Position Formula

To describe the motion of any object, we must be able to describe its position x. It reflects it at any particular time. In other words, we need to specify its position relative to the conventional frame of reference. A frame of reference is the arbitrary set of axes from which the position and motion of an object are described. We often use Earth as the frame of reference and describe the position of an object as it relates to stationary objects on Earth. In many cases, we use reference frames that are not stationary but are in motion relative to Earth. Thus, to describe the position of a person in an airplane we use the airplane for reference and not Earth. This article will explain the position formula and some examples. Let us learn it!

Position Formula

What is the position of an object?

The True Position of any object is its exact coordinate or location defined according to the basic dimensions or other means. In other words, Position means how far your features location can vary from its “True Position”.

A rectilinear movement is the one whose trajectory follows the straight line. Also, this movement is performed at constant acceleration. On the straight line, we may place the origin as \(x_0\), where there will be an observer who will measure the position x of the object in motion at the instant t.

Therefore, the position x of the mobile can be related to time t as in the following polynomial function.

position = \(initial position + initial velocity \times time + \frac{1}{2} \times acceleration \times (time)^2\)

Thus, something travels from a point to the other point, it is called displacement. Assuming the moving a ball from position \(x_1\) to position \(x_2\).

Formula for the Position:

The position change \(\Delta x (position formula)\) is determined as,

\(\Delta x = x_2 – x_1\)

Where,

\(x_2\)	first position
\(x_1\)	second position
\(\Delta x\).	change of displacement

If the body changes its position after some time t, then the change in position at any moment of time t will be represented as x(t). Its formula is:

x(t) = \(\frac{1}{2}a t^2 + v_0 t + x_0\)

Where,

x(t)	position of the body with respect to time t
\(x_0\)	the initial position of the body
\(v_0\)	the initial velocity of the body
a	acceleration the body possesses

Solved Examples

Q.1: A boy who has an initial velocity of \(3 m s^{-1}\), moves for a distance of 20 m. If it’s the angular acceleration is \(2 m s^{-2}\). Determine the position of the boy at the end of 5 sec.

Solution:

Known parameters:

\(V_0\)  (Initial velocity)  = \(3 m s^{-1}\),

\(X_0\)  (distance)  = 20 m,

a (angular acceleration)  = \(2 m s^{-2}\),

t (time) = 5s

The alteration in the position of the boy at instant of time t may be computed as:

X(t) = \(\frac{1}{2}a t^2 + v_0 t + X_0\)

Thus,

X(6) = \(0.5 \times 2 5^2 + 3 \times 5 +  20\)

= 25m + 15m + 20 m

= 60 m.

The position of the boy at the end of 5 sec will be 60 meters.