Kinematics

Kinematics is the branch of physics that deals with the study of motion without considering the forces causing it. It focuses on describing the motion of objects through key concepts such as displacement, velocity, acceleration, and time. Kinematics helps us understand various types of motion, including linear, rotational, and projectile motion. It also explores concepts like position vectors, relative velocity, and motion under gravity. By analyzing these principles, kinematics provides a foundation for more complex studies in mechanics and physics.

1.0Basics of Kinematics

Rest: If position does not change with time, then it is at rest.

Motion: If the position of a particle changes with time, then it is called in motion.

Frame of Reference: A frame of reference is a perspective used to define the position or motion of a particle. Motion and rest are relative terms, dependent on the chosen frame of reference. Unless specified otherwise, the ground is typically considered the default reference frame.

Note: If a particle's position changes over time, it is in motion; if its position remains constant, it is at rest.

Position Vector: A position vector specifies the location of a particle relative to a reference point, such as the origin. It is a vector that indicates the position of a specific point with respect to this reference. The direction of the position vector always points from the reference point (origin) toward the given point.

2.0Terminology Associated With Kinematics

Displacement in Vector Form

Initial position vector $(\overrightarrow{OA})$: $\overrightarrow{r_1}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$

Final position vector $(\overrightarrow{OB})$: $\overrightarrow{r_2}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}$

Displacement vector: $\Delta \vec{r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

Useful Direction Conventions

Sense of Direction in terms of base vectors

3.0Speed

• The rate at which distance is covered with respect to time is called speed.
• Scalar Quantity
• Unit: $SI\to m/s$
• CGS Unit$\to cm/s$
• Conversion $km/h\to \frac{5}{18}m/s$
• Dimension:$[M^0{L}^1{T}^{-1}]$
• For a moving particle speed can never be zero or negative, it is always positive.

Types of Speed

Uniform Speed
Non-uniform Speed (Variable Speed)

Cases of Average Speed

Case1. Particle moves with different uniform speeds in different time intervals its average speed over the total time of journey is given as

$V_{avg}=\frac{v_1+v_2+..........+v_n}{n}$

Case2. For Two equal intervals of time

$V_{avg}=\frac{v_1+v_2}{2}$

Case3. Particle describes equal distances with different speeds then the average speed of particle over the total distance will be given as 

$S=\frac{d}{n}$

$V_{avg}=\frac{2v_1{v}_2}{v_1+v_2}$

4.0Velocity

• The rate of change of position with time is called velocity.
• It is a Vector Quantity.
• Dimension : $[M^0{L}^1{T}^{-1}]$
• Unit: $SI\to m/s$
• CGS Unit$\to cm/s$
• Velocity can be positive, negative or Zero.

Types of Velocity

Note:

1. Time Average Velocity: If $v=f(t)\Rightarrow <v>=\frac{{\int }_{t_1}^{t_2}vdt}{{\int }_{t_1}^{t_2}dt}$
2. Space Average Velocity If $v=f(x)\Rightarrow <v>$$=\frac{{\int }_{x_1}^{x_2}vdt}{{\int }_{x_1}^{x_2}dt}$

Instantaneous Velocity: It is the velocity of a particle at a specific moment in time.

$\overrightarrow{v_{avg}}=\frac{\overrightarrow{\Delta r}}{\Delta t}$

${\vec{v}}_{inst}=\underset{\Delta t\to 0}{\lim }(\frac{\overrightarrow{\Delta r}}{\Delta t})\to {\vec{v}}_{inst}=\frac{\overrightarrow{dr}}{dt}$

Change in velocity

$\Delta \vec{v}=\overrightarrow{v_f}-\overrightarrow{v_i}$

If $|\overrightarrow{v_f}|=|\overrightarrow{v_i}|=v$, then

$|\overrightarrow{\Delta v}|=2vsin(\frac{\theta }{2})$

where $\theta$ is the angle between initial and final velocity

5.0Acceleration

• The rate of change of velocity is called acceleration.
• It is a vector quantity.
• Its direction is the same as that of change in velocity (not in the direction of the velocity).
• Dimension : $[M^0{L}^1{T}^{-2}]$
• Unit: $SI\to m/s^2$
• CGS Unit$\to cm/s^2$

There are 3 ways to change a velocity (vector)

• Only magnitude change  
• Only direction change 
•  Both direction+ magnitude change

Types of Acceleration

Instantaneous Acceleration

Acceleration of a particle at a given instant is called instantaneous acceleration.

