Kinematics

Syllabus

The frame of reference, motion in a straight line, Position- time graph, speed and velocity; Uniform and non-uniform motion, average speed and instantaneous velocity, uniformly accelerated motion, velocity-time, position-time graph, relations for uniformly accelerated motion, Scalars and Vectors, Vector. Addition and subtraction, zero vector, scalar and vector products, Unit Vector, Resolution of a Vector. Relative Velocity, Motion in a plane, Projectile Motion, Uniform Circular Motion.

Kinematics

The study of motion of objects without taking into account the cause of the motion is known as kinematics. In kinematics, time plays an important role.
In kinematics, we do not consider the mass or size of the moving objects. Hence, in kinematics all moving bodies, irrespective of their geometric sizes, are treated as particles or point masses.
Point object or Point mass: If the position of an object changes by distances much greater than its size in a considerable interval of time, then the object can be regarded as a point object. Example, The car under a journey of several kilometers, the moon for its motion around the earth.
Rest: If the position of a particle does not change with respect to time and surroundings, then the state of the particle is known as the state of rest. Example: A book lying on the table, An electric pole standing on the street.
Note: No object in the universe is in the state of absolute rest. So, the state of rest is a relative term.
Motion: If the position of a particle changes with respect to time and surroundings, then the state of the particle is known as the state of motion. Example: A car moving on the road, the earth revolving around the sun, the wind blow in the environment etc.
Note: To determine the absolute motion of an object, a fixed reference point is needed, but no object in the universe is at absolute rest. So, the absolute motion cannot be realised. Hence, rest and motion are relative.

Frame of reference

A set up of three co-ordinate axes X, Y and Z with origin O to describe the position of an object is called frame of reference. The origin ‘O’ is the place where the observer sits and observes the position of an object and is also called ‘reference point.’

Description: To locate the position of a particle we need a frame of reference. In order to fix up a frame of reference, we choose three mutually perpendicular axes named as X-, Y-, and Z axes called the coordinate frame. The coordinate i.e., x, y and z of the particle then specify the position of the particle with respect to that frame. Add a clock into the frame of reference to measure the time.

(i) Rest: If all the three coordinates x, y and z of the particle remain unchanged as time passes, we say that the particle is at rest with respect to the frame of reference.
(ii)Motion: If any one or more coordinates change with time, we say that the body is moving with respect to this frame.
Here, in figure, the point P with the coordinates (x, y, z) represents the position of the particle.

Notes:
(i) There is no rule and restriction on the selection of a frame.
(ii) A frame of reference can be chosen according to our convenience to describe a particular situation under study.
(iii) The change in position of the particle or object with respect to time means the change in values of x, y and z with time.
(iv) The position of an object may change due to a change in one or two or all the three coordinates. So, we define the motion accordingly as one dimensional motion or two dimensional motion or three dimensional motion.

Types of Frame of Reference:
There are two types of frame of reference:
(i) Inertial frame of reference: A frame of reference which is either at rest or moving with constant velocity is known as inertial frame of reference.
(ii) Non-inertial frame of reference: A frame of reference moving with some acceleration is known as non-inertial frame of reference.
Notes:
(i) Newton’s law is valid only on Inertial frame of Reference.
(ii) In Non-inertial frame, a pseudo (false) force is applied
in the opposite direction of the motion of the frame to make
them inertial and then Newton’s law is valid on it.

Motion in different directions

(i) One dimensional motion: The motion of a particle is said to be in one dimensional if only one out of the three coordinates specifying the position of the particle changes with time. Example, The motion of a train along a straight railway track, motion of a body moving upward or downward, etc.
(ii) Two dimensional motion: The motion of a particle is said to be in two dimensional if any two out of the three coordinates specifying the position of the particle
change with time. Example, An athlete runs on a circular track, projectile motion, etc.
(iii) Three dimensional motion: The motion of a particle is said to be in three dimensional when all the three coordinates specifying the position of the particle change with time. In such a motion, the object moves in a space. Example, The motion of a bird or kite flying in space, a flying aero plane, the random motion of gas molecules, etc.
Trajectory: The path followed by a point object during its motion is called its trajectory. Shape of path depends upon closer reference frame.

