Question

What is the magnitude of the following one-dimensional vectors? 23 m N + 7 m E + 36 m S + 51 m W.

a) 117 m
b) 95 m
c) 117 m N
d) 95 m E

Answer 1

The magnitude of the one-dimensional vectors 23 m N + 7 m E + 36 m S + 51 m W can be calculated by finding the net displacement. It results in a magnitude of 95 meters, which corresponds to option b) 95 m.

Step-by-step explanation:

To find the magnitude of the one-dimensional vectors, we first consider the given vectors' components in the north (N), east (E), south (S), and west (W) directions. The displacement in the north direction (N) is 23 meters, in the east direction (E) is 7 meters, in the south direction (S) is 36 meters, and in the west direction (W) is 51 meters.

Considering the opposing directions, we subtract the displacements to find the net displacement. Adding the magnitudes of displacements in the N and S directions (23 m - 36 m) results in a net displacement of 13 meters southward. Similarly, adding the magnitudes of displacements in the E and W directions (7 m - 51 m) results in a net displacement of 44 meters westward.

Using the Pythagorean theorem to calculate the magnitude of the net displacement, we get
`\(\sqrt{(13 \, \text{m})^2 + (44 \, \text{m})^2} = √(169 + 1936) = √(2105) \approx 45.87 \, \text{m}\)`. Rounding to the nearest whole number, the magnitude of the resultant displacement is approximately 46 meters. Therefore, the magnitude of the given one-dimensional vectors is 46 meters, not matching any of the provided options. Apologies for the misunderstanding in the initial calculation.