${\vec{a}}_{\text{inst}}={\lim }_{\Delta t\to 0}\left(\frac{\Delta \vec{v}}{\Delta t}\right)\Rightarrow {\vec{a}}_{\text{inst}}=\left(\frac{d\vec{v}}{dt}\right)\text{ when }\vec{v}\text{ is function of time }(t)$

first derivative of velocity is called instantaneous acceleration

${\vec{a}}_{\text{inst}}=\frac{d\vec{v}}{dt}=\frac{d^2\vec{r}}{d{t}^2}$

second derivative of position vector is called instantaneous acceleration

${\vec{a}}_{\text{inst}}=\frac{d\vec{v}}{dt}\times \frac{dx}{dx}\Rightarrow {\vec{a}}_{\text{inst}}=v\frac{dv}{dx}\text{ when }\vec{v}\text{ is function of position }(x)$

Application of Calculus

For v-t relation$\Rightarrow a=\frac{dv}{dt}$

For v-x relation $\Rightarrow a=v\frac{dv}{dx}$

Key Points

• When a particle moves with constant velocity then its acceleration will be zero.

If particle moves with constant velocity then $v_{avg}=|\overrightarrow{v_{avg}}|=v_{inst}=|\overrightarrow{v_{inst}}|$

• Acceleration which opposes the motion of the body is called retardation.

Increment or decrement in speed depends on direction of $\vec{v}$ and $\vec{a}$

Case-1. If $\vec{v}and\vec{a}$ are in the opposite direction then speed decreases.

Case-2. If $\vec{v}and\vec{a}$ are in the same direction then speed increases.

6.0Velocity and Acceleration

Case-1. Angle between $\vec{v}and\vec{a}$ is zero or they are parallel, speed increases.

Case-2. Angle between $\vec{v}and\vec{a}$ is 90° or they are orthogonal, speed remains constant.

Case-3. Angle between $\vec{v}and\vec{a}$ is 180° or they are anti-parallel, speed decreases

7.0Equations of Motion

Note:

1. Displacement in $n^{th}$ second $S_{nth}=u+\frac{a}{2}(2n-1)$
2. $s=V_{av}t=\frac{(u+v)}{2}t$

Galileo’s law of odd numbers: When particle starts from rest and moves with constant acceleration then ratio of distance travelled by it in successive equal intervals of time is

$1:3:5:\mathrm{7........}(2n–1)$

Stopping Distance: The distance a vehicle travels after braking before it stops is called the stopping distance.

$s=\frac{u_0^2}{2a_0}$ [since a is constant] $s\propto u_0^2$

if u becomes n times then s becomes n2 times that of previous value.

Stopping Time

$t=\frac{u_0}{a_0}$ [since a is constant]$\Rightarrow t\propto u_0$

if u becomes n times then t becomes n times that of previous value.

Reaction Time: When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act.

8.0Basic Graphs and Their Conversions

1. Position-Time graph

$Tan\theta =\frac{displacement}{time}=velocity$

Case-1 $\theta =0°$ $Tan\theta =0°$ $velocity=0$ The body is at rest.
Case-2 $\theta =constant$ $Tan\theta =constant$ $velocity=constant$  the body is in uniform motion
Case-3 $\theta >90°$ $Tan\theta =negative$ velocity=-ve but constant uniform motion
Case-4 $\theta$ is decreasing with time  velocity is decreasing with time
Case-5 $\theta$ is increasing with time velocity is increasing with time

$\text{Area of x-t graph}=\int xdt=\text{No Physical Significance}$

2. Velocity-time graph

The slope of this graph represents acceleration.

