

r1 = 190 m straight west

r1x= -190
r1y= 0

r2= 220 m in a direction 45° east of south

r2x = 220 * sin 45° east = +220 * sin 45° =155.5 East
r2y = 220 * cos 45 south = -220 * cos 45°= -155.5 South


r3= 270 m at 30° east of north.

r3x = 270* sin 30° east = +270 * sin 30° = 135 East
r3y= 270 * cos 30° north = +270 * cos 30° = 233.82North

Find sum of x and y coordinates
rx = -190 + 155.5 + 135 = +100.5 East
ry = 0 + -155.5 + 233.82 = +78.27 North

End point of sum of 3 vectors = (100.5, 78.27)

4th vector must go from end point of sum of 3 vectors, ((100.5, 78.27) back to the origin, (0, 0).

x = 0 – 100.5 = -100.5 m West
y = 0 – 78.27 = -78.27 m South

Magnitude of vector 4 is

r 4= root(-100.5)^2 + (-78.27)^2= 127.38 m

the angle is

theta = tan^-1 ( -78.27/ -100.5 ) = 37.91 west of south