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  • Last Modified 28-01-2023

Kinematics Formulae: Definition and Equations

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Kinematics Formulae: In physics, we have different formulae derived to make the calculation and analysis easier and faster. One such important topic is kinematics. In this topic, we have numerous formulas and concepts. Example: a uniform motion, circular motion, relative motion, projectile motion, projectile on an inclined plane.

Let us read further to get a brief idea about these topics and the formulae used in these topics.

What is Kinematics?

Kinematics is a branch of mechanics that deals with the study of the motion of a body without analysing the cause of the motion. For example, We mention that the body is moving with acceleration, but we don’t bother to explain the cause of the acceleration, the amount of force experienced by it, and we calculate different parameters such as displacement, velocity acceleration etc.

Learn about Motion here

Position: It is the location of the particle with respect to the chosen reference point. It is denoted by \(\overrightarrow r \)

Displacement: It is the change in position of the particle with respect to the reference point. We calculated it by taking the difference between the final and the initial position. It is denoted by \(\overrightarrow S \).

Distance: It is the total length of the path taken by the particle.

Velocity: The rate of change of displacement with respect to time is known as velocity. It is denoted by \(\overrightarrow v \) or \(\overrightarrow u \)

Acceleration: The rate of change of velocity with respect to time is known as acceleration. It is denoted by \(\overrightarrow a \).

Negative acceleration is also sometimes referred to as retardation.

First Equation of Motion

The First Equation of Motion

The first equation of motion gives us the relation between the initial velocity, final velocity, acceleration and time.
From the definition of the acceleration, we have,
The rate of change of the velocity with respect to time is known as acceleration.

\(\overrightarrow a  = \frac{{d\overrightarrow v }}{{dt}}\)

\(\overrightarrow a dt = d\overrightarrow v \)

Suppose a body started with some initial velocity \(\overrightarrow u \) and has a uniform acceleration \(\overrightarrow a \) and due to this acceleration, the final velocity after time \(t\) becomes \(\overrightarrow v \).
Thus integrating and putting in the limits we get,
\(\int\limits_0^t {\overrightarrow a dt = } \int\limits_{\overrightarrow u }^{\overrightarrow v } {d\overrightarrow v } \)
Solving we get the first equation of motion,
\(\overrightarrow a t = \overrightarrow v  – \overrightarrow u \)
\( \Rightarrow \,\,\overrightarrow v  = \overrightarrow u  + \overrightarrow a t\)

Second Equation of Motion

The second equation of motion gives the relation between the displacement, initial velocity, acceleration and the time interval.
From the definition of the velocity we have,
The rate of change of the displacement with respect to time is known as velocity,
\(\overrightarrow v  = \frac{{d\overrightarrow s }}{{dt}}\)
From the first equation of motion, we have the velocity at any time \(t\) is given by,

\(\overrightarrow v  = \overrightarrow u  + \overrightarrow a t\)
Substituting the above equation we get,
\(\overrightarrow u  + \overrightarrow a t = \frac{{d\overrightarrow s }}{{dt}}\)
\( \Rightarrow \,\,\left( {\overrightarrow u  + a\overrightarrow t } \right)dt = d\overrightarrow s \)

Integrating the above equation with proper limits that the body starts from \(0\) and takes time \(t\) to cover the distance \(\overrightarrow s \).

\(\int\limits_0^{\overrightarrow s } {d\overrightarrow s }  = \int\limits_0^t {\left( {\overrightarrow u  + a\overrightarrow t } \right)dt} \)

Solving and putting the proper limits we have,

\(\overrightarrow s  = \overrightarrow u t + \frac{1}{2}\overrightarrow a {t^2}\)

Third Equation of Motion

The third equation of motion gives us the relation between the final and the initial velocities, acceleration and displacement.
It is important to note that the third equation of motion doesn’t involve time.
So, From the definition of the acceleration, we have, The rate of change of the velocity with respect to time is known as acceleration.

\(\overrightarrow a  = \frac{{d\overrightarrow v }}{{dt}}\)

Now we take dot product on both sides with \(d\overrightarrow s \),

\( \Rightarrow \,\,\overrightarrow a  \cdot d\overrightarrow s  = d\overrightarrow s  \cdot \frac{{d\overrightarrow v }}{{dt}} = \frac{{d\overrightarrow s }}{{dt}} \cdot d\overrightarrow v \)

We know that,

\(\overrightarrow v  = \frac{{d\overrightarrow s }}{{dt}}\)

Thus,

\(\Rightarrow \overrightarrow a  \cdot d\overrightarrow s  = \overrightarrow v  \cdot d\overrightarrow v \)

Now integrating with proper limits, that is, the particle starts from zero and displaces by \(\overrightarrow s \) meanwhile, the velocity changes from \(\overrightarrow u \)to \(\overrightarrow v \).

