Equations of Motion

The equations of motion are the fundamental concept of classical mechanics. They provide a basic understanding of how objects move under the influence of forces. These equations, also known as the kinematic equations of motion, are derived from Newton’s laws of motion. They help in understanding and analysing the motion of an object’s position, velocity, and acceleration over time. These equations are the cornerstone in understanding the motion of an object not only in two dimensions but also in one and three dimensions. Here, we will explore these equations, a stepping stone in physics and science. 

1.0Equations of Motion Formula

First Equation of Motion

The first equation of motion states that the final velocity(v) of an object is equal to its initial velocity(u) plus the product of acceleration(a) and time(t). The equation applies when an object is moving with uniform acceleration. Mathematically, it can be expressed as: 

v=u+at

Second Equation of Motion

This equation gives the displacement(s) of an object when the acceleration(a) of an object remains constant. The formula for the second equation of motion: 

$s=ut+\frac{1}{2}a{t}^2$

Third Equation of Motion

The third equation in the typical three equations of motion establishes the relation between final velocity(v), initial velocity(u), acceleration(a), and displacement(s) without the use of time. The formula can be expressed as: 

$v^2=u^2+2as$

2.0Deriving Equations of Motion

The equations of motion derivation include different methods, the two of them that we are going to use to derive these equations here are simple algebraic methods and graphical methods. 

Notations: 

v = final velocity 

u = initial velocity 

t = time

a = acceleration 

s = displacement 

First Equation of Motion Derivation

To derive: v=u+at

First Equation of Motion by Algebraic Method

It is known that the acceleration of an object can be expressed as the rate of change in velocity, which can mathematically be written as: 

$a=\frac{v-u}{t}$

$at=v-u$

By rearranging the above equation, we will get:          

$v=u+at$

First Equation of Motion by Graphical Method

The graph shown above is a velocity-time graph with velocity at the y-axis and time (t) at the x-axis. By constructing a line perpendicular to OC from B, a line AD parallel to OC, and another perpendicular from B to OE.  

Now, 

BC = v = Final velocity 

AO = u = initial velocity 

AD = t = total time taken by an object to move from A to B. 

According to the graph, 

BC = BD + DC = v 

Since DC = AO = u

v = BD + AO 

According to graph 

v = BD + u …..(1)

BD = BC – DC 

BD = v – u 

We know that, at=v-u

Hence, BD=at …..(2) 

From equation 1 and 2 

v=u+at

2. Second Equation of Motion Derivation:

Second Equation of Motion by Algebraic Method:

To solve this equation, we use the concept of velocity. Velocity is defined as the rate of change of displacement that can be written as: 

$v=\frac{s}{t}ors=vt$

If not constant, we can rewrite v in the above equation as the measure of average velocity: 

$s=\left(\frac{u+v}{2}\right)\times t$

From the first equation of motion: 

$s=\left(\frac{u+(u+at)}{2}\right)\times t$

$s=\left(\frac{2u+at}{2}\right)\times t$

$s=(\frac{2u}{2}+\frac{at}{2})\times t$

$s=(u+\frac{at}{2})\times t$

$s=ut+\frac{1}{2}a{t}^2$

Second Equation of Motion by Graphical Method: 

In a Velocity-time graph, 

Distance travelled (s) = Area of figure ABCO = Area of rectangle ADCO + Area of triangle ADB.

$s=OA\times OC+\left(\frac{1}{2}\right)\times AD\times BD$

Here, OA = u, OC = t, AD = t, BD = at

$s=u\times t+\left(\frac{1}{2}\right)\times t\times at$

$s=ut+\frac{1}{2}a{t}^2$

3. Third Equation of Motion Derivation: 

Third Equation of Motion by Algebraic Method: 

We know from the above equation that: 

$s=\left(\frac{v+u}{2}\right)\times t$

Here, 

$a=\frac{v-u}{t}\text{or}t=\frac{v-u}{a}$

$s=\left(\frac{v+u}{2}\right)\times \frac{v-u}{a}$

$s=\frac{v^2-u^2}{2a}$

$2as=v^2-u^2$

$v^2=u^2+2as$

Third Equation of Motion by Graphical Method: 

In the graph, AOCB is a trapezium and the area under the curve AB is equal to the distance travelled by an object in a velocity-time graph. Hence, 

s = Area of trapezium 

$s=\frac{1}{2}\times \text{(Sum of Parallel Sides)}\times \text{Height}$

$s=\frac{1}{2}\times (AO+BC)\times AD$

$s=\frac{1}{2}\times (u+v)\times t$

As,

$t=\frac{v-u}{a}$

$s=\left(\frac{v+u}{2}\right)\times \frac{v-u}{a}$

$s=\frac{v^2-u^2}{2a}$

$2as=v^2-u^2$

$v^2=u^2+2as$

3.0Equations of Motion Numericals

Problem 1: A car starts from rest and accelerates at 2 m/s². How far does it travel in 8 seconds?

Solution: Given that u = 0m/s (as the car starts from rest)

a = 2 m/s², t = 8 seconds 

Now, using the second equation of motion

$s=ut+\frac{1}{2}a{t}^2$

$s=0+\frac{1}{2}\times 2\times 8^2=64m$

Problem 2: A ball is thrown upwards with an initial velocity of 10 m/s. Calculate the time it will take to reach its maximum height.

Solution: Given, 

u = 10m/s, a = –9.8m/s² (Negative because ball is thrown upward), 

v = 0m/s (the balls come to rest at the highest point)

Using the first equation: 

v=u+at

0=10+(-9.8)t

9.8t=10

t=1.02sec

Problem 3: A car is travelling at a velocity of 20 m/s and comes to a stop after applying the brakes. The car decelerates uniformly at a rate of 4 m/s². Calculate the distance the car travels before coming to a stop.

Solution: Given, 

u = 20m/s, v = 0m/s, a = –4m/s² (Negative because the car is decelerating) 

By using, 

$v^2=u^2+2as$

$0^2=20^2+2(-4)s$

8s=400

s=50meters

4.0Hamilton's Equations of Motion

The equations of motion mentioned in the previous sections are the fundamentals of mechanics but are not used for more complex systems. The problems related to such systems are solved with Hamilton's Equations of Motion. Hence, this equation adds another layer of solutions, making the conventional three to 4 equations of motion. The formula for the equation can expressed as: 

$\frac{d{q}_i}{dt}=\frac{\partial H}{\partial p_i}$

$\frac{d{p}_i}{dt}=-\frac{\partial H}{\partial q_i}$

Here, 

• q_i​ is the generalised coordinates (position coordinates in the system)
• p_i​ is the generalised momenta (which are the partial derivatives of the Lagrangian L concerning the velocities $p_i=\frac{\partial L}{\partial \dot{q_i}}$
• H is the Hamiltonian of the system, which is H = T+V (T = Kinetic energy and V = Potential energy)