Identify the individual displacement vectors.

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Here's the solution for the lady's displacement:

Identify the individual displacement vectors.
The lady walks 1 km due East. Let this be $\mathbf{D_1}$ .
The lady then walks 1 km due North. Let this be $\mathbf{D_2}$ .
Represent the displacements in component form.
Let East be the positive x-direction and North be the positive y-direction.
$\mathbf{D_1} = 1\mathbf{i} + 0\mathbf{j} \text{ km}$ $\mathbf{D_2} = 0\mathbf{i} + 1\mathbf{j} \text{ km}$
Calculate the total displacement vector.
The total displacement $\mathbf{D_{total}}$ is the sum of the individual displacements:
$\mathbf{D_{total}} = \mathbf{D_1} + \mathbf{D_2}$ $\mathbf{D_{total}} = (1+0)\mathbf{i} + (0+1)\mathbf{j} \text{ km}$ $\mathbf{D_{total}} = 1\mathbf{i} + 1\mathbf{j} \text{ km}$
Calculate the magnitude of the total displacement.
The magnitude of the total displacement is found using the Pythagorean theorem:
$|\mathbf{D_{total}}| = \sqrt{(1 \text{ km})^2 + (1 \text{ km})^2}$ $|\mathbf{D_{total}}| = \sqrt{1 + 1} \text{ km}$ $|\mathbf{D_{total}}| = \sqrt{2} \text{ km}$ $|\mathbf{D_{total}}| \approx 1.414 \text{ km}$
Calculate the direction of the total displacement.
The direction $\theta$ is found using the arctangent function:
$\theta = \arctan\left(\frac{\text{North component}}{\text{East component}}\right)$ $\theta = \arctan\left(\frac{1 \text{ km}}{1 \text{ km}}\right)$ $\theta = \arctan(1)$ $\theta = 45^\circ$
Since both components are positive, the direction is $45^\circ$ North of East.

The lady's displacement is $1.414 \text{ km at } 45^\circ \text{ North of East}$ .

Identify the individual displacement vectors.

Answer

Step-by-step solution

Identify the individual displacement vectors.

Represent the displacements in component form.

Calculate the total displacement vector.

Calculate the magnitude of the total displacement.

Calculate the direction of the total displacement.

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