while exploring a cave, a spelunker starts at the entrance and moves the following distances in a horizontal plane. she goes 75.o m north, 250 m east, 125 m at angle theta =30.0 north of east, and 150 m south. find her resultant displacement from the cave entrance.
the displacement for each leg of the journey is
(0,75)
(250,0)
(108.25,62.5)
(0,-150)
add 'em up to get
(358.25,-12.5)
358.5m final displacement.
how did you get (108.25,62.5) and (0,-150)
To find the resultant displacement from the cave entrance, we need to add up the individual displacements. We can do this by using vector addition.
Let's break down the given displacements into their x and y components:
The displacement of 75.0 m north can be represented as (0, 75.0) since it is entirely along the y-axis.
The displacement of 250 m east can be represented as (250, 0) since it is entirely along the x-axis.
The displacement of 125 m at an angle of 30.0° north of east can be split into its x and y components. The x-component is cos(30.0°) * 125 = 125 * √3/2 = 108.25 m. The y-component is sin(30.0°) * 125 = 125 * 1/2 = 62.5 m. So, this displacement can be represented as (108.25, 62.5).
The displacement of 150 m south can be represented as (0, -150) since it is entirely along the negative y-axis.
Now, we can add up these displacements:
Resultant displacement in the x-direction (horizontal): 250 + 108.25 + 0 = 358.25 m
Resultant displacement in the y-direction (vertical): 75.0 + 62.5 - 150 = -12.5 m
Thus, the resultant displacement from the cave entrance is (358.25, -12.5) m.