Kinematics Formulas

The study of the motion of points, objects, and groupings of things without examining their causes is known as kinematics. Kinematics is a part of classical mechanics that deals with the motion of points, objects, and systems made up of several items. Kinematics is sometimes referred to as the "geometry of motion" by certain experts.

Kinematics Formulas

Kinematics is the branch of classical mechanics that describes the motion of points, systems of bodies (groups of objects), bodies (objects) without consideration of the causes of motion. The four kinematics formulae deal with displacement, velocity, time, and acceleration, respectively.

In addition, below are the four physics kinematics formulas are as follows:

1. $$A=\frac{v_f-v_i}{t}$$

2. $$\frac{v_i+v_f}{2}=\frac{D}{t}$$

3. $$D=v_{i}t+\frac{1}{2}a{t}^2$$

4. $$v_f^2=v_i^2+2Ad$$

Where, 

D = displacement

a = acceleration

t = time

$$v_f$$= final velocity

$$v_i$$= initial velocity

Deriving kinematic Equations

First Equation Derivation

$$A=\frac{v_f-v_i}{t}$$

1.) Before we begin deriving the first kinematic equation, we must first define acceleration. $$a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$$

2.) We know $$\mathrm{\Delta }v=v_f-v_i$$ and when we plug that in, we get $$a=\frac{v_f-v_i}{\mathrm{\Delta }t}$$

3.) If we solve for $$v_f$$ then the equation becomes a $$v_f=v_i+a\mathrm{\Delta }t$$

4.) We can denote the time interval (t) to generate the first kinematic equation.

$$A=\frac{v_f-v_i}{t}$$

Hence it is solved.

Second Equation Derivation

$$\frac{v_i-v_f}{2}=\frac{D}{t}$$

1.) When we sum the red and blue areas, we get $$D=t(\frac{v_i-v_j}{2})$$

2.) When we distribute the factor $$\frac{t}{2}$$ we get, $$D=v_{i}t+\frac{1}{2}{v}_{f}t-\frac{1}{2}{v}_{i}t$$

3.) By combining the initial velocity terms we can simply get the equation. $$D=v_{i}t+\frac{1}{2}{v}_{f}t+\frac{1}{2}{v}_{i}t$$

4.) Finally, we reduce the equation even more to obtain our second kinematics equation.

$$\frac{v_i+v_f}{2}=\frac{D}{t}$$

Third Equation Derivation

$$D=v_{i}t+\frac{1}{2}a{t}^2$$

By putting the first kinematic formula into the second formula, we may get the third kinematic equation.

1.) Start with the second kinematic equation. $$\frac{v_i+v_f}{2}=\frac{D}{t}$$ 

2.) Substitute the first kinematic formula for $$v_f$$

$$v_f=v_i+at$$

3.) Once substituted, the equation becomes $$\frac{v_i+(v_i+at)}{2}=\frac{D}{t}$$

4.) Expand the equation into $$\frac{D}{t}=\frac{v_i}{2}+\frac{v_i}{2}+\frac{at}{2}$$ 

5.) We can combine terms to the equation into $$\frac{D}{t}+v_i+\frac{at}{2}$$

6.) To obtain our third kinematic formula, we may multiply both sides by the time, equation.

$$D=v_{i}t+\frac{1}{2}a{t}^2$$

Fourth Equation Derivation

$$v_f^2=v_i^2+2AD$$

The first and second kinematics equations may be used to obtain the fourth kinematics formula.

1.) Let’s start with the second kinematic formula. $$\frac{v_i+v_f}{2}=\frac{D}{t}$$

2.) We can use the first kinematic formula, $$v_f=v_i+at$$ solving for time (t), we get, $$t=\frac{v_f-v_i}{a}$$

3.) We can plug our expression for a time into the second kinematic formula. $$D=(\frac{v_i+v_f}{2})(\frac{v_f-v_i}{a})$$

4.) Next, we can multiply the fractions to get the equation.$$D=(\frac{v_i^2+v_f^2}{2a})$$

Kinematic Viscosity Formula

When studying liquid behaviour and fluid motion near solid barriers, viscosity is an important property to consider. A fluid's viscosity is a measurement of its resistance to progressive deformation caused by shear or tensile stress. 

