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# Vectors: Adding and subtracting vectors

## Algebraic Addition and Subtraction of Vectors

### Vectors in a Straight Line

Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique:

Method: Addition/Subtraction of Vectors in a Straight Line

1. Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative.
2. Next simply add (or subtract) the magnitude of the vectors using the appropriate signs.
3. As a final step the direction of the resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction).

Let us consider a few examples.

#### Exercise 1: Adding vectors algebraically I

A tennis ball is rolled towards a wall which is 10 m away from the ball. If after striking the wall the ball rolls a further 2,5 m along the ground away from the wall, calculate algebraically the ball's resultant displacement.

##### Solution
1. Step 1. Draw a rough sketch of the situation :

2. Step 2. Decide which method to use to calculate the resultant :

We know that the resultant displacement of the ball (xRxR) is equal to the sum of the ball's separate displacements (x1x1 and x2x2):

x R = x 1 + x 2 x R = x 1 + x 2
(1)

Since the motion of the ball is in a straight line (i.e. the ball moves towards and away from the wall), we can use the method of algebraic addition just explained.

3. Step 3. Choose a positive direction :

Let's choose the positive direction to be towards the wall. This means that the negative direction is away from the wall.

4. Step 4. Now define our vectors algebraically :

With right positive:

x 1 = + 10 , 0 m · s - 1 x 2 = - 2 , 5 m · s - 1 x 1 = + 10 , 0 m · s - 1 x 2 = - 2 , 5 m · s - 1
(2)
5. Step 5. Add the vectors :

Next we simply add the two displacements to give the resultant:

x R = ( + 10 m · s - 1 ) + ( - 2 , 5 m · s - 1 ) = ( + 7 , 5 ) m · s - 1 x R = ( + 10 m · s - 1 ) + ( - 2 , 5 m · s - 1 ) = ( + 7 , 5 ) m · s - 1
(3)
6. Step 6. Quote the resultant :

Finally, in this case towards the wall is the positive direction, so: xRxR = 7,5 m towards the wall.

#### Exercise 2: Subtracting vectors algebraically I

Suppose that a tennis ball is thrown horizontally towards a wall at an initial velocity of 3m·s-13m·s-1 to the right. After striking the wall, the ball returns to the thrower at 2m·s-12m·s-1. Determine the change in velocity of the ball.

##### Solution
1. Step 1. Draw a sketch :

A quick sketch will help us understand the problem.

2. Step 2. Decide which method to use to calculate the resultant :

Remember that velocity is a vector. The change in the velocity of the ball is equal to the difference between the ball's initial and final velocities:

Δ v = v f - v i Δ v = v f - v i
(4)

Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed.

3. Step 3. Choose a positive direction :

Choose the positive direction to be towards the wall. This means that the negative direction is away from the wall.

4. Step 4. Now define our vectors algebraically :
v i = + 3 m · s - 1 v f = - 2 m · s - 1 v i = + 3 m · s - 1 v f = - 2 m · s - 1
(5)
5. Step 5. Subtract the vectors :

Thus, the change in velocity of the ball is:

Δ v = ( - 2 m · s - 1 ) - ( + 3 m · s - 1 ) = ( - 5 ) m · s - 1 Δ v = ( - 2 m · s - 1 ) - ( + 3 m · s - 1 ) = ( - 5 ) m · s - 1
(6)
6. Step 6. Quote the resultant :

Remember that in this case towards the wall means a positive velocity, so away from the wall means a negative velocity: Δv=5m·s-1Δv=5m·s-1 away from the wall.

#### Resultant Vectors

1. Harold walks to school by walking 600 m Northeast and then 500 m N 4040 W. Determine his resultant displacement by using accurate scale drawings.
2. A dove flies from her nest, looking for food for her chick. She flies at a velocity of 2m·s-12m·s-1 on a bearing of 135135 and then at a velocity of 1,2m·s-11,2m·s-1 on a bearing of 230230. Calculate her resultant velocity by using accurate scale drawings.
3. A squash ball is dropped to the floor with an initial velocity of 2,5m·s-12,5m·s-1. It rebounds (comes back up) with a velocity of 0,5m·s-10,5m·s-1.
1. What is the change in velocity of the squash ball?
2. What is the resultant velocity of the squash ball?

Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. When vectors are not in a straight line, i.e. at an angle to each other, the following method can be used:

### A More General Algebraic technique

Simple geometric and trigonometric techniques can be used to find resultant vectors.

#### Exercise 3: An Algebraic Solution I

A man walks 40 m East, then 30 m North. Calculate the man's resultant displacement.

##### Solution
1. Step 1. Draw a rough sketch :

As before, the rough sketch looks as follows:

2. Step 2. Determine the length of the resultant :

Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let xRxR represent the length of the resultant vector. Then:

x R 2 = ( 40 m ) 2 + ( 30 m ) 2 x R 2 = 2 500 m 2 x R = 50 m x R 2 = ( 40 m ) 2 + ( 30 m ) 2 x R 2 = 2 500 m 2 x R = 50 m
(7)
3. Step 3. Determine the direction of the resultant :

Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle αα between the resultant displacement vector and East, by using simple trigonometry:

tan α = opposite side adjacent side tan α = 30 40 α = tan - 1 ( 0 , 75 ) α = 36 , 9 tan α = opposite side adjacent side tan α = 30 40 α = tan - 1 ( 0 , 75 ) α = 36 , 9
(8)
4. Step 4. Quote the resultant :

The resultant displacement is then 50 m at 36,936,9 North of East.

This is exactly the same answer we arrived at after drawing a scale diagram!

In the previous example we were able to use simple trigonometry to calculate the resultant displacement. This was possible since the directions of motion were perpendicular (north and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to one another. The following example illustrates this.

#### Exercise 4: An Algebraic Solution II [ADVANCED]

A man walks from point A to point B which is 12 km away on a bearing of 4545. From point B the man walks a further 8 km east to point C. Calculate the resultant displacement.

##### Solution
1. Step 1. Draw a rough sketch of the situation :

BA^F=45BA^F=45 since the man walks initially on a bearing of 4545. Then, AB^G=BA^F=45AB^G=BA^F=45 (parallel lines, alternate angles). Both of these angles are included in the rough sketch.

2. Step 2. Calculate the length of the resultant :

The resultant is the vector AC. Since we know both the lengths of AB and BC and the included angle AB^CAB^C, we can use the cosine rule:

A C 2 = A B 2 + B C 2 - 2 · A B · B C cos ( A B ^ C ) = ( 12 ) 2 + ( 8 ) 2 - 2 · ( 12 ) ( 8 ) cos ( 135 ) = 343 , 8 A C = 18 , 5 km A C 2 = A B 2 + B C 2 - 2 · A B · B C cos ( A B ^ C ) = ( 12 ) 2 + ( 8 ) 2 - 2 · ( 12 ) ( 8 ) cos ( 135 ) = 343 , 8 A C = 18 , 5 km
(9)
3. Step 3. Determine the direction of the resultant :

Next we use the sine rule to determine the angle θθ:

sin θ 8 = sin 135 18 , 5 sin θ = 8 × sin 135 18 , 5 θ = sin - 1 ( 0 , 3058 ) θ = 17 , 8 sin θ 8 = sin 135 18 , 5 sin θ = 8 × sin 135 18 , 5 θ = sin - 1 ( 0 , 3058 ) θ = 17 , 8
(10)

To find FA^CFA^C, we add 4545. Thus, FA^C=62,8FA^C=62,8.

4. Step 4. Quote the resultant :

The resultant displacement is therefore 18,5 km on a bearing of 062,8062,8.

#### More Resultant Vectors

1. A frog is trying to cross a river. It swims at 3m·s-13m·s-1 in a northerly direction towards the opposite bank. The water is flowing in a westerly direction at 5m·s-15m·s-1 . Find the frog's resultant velocity by using appropriate calculations. Include a rough sketch of the situation in your answer.
2. Sandra walks to the shop by walking 500 m Northwest and then 400 m N 3030 E. Determine her resultant displacement by doing appropriate calculations.

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