Question

Depict the following path: You drive 14 blocks East, 7 blocks North, and 2 blocks West. Use the vertex of the graph (coordinates 0,0) as you lay out the vectors. What is the displacement between your starting and end points? What is the angle of displacement? (Include a screenshot of the pathway in your response for this question.)

Answer 1

Answer:

Displacement = $12\hat{i} +7\hat{j}$

Magnitude of displacement = 13.89 units

Angle of displacement = 30.26°

Explanation:

Mark the co-ordinates of each of the end points of the vectors.

The marked co-ordinates are shown below.

Displacement is the shortest distance between two points. Here, displacement between the starting and end points is the vector $\overrightarrow{OA}$.

Displacement vector for a point A$(x,y)$ from origin O is given as:

$\overrightarrow{OA}=(x-0)\hat{i}+(y-0)\hat{j}$

Magnitude is given as:

$|\overrightarrow{OA}|=\sqrt{x^{2}+y^{2}}$

Direction of the vector is given as:

$\theta=tan^{-1}\frac{y}{x}$

Therefore, displacement vector is:

$\overrightarrow{OA}=(12-0)\hat{i}+(7-0)\hat{j}\\ \overrightarrow{OA}=12\hat{i}+7\hat{j}$

Magnitude of displacement is given as:

$|\overrightarrow{OA}|=\sqrt{x^{2}+y^{2}}\\ |\overrightarrow{OA}|=\sqrt{12^{2}+7^{2}}\\ |\overrightarrow{OA}|=\sqrt{144+49}\\\\ |\overrightarrow{OA}|=13.89$

Angle of displacement is:

$tan\theta=\frac{y}{x} =\frac{7}{12}\\\theta=tan^{-1}(\frac{7}{12})$

$\theta=30.26$°