# Kinematic Equations

## What are Kinematic Equations of Motion?

The equations governing the motions of an object along a straight line path or a curved path are called Kinematic equations.

The kinematic equations are derived from the theories developed by Newton’s Laws of motion. These equations are of four types –

1. Equation for velocity after certain time.
2. Equation for displacement covered after certain time.
3. Equation for velocity at a certain position.
4. Equation for displacement in ( n_{th} ) second.

### Derivation of Kinematic Equations

Kinematic equations are based on the Newton’s laws of motion. These equations are derived by two methods –

1. Analytical method for derivation of Kinematic equations.
2. Integration or Calculus method for derivation of Kinematic equations.

## Analytical method for Kinematic Equations

Equations for linear motion and circular motion, both are identical. To determine these equations, first know about the following points –

For Linear motion –

Consider about an object moving with uniform acceleration ( \alpha ) along positive direction of X axis as shown in figure.

In the figure, let –

• x_0 is the position of particle at instant ( t = 0 )
• x is the position of particle at instant ( t = t )
• v_0 is the initial velocity of particle at instant ( t = 0 )
• v is the final velocity of particle at instant ( t = t )
• a is the acceleration acting on the particle.
• S is the displacement in time ( t ) second.

For Circular motion –

Consider about an object moving with angular velocity ( \omega ) in anti-clockwise direction in a circular path of radius ( r ) as shown in figure.

In the figure, let –

• Initial velocity of particle is ( \omega_0 ) at time ( t = 0 ) .
• Final velocity of particle is ( \omega ) at time ( t = t ) .
• Total angular displacement is ( \theta ) in time ( t ) .
• Angular acceleration acting on the particle is ( \alpha ) .

Now applying Newton’s laws of motion, kinematic equations are derived in following ways –

### (1) Velocity after certain time

By definition, linear acceleration of a moving particle is given by –

a = \left ( \frac {\text {Change in velocity}}{\text {Time taken}} \right )

= \left ( \frac {v - v_0}{t - 0} \right )

Or, \quad a t = v - v_0

Or, \quad v = v_0 + a t ………… (1)

This equation gives the final velocity ( v ) of the object after time ( t ) when initial velocity ( v_0 ) is known.

In similar manner, for circular motion, this equation will be –

\omega = \omega_0 + \alpha t ……….. (1a)

### (2) Distance covered in certain time

Average velocity of the particle in time interval 0 to t is given by –

v_{av} = \left ( \frac {\text {Displacement}}{\text {Time}} \right ) = \left ( \frac {x - x_0}{t} \right )

Or, \quad ( x - x_0 ) = v_{av} \times t

Or, \quad S = v_{av} \times t

But, \quad v_{av} = \left ( \frac {v_0 + v}{2} \right )

Therefore, \quad S = \left ( \frac {v_0 + v}{2} \right ) \times t

Putting \left [ v = v_0 + a t \right ] from equation (1), we will get –

S = \left [ v_0 t + \left ( \frac {1}{2} \right ) \alpha t^2 \right ]

Therefore, \quad S = v_0 t + \left ( \frac {1}{2} \right ) \ a t^2 ………. (2)

In similar manner, for circular motion this equation will be –

\theta = \omega_0 t + \left ( \frac {1}{2} \right ) \alpha t^2 ……… (2a)

### (3) Velocity at a certain position

From equation (1), we have –

v = ( v_0 + a t )

Or, \quad ( v - v_0 ) = a t …………. (3)

Also, \quad \left [ S = v_{av} \times t \right ]

Or, \quad \left ( \frac {v + v_0}{2} \right ) \times t = S

Therefore, \quad ( v + v_0 ) = \left ( \frac {2S}{t} \right ) ………… (4)

Multiplying equations (3) with (4), we will get –

\left ( v + v_0 \right ) \left ( v - v_0 \right ) = \left ( \frac {2 S a t}{t} \right )

Or, \quad ( v^2 -  v^2_0 ) = 2 \alpha S

Therefore, \quad v^2 = ( v^2_0 + 2 a S ) ………. (5)

In similar manner, for circular motion this equation will be –

\omega^2 = ( \omega^2_0 + 2 \alpha \theta ) …….… (5a)

### (4) Distance covered in ( n_{th} ) second

Distance traveled in ( n_{th} ) second is obtained by the difference of distance traveled in ( n ) seconds and distance traveled in \left ( n - 1 \right ) seconds as shown in figure.

