Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.42 m due east, then 0.72 m at 32o north of due east. Beetle 2 also makes two runs; the first is 1.48 m at 47o east of due north. What must be the magnitude of its second run if it is to end up at the new location of beetle 1?

Question

Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.42 m due east, then 0.72 m at 32o north of due east. Beetle 2 also makes two runs; the first is 1.48 m at 47o east of due north. What must be the magnitude of its second run if it is to end up at the new location of beetle 1?

Answer 1

Don' t know the answer and I want the answer with steps

Answer 2

To find the magnitude of the second run for beetle 2, we need to determine the final position of beetle 1 and then calculate the distance between that position and the starting point of beetle 2.

Let's start by finding the final position of beetle 1:

First run of beetle 1: 0.42 m due east

Second run of beetle 1: 0.72 m at 32o north of due east

To find the final position, we can add the displacements of both runs in both the x (east) and y (north) directions:

For the x-direction:

Displacement of the first run: 0.42 m (due east)

Displacement of the second run: 0.72 m * cos(32°) ≈ 0.618 m (eastward component of the second run)

Total displacement in the x-direction: 0.42 m + 0.618 m = 1.038 m (east)

For the y-direction:

Displacement of the first run: 0 m (due north)

Displacement of the second run: 0.72 m * sin(32°) ≈ 0.378 m (northward component of the second run)

Total displacement in the y-direction: 0 m + 0.378 m = 0.378 m (north)

Therefore, the final position of beetle 1 is 1.038 m east and 0.378 m north of the starting point.

Now, let's calculate the distance between the final position of beetle 1 and the starting point of beetle 2:

First run of beetle 2: 1.48 m at 47o east of due north

To find the distance, we can use the Pythagorean theorem:

Distance = sqrt((change in x)^2 + (change in y)^2)

Change in x for beetle 2: 1.48 m * cos(47°) ≈ 1.48 m * 0.682 ≈ 1.008 m (eastward component)

Change in y for beetle 2: 1.48 m * sin(47°) ≈ 1.48 m * 0.731 ≈ 1.080 m (northward component)

Distance = sqrt((1.038 m - 1.008 m)^2 + (0.378 m - 1.080 m)^2)
= sqrt(0.03 m^2 + (-0.702 m)^2)
≈ sqrt(0.03 m^2 + 0.492404 m^2)
≈ sqrt(0.522404 m^2)
≈ 0.722 m

Therefore, the magnitude of the second run for beetle 2 should be approximately 0.722 meters in order to end up at the new location of beetle 1.