Motion in a straight line

The motion of a point object in a straight line is one dimensional motion. During such a motion the point object occupies definite position on the path at each instant of time. Different terms used to described motion are defined below:

Distance and Displacement

Distance: It is defined as the length of actual path travelled by a body during its motion in a given interval of time. It is denoted by ‘d’or ‘s’.
(i) Distance is a scalar quantity i.e., it does not take direction into consideration.
(ii) Distance between a given set of initial and final positions can have infinite values.

(iii) Distance covered is always a positive quantity.
(iv) For a moving particle, the distance covered can never be zero.
(v) Distance is never negative.
(vi) The distance covered by a particle never decreases with time, it always increases.
(vii) The unit of distance is the unit of length i.e., cm, m, mm, km, nm etc. Its SI unit is metre (m).
Example: If a person runs in a triangular track from initial point ‘O’ and returns to its initial position, then distance travelled is given by
Distance = OA + AB + BO.

Displacement: It is defined as the change in position (position vector) of a particle in a particular direction during a given interval of time.
It may also be defined as the vector drawn from initial position to final position and its magnitude is equal to the shortest distance between the initial and final positions. It is denoted by $\mathrm{\vec{s} \ or\ \vec{r} \ }$.
(i) Displacement is a vector quantity i.e., it has both magnitude and direction.
(ii) For a moving particle, the displacement can be positive, negative or zero.
(iii) Displacement of a particle does not give any idea about the nature of the path followed by the particle.
(iv) The displacement of a particle is along the unique path (which is the shortest path between the initial and final positions of the particle), which may or may not be the actual path travelled by the particle. This path is called straight path.
(v) The shortest distance between the initial and final positions of the particle gives the magnitude of displacement.
(vi) The displacement of a particle does not depend upon the choice of origin. That means the displacement remains the same, even if origin is shifted.
(vii) The unit of displacement is the unit of length i.e., cm, m, km, etc. and has dimensional formula [L]. SI unit of displacement is metre (m).
(viii) When the motion of the particle is along a straight line and in the same direction, magnitude of displacement = distance
In general, displacement ≤ distance.
(ix) When a particle returns to its starting point, its displacement is zero but distance is not zero.

Example:
Consider a person travelling along a circular path of radius ‘r’ from the initial point ‘A’

Case I: If he travels from point ‘A’ to point ‘B’ via ACB then Distance = ACB = πr and Displacement = AB = 2 r.
Case II: If he completes one complete round and returns to its initial point, then Distance = 2πr and Displacement = zero

(x) Displacement can decrease with time, can be zero, or even negative if body is returning to its initial position, reaches the initial position and moves back from the initial point.
Example: If a person travels in a straight line from initial point ‘O’ to point ‘A’ and returns to its initial point and again moves back to point ‘B’ in same straight line, then

Distance = OA + AO + OB = 5 cm + 5 cm + 5 cm = 15 cm
Displacement = – OB = – 5 cm

Speed and Velocity

Speed: The distance travelled by a body in unit time is called its speed.
$$\mathrm{Speed\ \ =\ \ \frac{Distance}{Time}}$$
(i) It is a scalar quantity. i.e., it has magnitude but no direction.
(ii) Speed is either positive or zero but never negative.
(iii) Speed is an indication of “how fast the particle is moving”. A fast moving particle has high speed and a slow moving particle has low speed. A particle at rest has zero speed.
(iv) The SI unit of speed is ms^-1. and its dimensional formula is [LT^-1].

Velocity: The time rate of change of displacement of the body is defined as velocity. In other words, The distance travelled by the body in a certain direction in unit time is called as velocity.
$$\mathrm{Velocity\ \ =\ \ \frac{Displacement}{Time}}$$

(i) It is a vector quantity. i.e., it has both magnitude and direction.
(ii) Velocity can be positive, zero or negative.
(iii) Velocity of a particle is an induction of the rate at which a particle is changing its position. If a person rapidly jumps one step forward and one step backward, it might result in a frenzy of activity but zero velocity, as there is no change in his position.
(iv) Velocity of a particle changes when its magnitude changes or its direction changes or both change.
(v) The direction of a velocity of a particle determines the direction of motion of the particle.
(vi) Its SI unit is ms^-1 and its dimensional formula is [LT^-1].
(vii) The magnitude of the velocity is equal to the speed of the body moving along a straight line.
(viii) The velocity of a body is uniform if both magnitude and direction do not change.
(ix) In general, |velocity| ≤ speed.
Example:
Consider a car moving round a circular track at 50 km/h. At every point of circular track, the speed is the same i.e., 50 km/h but the velocity is different as the direction changes.