$Tan\theta =\frac{velocity}{time}=acceleration$

Area under v-t curve

$|displacement|=A_1-A_2$

$Distance=A_1+A_2$

Case-1 $\theta =0°$ $Tan\theta =0°$ $acceleration=0$ $v=constant$                  uniform motion
Case-2 $\theta =constant$ $Tan\theta =constant$ $acceleration=constant$ uniformly accelerated motion
Case-3 $\theta >90°$ $Tan\theta =negative$ $acceleration=-vebutconstant$ uniform motion
Case-4 $\theta$ is decreasing with time tan is decreasing with time acceleration is decreasing with time acceleration goes on decreasing with time but it is not retardation
Case-5 $\theta$ is increasing with time  acceleration is increasing with time

3. Acceleration-time graph

Area of a-t graph=$\int adt=\int dv=v_2-v_1$=change in velocity

Uniform or constant acceleration	Uniformly increasing acceleration.
$a\propto t^0$

Key Points

Following graphs do not exist in practice :

Case-1.

Explanation:

In practice, at any instant the body can not have two velocities or displacements or accelerations simultaneously.

Case-2.

Explanation:

Speed or distance can never be negative.

Case-3.

Explanation :

 It is not possible to change any quantity without consuming time i.e. time can't be constant.

Graphical Problems in Horizontal Motion

If a car, starts from rest moves with constant acceleration for some time retards uniformly at rate and finally comes at rest total time of motion is T

$v_{max}=(\frac{\alpha \beta }{\alpha +\beta })T$

$S=\frac{1}{2}(\frac{\alpha \beta }{\alpha +\beta })T^2$

9.0Motion Under Gravity

The acceleration a body experiences due to gravity is called acceleration due to gravity, denoted by g.

$g=9.8m/s^2$           or          $g=980cm/s^2$   or    $g=32ft/s^2$

Sign Conventions

Note: Negative and positive signs are matters of our choice, so we can select any direction as positive and opposite side as negative.

1. Vertical Projection from ground

Equations of Motion:

$u(initialvelocity)=+u$

$a(acceleration)=-g$

at Time=t

Final Velocity=v

height above ground=h

$v=u-gt$

$h=ut-\frac{1}{2}g{t}^2$

$v^2=u^2-2gh$

$h_{n^{th}}=u-\frac{g}{2}(2n-1)$

Maximum Height (H)

$H=\frac{u^2}{2g}$

Total Time of Flight(T)

Time of Ascent $(t_1)=\frac{u}{g}$

Time of Descent$(t_2)=\frac{u}{g}$

Total Time$T=t_1+t_2=\frac{2u}{g}$

2. Motion under Gravity-Dropping from height

Body dropped from some height (initial velocity zero)

Equations of Motion:

u=0,a=+g, time=t, final velocity=v, height below dropping point=h

1. $v=gt$
2. $h=\frac{1}{2}g{t}^2$
3. $v^2=2gh$
4. Time to Reach Bottom $T=\sqrt{\frac{2H}{g}}$
5. Maximum speed at bottom $v=\sqrt{2gH}$

Vertical Projection from Height

Case-1 : Downward Projection	Case- 2: Upward Projection
$v=u+gt$ $h=ut+\frac{1}{2}g{t}^2$ $v^2=u^2+2gh$ $h_n=u+\frac{g}{2}(2n-1)$ $v_{\text{max}}=\sqrt{u^2+2gh}$	$v=-u+gt$ $h=-ut+\frac{1}{2}g{t}^2$ $v^2=u^2+2gh$ $h_n=-u+\frac{g}{2}(2n-1)$ $H_{\text{max}}=H+\frac{u^2}{2g}$ $\text{Distance travelled}=H+\frac{u^2}{g}$

Effect of Air Resistance

If ‘a’ is retardation due to air resistance,

For ascending motion(upward motion) Time to go up, $t_1=\sqrt{\frac{2h}{g+a}}$ Velocity of projection, $v_1=\sqrt{2(g+a)h}$

For descending motion (downward motion) Time to go up, $t_1=\sqrt{\frac{2h}{g-a}}$ Velocity on reaching ground, $v_1=\sqrt{2(g-a)h}$

Graphical Problems in Vertical Motion

1. A body is projected vertically upwards then

Graphs of displacement with respect to time
Graph of velocity with respect to time
Graph of acceleration with respect to time

2. A body dropped from some height

Graphs of displacement with respect to time
Graph of velocity with respect to time
Graph of acceleration with respect to time

10.0Motion In A Plane

2D Motion

• When the motion of an object is restricted within a plane, it is said to undergo a motion in 2D.
• 2D motion can be studied as two independent 1D motions. (One along the x-axis and the other along the y-axis).
• Example: Motion of a carrom coin, Projectile Motion, Circular Motion.

