Last Updated : 04 Feb, 2024
The motion of any object can’t be described without defining its position, denoted by ‘x’. Basically, it is needful to give specifications of the position being relative to the conventional frame of reference. The frame of reference is a set of values or rules through which judgements of physics and measurements can be made. We often consider the earth as a frame of reference and tend to specify the position of different things and objects as it relates to other motionless objects on the earth. We can also make use of objects which are in motion but are in relative motion with the earth. Therefore, to describe the position of a passenger in the train we use the train as a reference. Let us further know more about the position and its formula.
What is the position of an object?
Any object’s True Position is its exact coordinate or location as defined by its basic dimensions or other means. To put it another way, position refers to how far a feature’s location can deviate from its “True Position.”
The trajectory of a rectilinear movement is that of a straight line. This movement is also carried out at a constant rate of acceleration. On a straight line, the origin may be x0, with an observer measuring the position x of the moving object at time t. Hence,
A place or point where something or someone is located or has been positioned in relation to other objects is referred to as a position. In physics, we talk about where something is in relation to an x, y-axis.
Position can be specified simply, such as in front of a computer, or more precisely, by using the longitude and latitude of a specific location on Earth. A position where a house or structure is located is known as an address. The axis on which the position on Earth is given is the Earth’s latitude and longitude.
Position Formula
According to the equations of motion, when an accelerates with acceleration a, in time duration t, with initial velocity v0 and initial position of the object is x0, then the position of the object in time t is given by,
x (t) = 1/2 at2 + v0t + x0
Here is the formula to determine the position change of an object, that is
Position change = Final Position – Initial position
or
Δx = x2 – x1
where:
- x2 is the final position of the object,
- x1 is the initial position of the object, and
- Δx. change in the position of the object.
Sample Problems
Problem 1: A ball rolls with an initial velocity of 3 m/s for 20 m and angular acceleration of 2 m/s2. What will be the position of the ball after 5 s?
Solution:
Given, u = 3 m/s, x0 = 20 m, a = 2 m/s2 and t = 5 s.
Since, we want to know final position of the ball we will use,
x = x0 + ut + 0.5at2
= 20 + (3)(5) + 0.5(2)(5)(5).
= 20 + 15 + 25.
= 60 m
Therefore, the final position of the ball would be 60 m.
Problem 2: A vehicle with an initial velocity of 2 m/s covered a distance of 10 m with a constant acceleration of 2 m/s2. Find the position of the vehicle after 5 s.
Solution:
Given, u = 2 m/s, x0 = 10 m, a = 2m/s2 and t = 5 s.
x = x0 + ut + 0.5at2
= 10 + (2)(5) + 0.5(2)(5)(5).
= 10 + 10 + 25.
= 45 m
Therefore, the final position of the vehicle would be 45 m.
Problem3: A car starts from rest and accelerates uniformly over a time of 5 s for a distance of 100 m. Determine the acceleration of the car.
Solution:
Given, t = 5 seconds, x0 = 0, u = 0 and x = 100 m.
x = x0 + ut + 0.5at2
100 = 0 + 0 + 0.5(a)(5)(5)
100 = 0 + 12.5a
12.5a = 100
a = 8 m/s2
Therefore, the acceleration of the car would be 8 m/s2.
Problem 4: A ball rolls with an initial velocity of 10 m/s for 20 m and angular acceleration of 0.2 m/s2. What will be the position of the ball after 10 s?
Solution:
Given, u = 10 m/s, x0 = 20 m, a = 0.2 m/s2 and t = 10 s.
Since, we want to know final position of the ball we will use,
x = x0 + ut + 0.5at2
= 20 + (10)(10) + 0.5(0.2)(10)(10).
= 20 + 100 + 10.
= 130 m
Therefore, the final position of the ball would be 130 m.
Problem 5: A vehicle with an initial velocity of 20m/s covered a distance of 20 m with a constant acceleration of 0.2 m/s2. Find the position of the vehicle after 20 seconds.
Solution:
Given, u = 20 m/s, x0 = 20 m, a = 0.2 m/s2 and t = 20 s
x = x0 + ut + 0.5at2
= 20 + (20)(20) + 0.5(0.2)(20)(20).
= 20 + 400 + 40
= 460 m
Therefore, the final position of the vehicle would be 460 m.
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