Average speed and Average Velocity

Average Speed: The average speed for the given motion is defined as the ratio of the total distance travelled by the object to the total time taken i.e.,
$$\mathrm{Average\ speed\ \ =\ \ \frac{Total\ distance\ travelled}{Total\ time\ taken}}$$
$$\ \ \ \ \ \ \ \ \ v_{av}\mathrm{\ \ \ \ \ \ =\ \ \frac{s_{1} +s_{2} +s_{3} +…….}{t_{1} +t_{2} +t_{3} +……..}}$$
Case I: Distance in each time intervals remains same
Let us consider that the object travels distances s₁, s₂, s₃, etc. with speed v₁, v₂, v₃, etc. repectively in the same direction. Then
$$v_{av}\mathrm{\ \ =\ \ \frac{s_{1} +s_{2} +s_{3} +…….}{t_{1} +t_{2} +t_{3} +……..}} \ =\ \frac{s_{1} +s_{2} +s_{3} +…….}{\frac{s_{1}}{v_{1}} +\frac{s_{2}}{v_{2}} +\frac{s_{3}}{v_{3}} +…….}$$
For two distances s₁ = s₂ = s(say). Then,
$$v_{av}{\ \ =\ \ \frac{s+s}{\frac{s}{v_{1}} +\frac{s}{v_{2}}}} \ ={\frac{2s}{s\left(\frac{1}{v_{1}} +\frac{1}{v_{2}}\right)}} \ =\ \frac{2}{\frac{1}{v_{1}} +\frac{1}{v_{2}}}$$
$$v_{av}\mathrm{\ \ =} \ \frac{2v_{1} v_{2}}{v_{1} +v_{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
Therefore, average speed for equal distances is equal to harmonic mean of individual speeds.

Case II: Time intervals remains same

Let us consider that the object travels with speed v₁, v₂, v₃, etc. during time intervals t₁, t₂, t₃, etc. respectively in the same direction. Then
$$v_{av}\mathrm{\ \ =\ \ \frac{s_{1} +s_{2} +s_{3} +…….}{t_{1} +t_{2} +t_{3} +……..}} \ =\ \frac{v_{1} t_{1} +v_{2} t_{2} +v_{3} t_{3} +…….}{t_{1} +t_{2} +t_{3} +……..}$$
When t₁ = t₂ = t₃ = …….. = t(say)
$$v_{av}\mathrm{\ \ =\ \ }\frac{v_{1} t+v_{2} t+v_{3} t+…….}{t+t+t+………} \ =\frac{t( v_{1} +v_{2} +v_{3} +…….)}{nt}$$
$$v_{av}\mathrm{\ \ =} \ \frac{v_{1} +v_{2} +v_{3} +…….}{n} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
Therefore, the average speed for equal time intervals is equal to arithmetic mean of individual speeds.

Average Velocity: The average velocity of the body is defined as the ratio of the total displacement covered by the object to the time interval.
$$\mathrm{Average\ velocity\ \ =\ \ \frac{Total\ displacement\ covered}{Time\ interval }}$$
If a body moves from position s₁ to position s₂ in time interval t₁ to t₂, then its average velocity is given by

\[\vec{v}_{av} \ \ =\ \ \frac{\vec{s}_{2} -\vec{s}_{1}}{t_{2} -t_{1}}\]

Let a body be at point P at time t₁ and at point Q at time t₂. The displacement in this time interval is $\vec{PQ}$. The average velocity in this time interval is given by

\[\vec{v}_{av} \ \ =\ \ \frac{\overrightarrow{PQ}}{t_{2} -t_{1}}\]

Instantaneous speed and Instantaneous velocity

Under variable speed, the object possesses different speeds at different instants. Therefore, the instantaneous speed may be defined as the speed of the object at a given instant of time. Let Δs be the distance travelled in time interval Δt . Then the average speed in this time interval is

\[v_{av}\ =\ \frac{\Delta s}{\Delta t}\]

As Δt approaches zero, the distance Δs also approaches zero but the ratio $\frac{\Delta s}{\Delta t}$ has a finite limit. The instantaneous speed at time t is defined as

\[v_{ins} \ =\ \lim _{\Delta t\rightarrow 0}\frac{\Delta s}{\Delta t} \ =\ \frac{ds}{dt}\]

Note: Average speed is defined for a time interval while instantaneous speed is defined at a particular instant.

