Question

sep develop a model changing classes, you walk 20.0 m down a hall, turn left, and then walk 10.0 m down another hall. define a coordinate system and draw the displacement vector in the space that follows. also, draw the displacement components, and then determine the vector’s magnitude.

Explanation:

Step1: Define Coordinate System

Let’s define the initial direction (first hall) as the positive \( x \)-axis and the direction after turning left (second hall) as the positive \( y \)-axis. So, the first displacement vector \( \vec{d}_1 = 20.0\,\text{m}\,\hat{i} \) (along \( x \)-axis) and the second displacement vector \( \vec{d}_2 = 10.0\,\text{m}\,\hat{j} \) (along \( y \)-axis).

Step2: Find Resultant Displacement Vector

The resultant displacement vector \( \vec{d} = \vec{d}_1 + \vec{d}_2 = 20.0\,\text{m}\,\hat{i} + 10.0\,\text{m}\,\hat{j} \).

Step3: Calculate Magnitude of Resultant Vector

The magnitude of a vector \( \vec{d} = d_x\,\hat{i} + d_y\,\hat{j} \) is given by the Pythagorean theorem: \( |\vec{d}| = \sqrt{d_x^2 + d_y^2} \). Substituting \( d_x = 20.0\,\text{m} \) and \( d_y = 10.0\,\text{m} \), we get:
\[
|\vec{d}| = \sqrt{(20.0)^2 + (10.0)^2} = \sqrt{400 + 100} = \sqrt{500} \approx 22.4\,\text{m}
\]

Answer:

The magnitude of the displacement vector is approximately \( \boldsymbol{22.4\,\text{m}} \).

(Note: For the diagram part, you would draw a coordinate system with the origin at the starting point. Draw a vector along the \( x \)-axis of length 20.0 m, then from the end of that vector, draw a vector along the \( y \)-axis of length 10.0 m. The resultant vector is from the origin to the end of the second vector, and its components are the two individual vectors along the axes.)