How to Add Vectors?

We know vectors differ from scalar because it not only has a magnitude, but also a direction. On a graph, we will assume that the positive directions for the y axis is north (up), and the positive direction for the x axis is east (right).Therefore, directions of south and west symbolize negative vectors.
Now for adding vectors on a graph. In most cases, there were be more than 1 or 2 vectors heading into all directions. If you are a great mathematician like Mr. Dulmage, you would say, "Oh, I can solve this trignonmetry, EASILY, because the you first get the sine of blah blah, and use that to get the cosine of blah blah..."
Confusing isn't it? Most of us aren't great mathematicians, or are at some extent.Therefore, a much simpler to calculate the resultant is to simply "add" all the x values and y values of vectors. (Note: in some cases when vector is negative, we're theoritically adding vectors, but adding by a negative digit is also the same as subtracting the number).
So next we'll have to figure out how to find the x and y values for the vectors on the graph to add.
In one case, if the vector's x and y values are given, that's great because you wouldn't need to worry about calculating them. But when they're not given, we'll use some trignometry we learned in grade 10 math to solve this problem.

Opposite side is the y value, adjacent side is the x value. With basic sin and cos laws, we would be able to calculate the x and y values.
So next as mentioned already, we add all the x and y values. But keep in mind that even when some times you're adding a vector, you have to be careful because the direction of the vector determine if the number is positive or negative.

 For the x axis, vectors with direction east are positive. Vice versa, vectors with direction west are negative. When considering only the x value for adding two vectors, if both vectors have direction east, you add the 2 numbers, if both have direction east, it would be (-x1)+(-x2), so theoritically adding 2 negative numbers. When one is east and one west, it will be one positive subtract one negative.
As demonstrated by the diagram, adding y values for the resultant y would be the same way.

Finall, now after all that, we get our resultant x and y values. But we're not done yet because we have to calculate the hypotenuse. We can do this with the help of Pythagora and his theorem.

 

So now we are able to calculate the hypotenuse with the derived formula x2 + y2 = R2, but we still have to know which direction specifically the vector is heading towards. This can be determined by going back to the x and y values for the resultant.Once again, positive y value means the vector is heading towards north, and vice versa. Positve x value means the vector is heading towards east, and vice versa.
So after we get these 2 directions, we can imagine a little triangle inside our head: If the vector start from the origin (0,0), by what degree have it gone east or west from south or north?

This can be again calculated from the help of trignometry. We know that the tan is calculated from opposite over adjacent, so now we just have to see the triangle in different views to see which value (x or y) is the opposite side and adjacent side. For this calculation, make sure that you're reading the angle from the y axis to the x axis with respect to where the vector is located.
And now finally, we write the resultant vector as: the hypotenuse value [ydirection (degree) xdirection]

If you're still confused because I know I didn't explain very well, check out this video - it MIGHT help you :)
http://www.youtube.com/watch?v=UmLu3HIKBG8

Deriving equations from the graph but flipped because I got my trapezoid backward, yah :)

When given the velocity graph, we already know that by calculating the area of the shape in respect of the x axis, we will get the displacement.

Generally, the shapes are rectangles, triangles and the combination of both - trapezoid.

We know that the formula for calculating the area for rectangle is:

A = LW,

and formula for triangle is:

A = 1/2 BH

and trapezoid:

A=(a+b)H/2

But sometimes, the trapezoid can also be seen as a triangle and rectangle stacked on top of each other. Therefore, another way of calculating the area of the trapezoid can be: Area of triangle + Area of Rectangle

A = LW + 1/2BH

In this sketch I drew on paint (:, the graph is a typical velocity time graph. As ou can see, the trapezoid is divided into 2 parts - a triangle and a rectangle.

So first, the base of triangle as we can see, is time(t), the height is the difference between v1 and v2. So the area of the triangle is:

A = 1/2(t)(v1 - v2)

For the area of the rectangle, obviously the length is t, and the width is v2, so the area is:

A = v2 * t

By adding the sum of these areas, we get the area of the trapezoid, or otherwise as we know the displacement:

D = 1/2(t)(v1 - v2)+v2*t (2)

Now from previous knowledge, we already have equation 1):

v2=v1+at (1) => v1=v2-at (3)

Sub (3) into (2):

D=1/2t(v2-at-v2)+v2*t

D=(v2)(t)+1/2at² 

And Voila! Equation 3 from the Big Five!