Instantaneous Velocity: The velocity of the body at any particular instant of time at any point of its path is called as instantaneous velocity.
Let $\Delta \vec s$ be the displacement traversed in time interval Δt .Then the average velocity in this time interval is

\[\vec v_{av}\ =\ \frac{\Delta \vec s}{\Delta {t}}\]

As Δt approaches zero, the displacement $\Delta\vec{s}$ also approaches zero but the ratio $\frac{\Delta\vec{s}}{\Delta t}$ has a finite limit. The instantaneous velocity at time t is defined as

\[\vec{v}_{ins} \ =\ \lim _{\Delta t\rightarrow 0}\frac{\Delta \vec{s}}{\Delta t} \ =\ \frac{d\vec{s}}{dt}\]

Note:
(i) Curvilinear translation in Cartesian coordinate system.
Consider a particle moving on a three dimensional curvilinear path AB. At an instant of time t it is at P (x, y, z) moving with velocity $\vec{v}$. Its position vector is defined as

\[\vec{r} =x\hat{i} +y\hat{j} +z\hat{k}\] \[\vec{v}\ \ =\ \frac{d\vec{r}}{dt}\ \ = \ \frac{d}{dt}(x\hat{i} +y\hat{j} +z\hat{k})\] \[\vec{v}\ \ =\ \frac{d\vec{x}}{dt}\hat{i} + \frac{d\vec{y}}{dt}\hat{i} + \frac{d\vec{z}}{dt}\hat{i}\] \[\vec{v}\ \ =\ \vec{v}_{x}\hat{i} +\vec{v}_{y}\hat{j} +\vec{v}_{z}\hat{k}\]

where $v_{x} = \frac{dx}{dt}$, $v_{y} = \frac{dy}{dt}$ and $v_{z} = \frac{dz}{dt}$.

(ii) In one dimensional motion we need not use the vector sign for the displacement, velocity, acceleration etc. Instead, we will use + or – sign wherever necessary.
(iii) The distance can be calculated as

\[d\ =\ \int |\vec{v} |dt\ =\ \int \left(\sqrt{v_{x}^{2} +v_{y}^{2} +v_{z}^{2}}\right) dt\]

(iv) The displacement can be as

\[\vec{s} \ =\ \int \vec{v} dt\ =\ \int ( v_{x}\hat{i} +v_{y}\hat{j} +v_{z}\hat{k}) dt\]

Uniform motion and non-uniform motion

Uniform Motion: When an object undergoes equal displacements in equal interval of time, however small these intervals may be, then the object is said to be in uniform motion.

Expression for velocity:
Consider an object moving with uniform velocity ‘v’ along a straight line OX. Let ‘O’ be taken as the origin for the measurement of position as well as time of the object. Suppose ‘A’ and ‘B’ be the positions of the object at instant of time t₁ and t₂ respectively.

We have, OA = x₁ and OB = x₂. The displacement of the object in time interval (t₂ – t₁) is given by
$$AB \ \ = \ \ OB \ – \ OA \ = \ x_{2}\ -\ x_{1}$$
Therefore, the velocity of uniform motion is given by
$$v \ =\ \frac{x_{2}\ -\ x_{1}}{t_{2}\ -\ t_{1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ …………………………………..(1)$$

Expression for displacement:
From equation (1), we can write
$$x_{2}\ -\ x_{1}\ =\ v(t_{2}\ -\ t_{1})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ …………………………………..(2) $$
$$x_{2}\ =\ x_{1}\ +\ v(t_{2}\ -\ t_{1})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ …………………………………..(3) $$
If $x_{2}\ -\ x_{1}\ =\ s$ and $(t_{2}\ -\ t_{1})\ =\ t$ then
$$s\ =\ vt \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ …………………………………..(4)$$

Equation (4) can be used to determine the distance travelled by an object in uniform motion.