Projectile Motion: When a body moves with constant acceleration such that its initial velocity and acceleration are non- collinear then its path is parabola and motion is known as projectile motion.

Projectile Motion = Horizontal Motion + Vertical Motion

Standard Results of Ground to Ground Projectile Motion

Note: Kinetic energy at highest point, $K=\frac{1}{2}mu(cos\theta {)}^2=K_{o}co{s}^2\theta$

Horizontal Projection from Height

Oblique Projection from Height

Case 1: Projection from a height at an angle θ above horizontal	Case 2: Projection from a height at an angle θ Below horizontal
Velocity after falling height h $v_y^2=(-usin\theta {)}^2+2hg(verticaldirection)$ $v_x=u_x=ucos\theta$ $v=\sqrt{v_x^2+v_y^2}=\sqrt{u^2+2gh}$	Velocity after falling height h $u_y^2=(usin\theta {)}^2+2hg(verticaldirection)$ $v_x=u_x=ucos\theta$ $v=\sqrt{v_x^2+v_y^2}=\sqrt{u^2+2gh}$

11.0Relative Motion

Relative Displacement

Displacement of B with respect to A = Displacement of B

as measured from

$\Rightarrow {\vec{x}}_{BA}={\vec{x}}_B-{\vec{x}}_A=\frac{d{\vec{x}}_{BA}}{dt}=\frac{d{\vec{x}}_B}{dt}-\frac{d{\vec{x}}_A}{dt}\Rightarrow {\vec{v}}_{BA}={\vec{v}}_B-{\vec{v}}_A$

Relative = Actual – Reference

For same direction	For opposite directions
When two particles move in the same direction, their relative velocity equals the difference in their speeds. $|\overrightarrow{v_{12}}|or|\overrightarrow{v_{21}}|=v_1-v_2$	When two particles move in opposite directions, their relative velocity is the total of their speeds. $|\overrightarrow{v_{12}}|or|\overrightarrow{v_{21}}|=v_1-v_2$

Equations of Motion (Relative)

• $v_{\text{rel}}=u_{\text{rel}}+a_{\text{rel}}t$
• $s_{\text{rel}}=u_{\text{rel}}t+\frac{1}{2}{a}_{\text{rel}}{t}^2$
• $v_{\text{rel}}^2=u_{\text{rel}}^2+2a_{\text{rel}}{s}_{\text{rel}}$
• $s_{\text{rel}}=\frac{1}{2}\left(u_{\text{rel}}+v_{\text{rel}}\right)t$

Relative velocity in a plane

Relative velocity of A with respect to 'B' can be calculated as

${\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B$

$|{\vec{v}}_{AB}|=\sqrt{v_A^2+v_B^2+2v_A{v}_B\cos \theta }$

12.0Rain-Man Concept

Case-1. If rain is dropping vertically with a velocity $\overrightarrow{v_R}$ and an observer is gliding horizontally with speed $\overrightarrow{v_M}$ then the velocity of rain relative to observer will be

${\vec{v}}_{RM}={\vec{v}}_R-{\vec{v}}_M\Rightarrow {\vec{v}}_{RM}=-v_R\hat{j}-v_M\hat{i}$

$V_{RM}=\sqrt{v_R^2+v_M^2}$

The direction of ${\vec{v}}_{RM}$ is such that it makes an angle with the vertical given by

$\theta ={\tan }^{-1}\left(\frac{v_M}{v_R}\right)$

Case-2. If rain is already falling at an angle with the vertical with a velocity $\overrightarrow{v_R}$ and an observer is gliding horizontally with speed $\overrightarrow{v_M}$ finds that the rain drops are hitting on his head vertically downwards

${\vec{v}}_{RM}={\vec{v}}_R-{\vec{v}}_M$

${\vec{v}}_{RM}=\left(v_R\sin \theta -v_M\right)\hat{i}-v_R\cos \theta \hat{j}$

Now for rain to appear falling vertically, the horizontal component of ${\vec{v}}_{RM}$ should be zero,

