## What is a Position in physics?

**Definition of position in physics:** Position in physics simply means the location of an object in a plane or space. We can determine the position of an object by considering it is distance and direction from a specific frame of reference. Therefore, this is to say that we can find the position of an object, once we know its distance and direction from a specific point of reference.

The location of your television stand in your room is the position of that television stand. The location of your school from your home is the position of that school. A football coach wants his players to occupy a specific position on the field during a match competition. A man driving a car occupies a certain position as he continues to move.

Therefore, finding the position of an object is very important. To find the position of an object, we need to consider the x and y coordinates on a plane. In simple terms, we need to draw two lines that meet each other in the middle (we call it **origin**), and make sure that the two lines are perpendicular to each other.

We start measuring the position of an object from the origin of that object on a plane. The origin of an object on an x-y plane is the point where it starts from zero, it is the reference point where the object is seen or observed.

The vertical reference line is along the y-axis, while the horizontal reference line is along the x-axis.

To find the position of an object, we need to understand whether it belongs to a space or a plane.

### A Plane

A plane is due to two frames of reference, in form of an x and y-axis. We have two coordinates on a plane, the x-coordinate is known as abscissa, while the y-coordinate is known as ordinate. The horizontal component consists of the right and left sides starting from the origin. When you start from the origin and move to the left, it means that you are moving along the negative x direction.

When you start from the origin of the horizontal plane and move to the right, it means that you are heading toward the positive direction of the x-axis.

We also have a vertical component of the plane called the y-axis. The origin intersects the y-axis at the middle making the upper side a positive side of the y-axis, and the lower side a negative y-axis.

A plane is in the form of (x,y) coordinate.

### A space

Space consists of a three-dimensional coordinate system. We can find the location of an object in space by considering (x,y,z) coordinates. We have three coordinates in space, and they are x, y, and z.

## How to Find a Position in Physics

Let’s assume we are dealing with a plane in form of x-y coordinates. We write an ordered pair of values as (x,y). We measure values on a vertical line along the y direction, and we also measure values on the horizontal line which is the x direction.

Assuming the position of an object is on a plane (5,7) as our values, here is how to find the position of the object on a plane.

- Starting from the origin (zero), count 5 units along the positive x-axis
- Additionally, count 3 units along the positive y-axis
- On the vertical line, draw a horizontal line starting from 7 units and move toward the right.
- on the horizontal axis, draw a line starting from 5 units and move upward until the two lines intersect
- Draw a line from both axis so that they meet at the center (5,7)
- Let’s assume we have A(x
_{1},y_{1},z_{1}) and B(x_{2},y_{2},z_{2})- To calculate the distance between A and B
- We apply the formula AB
^{2}= (x_{2}– x_{1})^{2}+ (y_{2}– y_{1})^{2}+ (z_{2}– z_{1})^{2}- Additionally, to find the direction of AB. We apply [ tanθ=Opposite/Adjacent ]

## Examples of Position in Physics

Here are some examples to guide you on finding an object’s position in a plane or a space.

### Example 1

Calculate the distance between two points M (5,6) and N (12,8) on a plane.

**Solution:**

**Data**

We can see that the coordinates belong to a plane in the form of (x,y).

Therefore, from the question:

x_{1} = 5

x_{2} = 12

y_{1} = 6

y_{2} = 8

MN^{2} = (x_{2} – x_{1} )^{2} + (y_{2} – y_{1} )^{2}

We now substitute our data into the above formula

MN^{2} = (12 – 5 )^{2} + (8 – 6 )^{2}

After squaring the right-hand side, we have

MN^{2} = 7^{2} + 2^{2}

which is equal to

MN^{2} = 49 + 4

Now, we sum up our right-hand side(RHS)

MN^{2} = 53

By making MN the subject of the formula, and transferring the square to the RHS so that it can become the square root

MN = √53 = 7 units

our final answer is now

**MN = 7 units**

### Example 2

What is the distance between P(3,5,7) and Q(-4,4,-3)