To derive equation 4, the idea is about the same, except this time, you look at the graph a little differently

You treat the trapezoid as a bigger rectangle, but without the shaped triangle. So another way to calculate its area is by calculating the area of the rectangle, and subtract the area of the imaginary triangle from it:

A = LW - 1/2(BH)

First, the length of the rectangle as we can see is t, the width is v1:

Area of Rectangle: v1 * t

Secondly, the base of the triangle is again t, the base as you can see by observing more closely, is (v1-v2).

Area of Triangle: 1/2(t) * (v1-v2)

Therefore, area of trapezoid: v1 * t - 1/2(t) * (v1-v2) (1)

Once again we apply the first equation that we already know of:

v2 = v1 + at (2)

Sub (2) into (1)

D = v1 * t - 1/2(t) * (v1 - v1 - at)

D = (v1)(t) - 1/2at²

And Ta da! Equation 4!

(Please note that v2 and v1 in the equations are opposite of the ones in the textbook, this is because the longer and shorter side of the trapezoid I have is directly opposite of the ones we did in class)

 

• This is a distance and time graph.
• When line is constant, it represents immobility, so we stay at one spot.
• When the line has a postive or inclining slope, it means moving East, in this case away from the sensor
• When the line has a negative or declining slope, it means moving towards the opposite of East - West, in this case towards the sensor.   

1. Stay at 1m for 1 second.
2. Walk 1.5m in 2 seconds (0.75 m/s) east.
3. Stay at 2.5m for 3 seconds.
4. Walk 0.75m in 1.5 seconds (0.5 m/s) west.
5. Stay at 1.75m for 2.5 s.

• This is also a distance and time graph
• Alike the first graph,line of + slope means moving East and away from the sensor.
• Vice versa, line of - slope means moving West and toward the sensor.

1. Walk from 3m east away from origin west at 1.5m in 3 s (0.5 m/s)

2. Stay at 1.5m for 1 s.

3. Walk west at 1m in 1 second (1.0m/s).

4. Stay at 0.5m for 2 s.

5. Walk east at 2.5m in 3 seconds (0.83 m/s)

 

• Different from the first two graphs, this is a velocity time graph.
• When the line is at 0, it means 0 velocity thus unmoving.
• When line is constant and +(above 0), it represents a constant velocity moving East (away from the sensor)
• When line is constant and -(below 0), it represents a constant velocity moving West (toward the sensor)

1. Stay for 2 seconds.
2. Walk east at 0.5 m/s for 3 seconds
3. Stay still for 2 seconds.
4. Walk west at 0.5 m/s for 3 seconds.

 

• This is also a velocity and time graph
• Alike the previous graph, line at 0 means 0 velocity (unmoving), constant line above or below zero means moving in East or West directions
• When line is linear (with a slope) and constant, it means a constant but gradual increase in velocity. To walk this, we have to gradually speed up in the time span

1. Run and accelerate east at 0.5 m/s in 4 seconds.

2. Walking east at 0.5 m/s for 2 seconds.

3. Walk west at 0.4 m/s for 3 seconds.

4. Stay still for 1 second.

 

• This is a distance and time graph
• Alike graph 1 & 2, line of + slope means moving east constantly. (vice versa for negative slope)
• Line of constant (horizontal line) represents 0 distance in a time span, thus 0 velocity. 

1. Walk east from 0.8m east of origin at 1m in 3.5s (0.29 m/s)

2. Stay still at 1.8m for 3.25 s.

3. Walk east at 1.4m in 2.25 s (0.62 m/s)

 

• This is a velocity and time graph
• line of constant and positive represents a constant velocity moving East (away from the sensor)
• Line of constant and negative represents a constant velocity moving West (toward the sensor)
• Sudden increase or decrease in the value of the line means a dramatic acceleration or slow down

1. Walk east at 0.35 m/s for 3 seconds.

2. Walk west 0.35 m/s for 3.5 seconds.

3. Stay still for 3.5 seconds.