In the Vector Addition Lab, the following data was collected for determining the displacement from the door of the Physics classroom to another location in the building: 2 m, West; 14.0 m, South; 25.0 m, East; 11.0 m, North; and 2 m, West. The magnitude of the resultant displacement from the physics classroom to the assigned location is closest to ___ meters. (Round to the nearest integer.)

21 m

Explanation:

Since the displacement from the door is 2 m West, we have our vector as -2i. The vector representing 14.0 m South is -14.0j. The vector representing 25.0 m East is 25.0i. The vector representing 11.0 m North is 11.0j. And, the vector representing 2.0 m West is -2.0i.

So, to get our position vector at the other location, we add all the vectors together.

So, r = -2i + (-14.0j) + 25.0i + 11.0j - 2i

= -4i + 25.0i - 14.0j + 11.0j

= 21.0i -3j m

Now, if we assume the position vector for the door is at the origin, we have r₀ = 0i + 0j m

So, our displacement from the door is r - r₀ = 21.0i - 3.0j - (0i + 0j) =  21.0i - 3.0j

So, the magnitude of the resultant displacement |r - r₀| = √(21.0² + 3.0²)

= 3.0√(7.0² + 1)

= 3.0√(49 + 1)

= 3.0√50

= 3.0 × 7.0711

= 21.2

≅ 21 m to the nearest integer

the answer is a. hope this .