First, distance traveled in first ( n ) seconds is obtained from equation (2), will be –

S_n = v_0 n + \left ( \frac {1}{2} \right ) \ a n^2

And, distance traveled in first \left ( n - 1 \right ) seconds will be –

S_{( n - 1 )} = v_0 ( n - 1 ) + \left ( \frac {1}{2} \right ) \ a ( n - 1 )^2

Hence, distance traveled in ( n_{th} ) second will be –

S_{n_{th}} = S_n - S_{( n - 1 )}

Or, \quad S_{n_{th}} = \left [ v_0 n + \left ( \frac {1}{2} \right ) \ a n^2 \right ] - \left [ v_0 ( n - 1 ) + \left ( \frac {1}{2} \right ) \ a ( n - 1 )^2 \right ]

By simplification, we get –

S_{n_{th}} = v_0 n + \left ( \frac {a}{2} \right ) \left ( 2n - 1 \right ) ……… (6)

In similar manner, for circular motion this equation will be –

\theta_{n_{th}} = \omega_0 + \left ( \frac {\alpha}{2} \right ) \left ( 2n - 1 \right ) ………. (6a)

## Integration or Calculus method for Kinematic Equations

Kinematic equations can be evaluated by calculus method as follows –

### (1) Velocity after certain time

By definition of linear acceleration, we have –

a = \left ( \frac {dv}{dt} \right )

Or, \quad dv = a dt …….. (1)

Let, when time ( t = 0 ) , initial velocity is ( u )

And when time ( t = t ) , final velocity is ( v )

Integrating equation (1) within the above limits of time and velocity, we will get –

\int\limits_{u}^{v} dv = \int\limits_{0}^{t} a dt

= a \int\limits_{0}^{t} dt

So, \quad \left [ v \right ]_{u}^{v} = a \left [ t \right ]_{0}^{t}

Or, \quad ( v - u ) = a \left ( t - 0 \right )

Therefore, \quad v = ( u + a t ) ………… (2)

### (2) Distance covered in a certain time

By definition of Velocity

\quad v = \left ( \frac {dS}{dt} \right )

Or, \quad dS = v dt = ( u + a t ) dt ……. (3)

When time ( t = 0 ) , distance traveled is ( 0 )

When time ( t = t ) , distance traveled is ( S )

Integrating equation (3) within above limits of time and distance, we get –

\int\limits_{0}^{S} dS = \int\limits_{0}^{t} \left ( u + a t \right ) dt

Or, \quad \int\limits_{0}^{S} dS = u \int\limits_{0}^{t} + a \int\limits_{0}^{t} t dt

Or, \quad \left [ S \right ]_{0}^{S} = u \left [ t \right ]_{0}^{t} + a \left [ \frac {t^2}{2} \right ]_{0}^{t}

Or, \quad ( S - 0 ) = u ( t - 0 ) + a \left [ \frac {t^2}{2} - 0 \right ]

Therefore, \quad S = \left [ u t + \left ( \frac {1}{2} \right ) a t^2 \right ] ……. (4)

### (3) Velocity at a certain position

By definition of acceleration and velocity, we have –

a = \left ( \frac {dv}{dt} \right )

= \left ( \frac {dv}{dS} \right ) \times \left ( \frac {dS}{dt} \right )

= \left ( \frac {dv}{dS} \right ) v

Or, \quad a \ dS = v \ dv ………. (5)

When time ( t = 0 ) , initial velocity is ( u ) and distance traveled is ( 0 )

When time ( t = t ) , final velocity is ( v ) and distance traveled is ( S )

Integrating equation (5) within above limits of velocity and distance, we will get –

\int\limits_{0}^{S} \alpha dS = \int\limits_{u}^{v} v dv

Or, \quad a \int\limits_{0}^{S} dS = \int\limits_{u}^{v} v dv

Or, \quad a [ S ]_{0}^{S} = \left [ \frac {v^2}{2} \right ]_{u}^{v}

Or, \quad a \left [ S - 0 \right ] = \left ( \frac {v^2}{2} - \frac {u^2}{2} \right )

Or, \quad 2 a S = ( v^2 - u^2 )

Therefore, \quad v^2 = ( u^2 + 2 a S ) …..…… (6)

### (4) Distance covered in  ( n_{th} )  second

By definition of velocity, we have –

v = \left ( \frac {dS}{dt} \right )

Or, \quad dS = v dt = ( u + a t ) dt …….. (7)

When time [ t = ( n - 1 ) ] , distance traveled is [ S_{(n - 1)} ]

When time ( t = n ) , distance traveled is ( S_{n} )

Integrating equation (7) within the limits of time and distance, we get –

\int\limits_{S_{(n-1)}}^{S_n} dS = \int\limits_{n}^{(n - 1)} ( u + a t ) dt

Or, \quad [ S ]_{S_{(n-1)}}^{S_n} = u [ t ]_{n}^{(n - 1)} + a \left [ \frac {t^2}{2} \right ]_{n}^{(n - 1)}

Or, \quad S_n - S_{(n - 1)} = u [ n - ( n - 1 ) ] + \frac {a}{2} [ n^2 - ( n - 1 )^2 ]

Or, \quad S_n - S_{(n - 1)} = u + \frac {a}{2} [ n^2 - ( n^2 - 2n + 1 ) ]

Therefore, \quad S_{n_{th}} = u + \frac {a}{2} ( 2n - 1 ) …….. (8)