Physics Formulas

Physics Motion Formula

In physics motion of any object is an important and very general phenomenon. We many times have to deal with the motion of the objects. Various terms of the physics like velocity, time, distance, etc. are related and explained in various laws and equations of the motion. Sir Isaac Newton has given three laws of motion to describe the motion of massive bodies and how they interact. These laws and equations may seem obvious to us today, but more than three centuries ago they were considered revolutionary. In this article, the student will learn some important physics motion formula with suitable examples. Let us learn it!

Physics Motion Formula

What is the motion?

Newton’s laws of motion are for the motion of massive bodies in a given inertial reference frame. This inertial frame of reference can be described as a 3-dimensional coordinate system that is either stationary or in uniform linear motion. These motions within such an inertial reference frame can be described easily by three simple laws.

These equations of motion relate the displacement of the moving object with its velocity, acceleration and time. The motion of a particle may follow many different paths. But in the here, we will focus on motion in a straight line, i.e. in a single dimension.

Since this motion is in one dimension, we will simply use signed magnitudes of the displacement, velocity, and acceleration. Also, negative values will be used to denote vectors in the opposite direction to positive quantities.

In simple words, if there is no acceleration, we will have the familiar formula:

s = v × t


s displacement
v constant velocity
t time interval

For a constant acceleration a, an initial speed u and an initial position of zero:

  1. Velocity:

v = \(v_0\) + at

  1. Displacement with positive acceleration

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

  1. Displacement with negative acceleration

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

  1. Velocity squared

\(v^2 = {v_0}^2 + 2a (x- x_0)\)


x displacement
\(x_0\) constant velocity
t time interval
v velocity
a acceleration
\(v_0\) initial velocity

Solved Examples

Q.1: A massive body slides along a frictionless surface with a constant acceleration of \(2 m s^{-2}\). At time t = 0 the block is at x = 5 m and traveling with a velocity of \(3 m s^{-1}\). Then determine,

a) Where is the block at t = 2 seconds?

b) What is the block’s velocity at 2 seconds?


The given parameters are:

\(x_0\) = 5 m

\(v_0 = 3 m s^{-1}\)

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

(a)        Formula is

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

Substitute t = 2 seconds and the appropriate values of \(x_0 and v_0\).

\(x = 5 + 3 \times 3 + \frac{1}{2} \times 2 \times 2^2\)

= 5 m + 6 m + 4 m

= 15 m

The body will be at the 15 meter mark at t = 2 seconds.

(b)        This time, Equation 2 is the useful equation.

v = v_0 + at

= 3 + 2 × 2

= \(5 m s^{-1}\)

The body will be travelling \(7 m s^{-1}\) at t = 2 seconds.

Share with friends

Customize your course in 30 seconds

Which class are you in?
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Dr. Nazma Shaik
Gaurav Tiwari
Get Started

5 responses to “Spring Potential Energy Formula”

  1. Typo Error>
    Speed of Light, C = 299,792,458 m/s in vacuum
    So U s/b C = 3 x 10^8 m/s
    Not that C = 3 x 108 m/s
    to imply C = 324 m/s
    A bullet is faster than 324m/s

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

Leave a Reply

Your email address will not be published. Required fields are marked *

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

No thanks.