\( \Rightarrow \int\limits_0^{\overrightarrow s } {\overrightarrow a  \cdot d\overrightarrow s }  = \int\limits_0^{\overrightarrow v } {\overrightarrow v  \cdot d\overrightarrow v } \)

\(\overrightarrow v  \cdot \overrightarrow v  – \overrightarrow u  \cdot \overrightarrow u  = 2\overrightarrow a  \cdot \overrightarrow s \)

\({v^2} – {u^2} = 2\overrightarrow a  \cdot \overrightarrow s \)

Relative Motion

We always have to define the position of any particle or object with respect to some reference point to measure any change or variation in it. For this purpose, we generally use a stationary reference point, but how does the parameter change if we have a non-stationary frame of reference. Let us visualise with an example.

Suppose two cars are moving in the same direction with the same velocities, then the difference between the cars remain constant; that is, if an observer is in the car, it will not see any change in the distance between the cars; therefore, the observer will observe the second car to be stationary, this gives us the idea that any non-stationary observer will observe all the objects moving with the same velocity as the observer with respect to a stationary frame will appear to be at rest. Thus, the observer will always observe itself to be at rest.

Also, if the second car has the velocity towards the observer, it will appear to move towards the observer with double the speed which it had with respect to the stationary frame, and if the second car is moving in the same direction with speed slightly greater than the observer, then the observer will observe the car to move with a slower speed. Therefore, we can conclude that the relative velocity of a body is given by subtracting the observer’s velocity from the velocity of the body with respect to the stationary frame.
\({\overrightarrow v _{{\rm{rel}}}} = {\overrightarrow v _1} – {\overrightarrow v _{{\rm{obs}}}}\)

relative motion

Projectile Motion

Projectile motion is a type of two-dimensional motion in which the particle is thrown or projected at some angle with the horizontal. We can analyse it by breaking it into a combination of two one dimensional motions. That is, along the horizontal and along the vertical. It should be noted that acceleration (gravity) only exist in the vertical direction if we chose the x-axis along the horizontal and the y-axis along the vertical.

Projectile Motion

The velocity of projection: It is the velocity with which the body is projected. It is denoted by \(u\).
The angle of projection: It is the angle with the horizontal at which the body is projected. It is denoted by \(\theta \).
Range: It is the maximum distance covered by the projectile in the horizontal direction or x-axis.
The range of a projectile is given by,
\(R = \frac{{{u^2}\sin \left( {2\theta } \right)}}{g}\)
For maximum range, the angle of projection should be \(45^\circ \).
Maximum height: It is the maximum height reached by the projectile.
The maximum height of the projectile is given by,
\(H = \frac{{{u^2}{{\sin }^2}\left( \theta  \right)}}{{2g}}\)
Time of flight:  It is the total time for which the ball remains in the air.
\(T = \frac{{2u\sin \left( \theta  \right)}}{\theta }\)
Equation of path of the projectile is given by,
\(y = x\tan \left( \theta  \right) – \frac{{g{x^2}}}{{2{v^2}\cos \left( \theta  \right)}}\)
From the above equation, we have the path of the projectile follows a parabolic trajectory.

Projectile on an Incline

Projectile on an Incline

Circular Motion

For circular motion, we have a different set of angular variables.

(i) Angular displacement:  It is the angle displaced by the body. It is denoted by \(\theta \).
(ii) Angular velocity: It is the rate of change of angular displacement with respect to time. It is denoted by \(\omega \).
Angular acceleration: It is the rate of change of angular velocity with respect to time. It is denoted by \(\alpha \).

Circular Motion

The equation of motion is the same for the angular variables, and just linear parameters are changed with the corresponding angular parameters.

\(\overrightarrow s  \to \overrightarrow \theta  \)
\(\overrightarrow u  \to {\overrightarrow \omega  _i}\)
\(\overrightarrow v  \to {\overrightarrow \omega  _f}\)
\(\overrightarrow a  \to \overrightarrow \alpha  \)
This gives us the equation of motion for angular variables to be.
The first equation of motion
\({\overrightarrow \omega  _f} = {\overrightarrow \omega  _i} + \overrightarrow \alpha  t\)
The second equation of motion
\(\overrightarrow \theta   = {\overrightarrow \omega  _i}t + \frac{1}{2}\overrightarrow \alpha  {t^2}\)
Third equation of motion
\(\overrightarrow \omega  _f^2 – \overrightarrow \omega  _i^2 = 2\overrightarrow \alpha  .\overrightarrow \theta  \)

It is important to note that the direction of the angular variable is perpendicular to the plane of motion given by the direction of the thumb of the right hand when fingers are curled according to the body’s motion.