Notes:
(i) The velocity of uniform motion is independent of the choice of the origin.
(ii) For an object in uniform motion, no cause or effect i.e., force is applied.
(iii) The average and instantaneous velocities have same value in uniform motion.
(iv) The positive value of velocity means that the object is moving towards right of origin while negative velocity means that the object is moving towards left of the origin.
(v) In uniform motion the velocity does not depend on time interval (t₂ – t₁).
(vi) When the uniform motion takes place along a straight line in a particular direction, then the magnitude of the displacement is equal to the actual distance covered by the object.
(vii) In uniform motion along a straight line without change in direction of motion,

\[\frac{d|\vec v|}{dt}\ =\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left|\frac{d\vec v}{dt}\right|\ =\ 0\]

(viii) If body moves uniformly but its direction of motion changes, then

\[\frac{d|\vec v|}{dt}\ =\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left|\frac{d\vec v}{dt}\right|\ \neq\ 0\]

Non-uniform Motion or Accelerated Motion:
If the velocity of the body changes with time, then the motion is said to be non-uniform or accelerated motion.
Example: When we go to walk our velocity is sometime slow and sometime fast.

Acceleration and its types

Acceleration: The rate of change of velocity of a body is called its acceleration. It is denoted by ‘a’. Thus,

\[Acceleration\ =\ \frac{Change\ in\ velocity}{time \ interval}\]

If a body moves with initial velocity ‘u’ at time t = 0, and after time t, its final velocity becomes ‘v’ then
Change in velocity = v – u, and
Time interval = t – 0 = 0 , then

\[a\ =\ \frac{v\ -\ u}{t}\]

(i) Acceleration is a vector quantity.
(ii) It can be either positive, zero or negative.
(iii)In case of uniform motion or at rest, a = 0.
(iv) If the magnitude of the particle’s velocity increases with time then the motion is called accelerated motion i.e., the acceleration is positive.
(v) If the magnitude of the particle’s velocity decreases with time then the motion is known as decelerated or retarded motion i.e., the acceleration is negative.
In general, Retardation = – acceleration
(vi) Its SI unit is ms-2.Its dimensional formula is [LT^-2].
(vii) If a particle is moving with uniform speed, it may or may not have acceleration but a particle moving with uniform velocity has no acceleration.

Uniform or Constant Acceleration: If the magnitude and direction of the acceleration of a particle remains constant at all time, the particle is said to be moving with constant acceleration. The change in magnitude of the velocity of a particle in uniform acceleration (either increase or decrease but not both) per unit time remains constant and this type of change is always in the same direction.
Non -uniform Acceleration: If the magnitude of the acceleration changes or direction of the acceleration changes or both magnitude and direction of acceleration of a particle change, then the particle is said to be moving with non-uniform acceleration.
Average Acceleration: The average acceleration of a body between any two time intervals is defined as the ratio of change in velocity to the time interval.

\[Average\ acceleration\ =\ \frac{Total\ change\ in\ velocity}{time \ interval}\]

Let the velocity of the body at time t₁ is $\vec {v_{1}}$ and at time t₂ is $\vec {v_{2}}$, Then
Change in velocity = $\vec {v_{2}}\ -\ \vec {v_{1}}$ in given time interval = t₂ – t₁.
Hence, the average acceleration is given by
$$\vec a_{av}\ =\ \frac{\vec {v_{2}}\ -\ \vec {v_{1}}}{t_{2}\ -\ t_{1}}\ =\ \frac{\Delta\vec v}{\Delta{t}}$$

Note: Average acceleration depends only on the velocities at time t₁ and t₂ and does not depend that how the velocities have changed.

Instantaneous Acceleration: The instantaneous acceleration is defined as the acceleration of the body at any instant or at any point of its path.
Let $\Delta \vec v$ be the small change in velocity in time interval $\Delta {t}$. Then, the average acceleration in this time interval is

\[\vec a_{av}\ =\ \frac{\Delta \vec v}{\Delta {t}}\]

As Δt approaches zero, the distance Δs also approaches zero but the ratio $\frac{\Delta\vec{v}}{\Delta t}$ has a finite limit. The instantaneous acceleration at time t is defined as

\[\vec a_{ins} \ =\ \lim _{\Delta t\rightarrow 0}\frac{\Delta \vec v}{\Delta t} \ =\ \frac{d\vec v}{dt}\]

(i) We know that, $\vec v\ =\ \frac{d\vec s}{dt}$, therefore we get

\[\vec a\ =\ \frac{d}{dt} \left(\frac{d\vec s}{dt}\right)\ =\ \frac{d^{2}\vec s}{dt^{2}}\]

Thus, the instantaneous acceleration may also be defined as the second time derivative of the position of the body at a given instant.