**Solution**

**Data**

From the above question, we have our data as follows:

x_{1} = 3

x_{2} = -4

y_{1} = 5

y_{2} = 4

z_{1} = 7

z_{2} = -3

Now we apply the formula that says

PQ^{2} = (x_{2} – x_{1} )^{2} + (y_{2} – y_{1} )^{2} + (z_{2} – z_{1} )^{2}

By substituting our values into the above equation, we now have

PQ^{2} = (-4 – 3)^{2} + (4 – 5)^{2} + (-3 – 7)^{2}

This is now equal to

PQ^{2} = (-7)^{2} + (- 1)^{2} + (-10)^{2}

Thus, it is now

PQ^{2} = 49 + 1 + 100

After summing the right-hand side, it becomes

PQ^{2} = 150

we now take the square root of both sides to eliminate the square on the left-hand side

√PQ^{2} = √150

This is now equal to

PQ = √150 = 12.3 units = 12 units

### Example 3

Two players were perfectly positioned at A(2,3,5) and B(6,0,8)

a. Determine the distances OA and OB if O is the point in a field represented by O(1,0,2)

b. Find the distance between the two players, BA

**Solution**

**Data:**

By extracting our data from the above question, we have

Point A(2,3,5) and B(6,0,8) with an origin O(1,0,2)

and if O(x_{1},y_{1},z_{1}), A(x_{2},y_{2},z_{2}) and A(x_{3},y_{3},z_{3})

Thus,

x_{1} = 1, x_{2} = 2, and x_{3} = 6

y_{1} = 0, y_{2} = 3, and y_{3} = 0

z_{1} = 2, z_{2} = 5, and z_{3} = 8

##### a. To find OA, we apply the formula

OA^{2} = (x_{2} – x_{1} )^{2} + (y_{2} – y_{1} )^{2} + (z_{2} – z_{1} )^{2}

By inserting our data into the above equation, we will have

OA^{2} = (2 – 1)^{2} + (3 – 0)^{2} + (5 – 2)^{2}

This is equal to

OA^{2} = 1^{2} + 3^{2} + 3^{2}

After squaring the right-hand side(RHS), it becomes

**OA ^{2} = 1 + 9 + 9**

We now add the values in the RHS and then apply the square root to both sides in order to make the OA subject of the formula

OA = √19 = 4.4 = 4 units

Therefore, the distance OA is 4 units or 4 meters

To calculate OB, we apply the same method for OA

Thus

OB^{2} = (x_{3} – x_{1} )^{2} + (y_{3} – y_{1} )^{2} + (z_{3} – z_{1} )^{2}

after substituting our data into the above formula for calculating distance, we now have

OB^{2} = (6 – 1)^{2} + (0 – 0)^{2} + (8 – 2)^{2}

This is equal to

OB^{2} = (5)^{2} + (0)^{2} + (6)^{2}

we will now have

OB^{2} = 25 + 0 + 36

after summing up the RHS, we now have

OB^{2} = 61

by making OB the subject of the formula

OB = √61 = 7.8 = 8 units

Therefore, the distance between OB is 8 units or 8 meters.

##### b. To Find the distance between the two players A and B, we apply the formula

AB^{2} = (x_{2} – x_{3} )^{2} + (y_{2} – y_{3} )^{2} + (z_{2} – z_{3} )^{2}

After substituting our data into the above equation, we will have

BA^{2} = (x_{3} – x_{2} )^{2} + (y_{3} – y_{2} )^{2} + (z_{3} – z_{2})^{2}

Thus,

BA = √[(6 – 2)^{2} + (0 – 3)^{2} + (8 – 5)^{2}]

We now have

BA = √[(4)^{2} +(- 3)^{2} + (3)^{2}]

it will now become

BA = √16 + 9 + 9

Thus,

BA = √16 + 9 + 9 = √34 = 5.8 = 6 units

Therefore the distance between the two players, BA is 6 units or 6 meters

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SS1 Lesson Note: Introduction to Physics For First Term

**Reference:**