Question

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how do i work with vectors graphically?
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• solve vector problems using graphing

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draw a vector map of the following story.
nevaeh and dayra decided to meet up for lunch. they started by walking 10 meters east on main st. at the corner, they made a left and continued walking for 8 meters until they came to a store. after shopping, they walked 5 meters east and 12 meters south before arriving at a bistro for lunch. after lunch, they walked 9 m west to the park to enjoy the sunshine.

Explanation:

Step1: Define directions

Let East - West be along the x - axis (East is positive x - direction, West is negative x - direction) and North - South be along the y - axis (North is positive y - direction, South is negative y - direction).

Step2: Analyze first part

They walk 10 meters East. So the first vector $\vec{v}_1=(10,0)$.

Step3: Analyze second part

They make a left (which is North) and walk 8 meters. So the second vector $\vec{v}_2=(0,8)$.

Step4: Analyze third part

They walk 5 meters East. So the third vector $\vec{v}_3=(5,0)$.

Step5: Analyze fourth part

They walk 12 meters South. So the fourth vector $\vec{v}_4=(0, - 12)$.

Step6: Analyze fifth part

They walk 9 meters West. So the fifth vector $\vec{v}_5=(-9,0)$.

Step7: Calculate net displacement in x - direction

$x = 10+0 + 5+0-9=6$

Step8: Calculate net displacement in y - direction

$y=0 + 8+0-12+0=-4$
The net displacement vector $\vec{R}=(6,-4)$

To draw the vector map:

1. Start at the origin. Draw an arrow 10 units long in the positive x - direction for the first vector.
2. From the tip of the first vector, draw an arrow 8 units long in the positive y - direction for the second vector.
3. From the tip of the second vector, draw an arrow 5 units long in the positive x - direction for the third vector.
4. From the tip of the third vector, draw an arrow 12 units long in the negative y - direction for the fourth vector.
5. From the tip of the fourth vector, draw an arrow 9 units long in the negative x - direction for the fifth vector. The end - point of the last vector represents the final position relative to the starting point.

Answer:

The net displacement vector is $(6,-4)$ and the vector map is drawn as described above.