(ii)

\[a\ =\ \frac{dv}{dt}\times\frac{ds}{ds}\ =\ \frac{ds}{dt}\times\frac{dv}{ds}\ =\ v\frac{dv}{ds}\]

(iii) curvilinear translation in Cartesian coordinate system:
We have

\[\vec{r} =x\hat{i} +y\hat{j} +z\hat{k}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ……………………………………….\ \ \ (i) \]

and

\[\vec{v}\ \ =\ \vec{v}_{x}\hat{i} +\vec{v}_{y}\hat{j} +\vec{v}_{z}\hat{k}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ……………………………………….\ \ \ (ii)\]

Acceleration can be obtained by differentiating velocity vector from equation (ii), we get

\[\vec{a}\ \ =\ \frac{d\vec {v_{x}}}{dt}\hat{i} + \frac{d\vec {v_{y}}}{dt}\hat{i} + \frac{d\vec {v_{z}}}{dt}\hat{i}\ =\ \vec{a}_{x}\hat{i} +\vec{a}_{y}\hat{j} +\vec{a}_{z}\hat{k}\ \ \ \ \ \ \ \ \ ………………….\ \ \ (iii) \]

where $a_{x} = \frac{dv_{x}}{dt}$, $a_{y} = \frac{dv_{y}}{dt}$ and $a_{z} = \frac{dv_{z}}{dt}$.

Equations of Motions (Relations of uniformly accelerated motion)

(A) Derivation by formula:
Consider an object moving in uniformly accelerated motion along a straight line OX with origin at ‘O’. Let its initial velocity at time t = 0 is u and after a time t it has final velocity v and displacement s.

1. Velocity-time relation: $v\ =\ u\ +\ at$

By the definition of acceleration, we have

\[a\ =\ \frac{v\ -\ u}{t}\] or, \[at\ =\ v\ -\ u\] or, \[v\ =\ u\ +\ at\ \ …..(a)\]

2. Displacement-time relation: $s\ =\ ut\ +\ \frac{1}{2}at^2$

By the definition of average velocity, we have

\[v_{av}\ =\ \frac{s}{t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \ \ \ v_{av}\ =\ \frac{u\ +\ v}{2}\] Therefore, \[\frac{s}{t}\ =\ \frac{u\ +\ v}{2}\ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ s\ =\ \frac{(u\ +\ v)t}{2}…..(i)\]

Substituting the value of ‘v’ from equation (a) in equation (i), we get

\[s\ =\ \frac{(u\ +\ u\ +\ at)t}{2}\ =\ \frac{2u\ +\ at)t}{2}\] or, \[s\ =\ ut\ +\ \frac{1}{2}at^2\ \ ……..(b)\]

If $s\ =\ x\ -\ x_{0}$, then

\[x\ -\ x_{0}\ =\ ut\ +\ \frac{1}{2}at^2\ \ \ \ \ or\ \ \ \ \ \ \ \ x\ =\ x_{0}\ +\ ut\ +\ \frac{1}{2}at^2\ \ \ \ ……….(c)\]

3.Velocity-displacement relation: $v^2\ =\ u^2\ +\ 2as$

From equation (a), we get

\[t\ =\ \frac{v\ -\ u}{a}\]

Putting the value of t in equation (i), we get

\[s\ =\ \frac{(v\ +\ u)}{2}\times\frac{(v\ -\ u)}{a}\ =\ \frac{(v^2\ -\ u^2)}{2a}\] \[2as\ =\ v^2\ -\ u^2\ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ v^2\ =\ u^2\ +\ 2as\ \ ……..(d)\]

f $s\ =\ x\ -\ x_{0}$, then

\[v^2\ =\ u^2\ +\ 2a(x\ -\ x_0)\ \ ……… (e)\]

Equation (a), (b) and (d) are called the equations of motion for uniformly accelerated motion.