Sample Problems

1. A motor car is going due north at a speed of \(50\,{\rm{km}}\,{{\rm{h}}^{ – 1}}\). It makes a \(90^\circ \) left turn without changing the speed. The change in the velocity of the car is
Let us assume direction conventions
\(\,\,\overrightarrow {{v_i}}  = 50\widehat j;\,\,{\overrightarrow v _f} =  – 50\widehat i\)
(As \(90^\circ \) west of north is west direction).
Change in velocity
\(\Delta \overrightarrow v = {\overrightarrow v _f} – {\overrightarrow v _i}\)
\(\Delta \overrightarrow v = \left( { – 50\widehat i} \right) – \left( {50\widehat j} \right)\)
\(\Delta \overrightarrow v = – 50\widehat i – 50\widehat i\)
\(\left| {\Delta \overrightarrow v } \right| = \sqrt {{{\left( {50} \right)}^2} + \left( {{{50}^2}} \right)}  = 50\sqrt 2  \approx 70\,{\rm{km}}\,{{\rm{h}}^{{\rm{ – 1}}}}\)
It is towards the southwest, as shown in the figure.

velocity of the car

2. A person travelling on a straight line moves with a uniform velocity \({v_1}\) for some time and with uniform velocity \({v_2}\) for the next equal time. The average velocity \(v\) is given by,
As velocity is uniform in both cases.
So, we can write that 
\({d_1} = {v_1}t\)
 where \({d_1} = \) Distance travelled in the first half 
and \({d_2} = {v_2}t\)
Where \({d_2} = \) Distance travelled in the second half.
Total displacement
\(d = {d_1} + {d_2}\)
The total time is given by,
\({t_{{\rm{net}}}} = t + t = 2t\)
\({\rm{Average}}\,{\rm{velocity = }}\frac{{{\rm{total}}\,{\rm{distance}}}}{{{\rm{total}}\,{\rm{time}}}}\)
Putting the above values,
\(v = \frac{{{d_1} + {d_2}}}{{2t}} = \frac{{{v_1} + {v_2}}}{2}\)

Learn about Projectile Motion here

Summary

Kinematics is a topic in which we study the motion of a body without worrying about the cause of the motion.
The first equation of motion for linear variables is given by,
\(\overrightarrow v  = \overrightarrow u  + \overrightarrow a t\)
The second equation of motion for a linear variable is given by,
\(\overrightarrow s  = \overrightarrow u t + \frac{1}{2}\overrightarrow a {t^2}\)
The third equation of motion for linear variables is given by,
\({\overrightarrow v ^2} – {\overrightarrow u ^2} = 2\overrightarrow a  \cdot \overrightarrow s \)
The first equation of motion for the angular variable is given by,
\({\overrightarrow \omega  _f} = {\overrightarrow \omega  _i} + \overrightarrow \alpha  t\)
The second equation of motion for the angular variables is given by,
\(\overrightarrow \theta   = {\overrightarrow \omega  _i}t + \frac{1}{2}\overrightarrow \alpha  {t^2}\)
The third equation of motion for the angular variable is given by,
\(\overrightarrow \omega  _f^2 – \overrightarrow \omega  _i^2 = 2\overrightarrow \alpha  .\overrightarrow \theta  \)

FAQs

Q.1. What is kinematics?
Ans:
Kinematics is the study of the parameters related to motion and their relation among themselves.

Q.2. What is the first equation of motion?
Ans:
The first equation of motion gives the relation between the initial velocity, final velocity, acceleration and the time interval.
The first equation of motion for the linear variable is given by,
\(\overrightarrow v  = \overrightarrow u  + \overrightarrow a  t\)
The first equation of motion for the angular variable is given by,
\({\overrightarrow \omega  _f} = {\overrightarrow \omega  _i} + \overrightarrow \alpha  t\)

Q.3. What is the second equation of motion?
Ans:
The second equation of motion gives the relation between the displacement, initial velocity, acceleration and the time interval.
The second equation of motion for the linear variable is given by,
\(\overrightarrow s  = \overrightarrow u t + \frac{1}{2}\overrightarrow a  {t^2}\)
The second equation of motion for the angular variable is given by,
\(\overrightarrow \theta   = {\overrightarrow \omega  _i}t + \frac{1}{2}\overrightarrow \alpha {t^2}\)

Q.4. What is the third equation of motion?
Ans:
The third equation of motion gives us the relation between the final and the initial velocities, acceleration and displacement. The third equation of motion for the linear variable is given by,
\({\overrightarrow v ^2} – {\overrightarrow u ^2} = 2\overrightarrow a  \cdot \overrightarrow s \)
The third equation of motion for the angular variable is given by,
\(\overrightarrow \omega  _f^2 – \overrightarrow \omega  _i^2 = 2\overrightarrow \alpha   \cdot \overrightarrow \theta  \)

Q.5. What is the expression for relative velocity?
Ans:
The relative velocity of a body is given by subtracting the observer’s velocity from the body’s velocity with respect to the stationary frame.
\({\overrightarrow v _{{\rm{rel}}}} = {\overrightarrow v _1} – {\overrightarrow v _{{\rm{obs}}}}\)

Practice Kinematics Questions with Hints & Solutions