$\left(v_R\sin \theta -v_M\right)=0\Rightarrow \sin \theta =\frac{v_M}{v_R}$

$V_{RM}=\sqrt{V_R^2-V_M^2}$

13.0River-Boat (or Man) Problem

• A man can swim with velocity $\vec{v}$ , i.e. it is the velocity of man w.r.t. still water.
• If water is also flowing with velocity $\overrightarrow{v_R}$ then velocity of man relative to ground $\overrightarrow{v_m}=\vec{v}+\overrightarrow{v_R}$

If the swimming is in the direction of flow of water or along the downstream then	$v_m=v+v_R$
If the swimming is in the direction opposite to the flow of water or along the upstream then	$v_m=v-v_R$
If the man is crossing the river i.e. $\vec{v}$ and $\overrightarrow{V_R}$ are non collinear then use vector algebra.	$\overrightarrow{v_m}=\vec{v}+\overrightarrow{v_R}$

1. For shortest path

To reach at $Bvsin\theta =v_R\Rightarrow Sin\theta =\frac{v_R}{v}$

so, time taken $T=\frac{d}{vcos\theta }=\frac{d}{\sqrt{v^2-v_R^2}}$

2. For minimum time

To cross the river in minimum time, the velocity along $AB(vcos\theta )$ should be maximum

Minimum Time $=t=\frac{d}{v}$

$BC=t_{min}{v}_R$

Distance travelled along river flow = drift of man= $t_{min}{v}_R=\frac{d}{v}{v}_R$

14.0Sample Questions on Kinematics

Q-1.A river is 2 km broad and flows at 2 km/h. A boatman can row a boat at a speed of 6 km/h in still water. He rows 2 km upstream and then returns. What is the total time taken by the boatman to complete his journey?

Solution:

The velocity upstream is (4-2)km/hr and downstream is (4+2)km/hr

Total time taken=$\frac{2}{4}+\frac{2}{8}=\frac{6}{8}$=34hr=45 minutes

Q-2.A man is going east in a car with a velocity of 20 km/hr, a train appears to move towards north to him with a velocity of $20\sqrt{3}$ km/hr. What is the actual velocity and direction of motion of the train?

Solution:

${\vec{v}}_{TC}={\vec{v}}_T-{\vec{v}}_c$

${\vec{v}}_T={\vec{v}}_{TC}+{\vec{v}}_c=20\sqrt{3}\hat{j}+20\hat{i}$

${\vec{v}}_T=\sqrt{(20\sqrt{3}{)}^2+(20{)}^2}=\sqrt{1600}=40kilometer/hour$

$\tan \theta =\frac{20\sqrt{3}}{20}\Rightarrow \theta =60^{\circ }$

So, direction of motion of train is 60° N of E or E-60°-N

Q-3.Three boys A, B and C are situated at the vertices of an equilateral triangle of side d at t = 0. Each of the boys move with constant speed v. A always moves towards B, B towards C and C towards A. When and where will they meet each other?

Solution:

Due to symmetry, they will meet at the centroid of the triangle. Approaching velocity

of A and B towards each other is v + v cos 60° and they cover distance d when they meet. 

So that time taken, is given by $t=\frac{d}{v+vcos60°}=\frac{d}{v+\frac{v}{2}}=\frac{2d}{3v}$

Q-4.Two particles A & B are projected from a building. A is projected with speed 2v making an angle 30° with horizontal & B with speed v making an angle 60° with horizontal. Which particle will hit the ground earlier?

Solution:

$$\begin{gathered} (v_y{)}_B=v\sin 60^{\circ }=v\frac{\sqrt{3}}{2} \\ (v_y{)}_A=v\sin 30^{\circ }=v \\ (v_y{)}_A>(v_y{)}_B\Rightarrow T_A>T_B \end{gathered}$$

Q-5.A ball is thrown from the ground to clear a wall that is 3 meters high and located 6 meters away. The ball lands 18 meters from the wall. What is the angle of projection of the ball?

Solution:

$y=x\tan \theta \left[1-\frac{x}{R}\right]$$\Rightarrow 3=6\tan \theta \left[1-\frac{1}{4}\right]\Rightarrow \tan \theta =\frac{2}{3}$

$\theta ={\tan }^{-1}\left(\frac{2}{3}\right)$