In addition, Kinematic viscosity is a type of viscosity calculated by dividing the fluid mass density by the dynamic fluid, viscosity, or absolute fluid viscosity. It's also known as momentum diffusivity from time to time. Kinematic viscosity is measured in terms of time and fluid area.

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$$\text{Kinematic Viscosity}=\frac{\text{Dynamic viscosity}}{\text{Fluid mass density}}$$

$$v=\frac{\mu }{\rho }$$

Where, 

v = kinematic viscosity (square m per second)

μ = absolute or dynamic viscosity ( unit is $$Ns{m}^{-2}$$)

ρ = density (unit is kg per cubic m)

In addition, the dimensional formula of kinematic viscosity is $$M{L}^2{T}^{-1}$$

Hence  the dimension of kinematic viscosity is $$L^2{T}^{-1}$$ and the unit is $$m^2{s}^{-1}$$

Kinematic Rotational Motion Formulas

The study of motion is known as kinematics.

The relationships between rotation angle, angular velocity, angular acceleration, and time are described by rotational motion kinematics. Let's begin by defining a relationship between ω, ⍺, and t. This equation may be found by considering a well-known kinematic equation for translational, or straight-line, kinetic equation of motion is, 

$$v=v_0+at$$

Where a is constant

Note that in rotational motion, a = at, and from here on, we'll refer to tangential or linear acceleration by the symbol a. We assume that an is constant, as in linear kinematics, which indicates that angular acceleration is also constant, because a = r⍺. Let us now put v = rω and a = r⍺  into the following linear equation:

$$v=v_0+at$$

$$r\omega =r{\omega }_0+r\alpha t$$

The radius r is cancelled in the equation, yielding

$$\omega ={\omega }_0+\alpha t$$

One-Dimensional Kinematic Formulas

One-dimensional motion can be described using formulas that relate velocity, displacement, and acceleration. The rate of change of displacement with respect to time is known as velocity. The rate of change of velocity (v) for time (t) is called acceleration. The acceleration is considered to be constant in these formulae. The metre (m), metres per second (m/s), and metres per second squared ($$m/s^2$$) are the units of displacement, velocity, and acceleration, respectively. The motion in one dimension formulas are as follows,

Velocity 

$$\text{final velocity}=\text{(initial velocity)}+\text{(acceleration)(time)}$$

$$v_f=v_i+at$$

Displacement

$$\text{final displacement}=\text{(initial displacement)}+\text{(initial velocity)(time)}+\frac{1}{2}\text{(acceleration)}$$

$$D_f=D_i+v_{i}t+\frac{1}{2}a{t}^2$$

$$v_f$$: Final velocity

$$v_i$$: Initial velocity

$$a$$: Acceleration

$$t$$: Time

$$D_f$$: Final displacement

$$D_i$$: Initial displacement

Solved Examples

Ex.1. A man is riding his bike to the store at a velocity of 8m/s. Then, suddenly a cat runs out in front of him. He rapidly breaks to a complete stop, with an acceleration of   $$-6m/s^2$$. Find out his displacement?

Solution:

Here, the final velocity, $$v_f=0$$. Also, the initial velocity, $$v_i=8m/s$$. Furthermore, the acceleration, $$a=-6m/s^2$$ . Therefore, one must use the following formula:

$$v_f^2=v_i^2+2Ad$$

$$0=(8{)}^2+2\times (-6)\times d$$

$$d=\frac{-64}{-12}$$

$$d=5.3334$$

Ex.2. A car with initial velocity zero experiences a uniform acceleration of $$14m/s^2$$. for the time interval $$t=12s$$ Calculate its distance covered?

Solution:

Given parameters

$$v_i=0m/s$$

$$t=5s$$

$$a=14m/s^2$$.

To find the Distance covered D.

By using the Kinematic Equation,

$$D=v_{i}t+\frac{1}{2}a{t}^2$$

$$D=0\times 5+\frac{1}{2}\times 14\times (5{)}^2$$

$$D=175m/s^2$$