4.Distance covered in nth second: $s_{n^{th}}\ =\ u\ +\ \frac{1}{2}a(2n\ -\ 1)$

From equation (b), we have

\[s\ =\ ut\ +\ \frac{1}{2}at^2\]

Distance travelled in first n second is

\[s_n\ =\ un\ +\ \frac{1}{2}an^2\]

and, distance travelled in first (n – 1) second is

\[s_{n\ -\ 1}\ =\ u(n\ -\ 1)\ +\ \frac{1}{2}a(n\ -\ 1)^2\] or \[s_{n\ -\ 1}\ =\ u(n\ -\ 1)\ +\ \frac{1}{2}a(n^2\ -\ 2n\ +\ 1)\]

Hence, the distance travelled in n^th second is

\[s_{n^{th}}\ =\ s_n\ -\ s_{(n\ -\ 1)}\]

or,

\[s_{n^{th}}\ =\ un\ +\ \frac{1}{2}an^2\ -\ \left\{u(n\ -\ 1)\ +\ \frac{1}{2}a(n^2\ -\ 2n\ +\ 1)\right\}\] \[ =\ un\ +\ \frac{1}{2}an^2\ -\ u(n\ -\ 1)\ -\ \frac{1}{2}a(n^2\ -\ 2n\ +\ 1)\ \ \ \] \[=\ u(n\ -\ n\ +\ 1)\ +\ \frac{1}{2}a(n^2\ -\ n^2\ +\ 2n\ -\ 1)\ \ \ \ \ \ \ \ \ \ \ \] \[s_{n^{th}}\ =\ u\ +\ \frac{1}{2}a(2n\ -\ 1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ …(f) \]

(B) Derivation by Calculus method:

1.Velocity-time relation: $v\ =\ u\ +\ at$

By the definition of acceleration, we have

\[a\ =\ \frac{dv}{dt}\ \ \ \ \ or\ \ \ \ \ dv\ =\ adt\ \ \ \ \ \ ….(i)\]

Integrating equation (i) with proper limits, we get

\[\int _{u}^{v} dv\ =\ a\int _{0}^{t} dt\] \[\left\{ v \right\}_{u}^{v}\ =\ a\left\{ t \right\}_{0}^{t}\] \[(v\ -\ u)\ =\ a(t\ -\ 0)\] \[v\ =\ u\ +\ at\ …(a)\]

2.Displacement-time relation: $s\ =\ ut\ +\ \frac{1}{2}at^2$

By the definition of instantaneous velocity, we have

\[v\ =\ \frac{ds}{dt}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \] \[ds\ =\ vdt\ =\ (u\ +\ at)dt\ \ \ \ \ ..(ii)\]

Integrating equation (ii) with proper limits, we get

\[\int _0^s ds\ =\ \int _0^t (u\ +\ at)dt\ =\ u\int _0^t dt\ +\ \int _0^t tdt\] \[\{s\}_{0}^{s} \ =\ u\{t\}_{0}^{t} \ +\ a\left\{\frac{t^{2}}{2}\right\}_{0}^{t}\] \[(s\ -\ 0)\ =\ u(t\ -\ 0)\ +\ \frac{1}{2}a(t^2\ -\ 0)\] \[s\ =\ ut\ +\ \frac{1}{2}at^2\ \ \ \ \ \ \ \ \ \ \ ..(iii)\]

3.Velocity-displacement relation: $v^2\ =\ u^2\ +\ 2as$

By the definition of acceleration, we have

\[a\ =\ \frac{dv}{dt}\ =\ \frac{dv}{dt} \times \frac{ds}{ds}\ =\ \frac{ds}{dt} \times \frac{dv}{ds}\ =\ v\frac{dv}{ds}\] \[vdv\ =\ ads\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ……(iii)\]

On integrating equation (iii) with proper limits, we get

\[\int _{u}^{v} vdv\ =\ a\int_{0}^{s} ds\] \[\left\{\frac{v^{2}}{2}\right\}_{u}^{v}\ =\ a\{s\}_{0}^{s}\] \[\frac{1}{2}(v^2\ -\ u^2)\ =\ as\] \[v^2\ -\ u^2\ =\ 2as\] \[v^2\ =\ u^2\ + 2as\ \ \ \ …..(c)\]

4.Distance covered in n^th second: $s_{n^{th}}\ =\ u\ +\ \frac{1}{2}a(2n\ -\ 1)$

By the definition of instantaneous velocity, we have

\[v\ =\ \frac{ds}{dt}\] \[ds\ =\ vdt\ =\ (u\ +\ at)dt\ \ \ \ \ \ \ …….(iv)\]

On integrating equation (iv) with proper limits, we get

\[\int_{s_{(n\ -\ 1)}}^{s_{n}} ds\ =\ \int_{(n\ -\ 1)}^{n} (u\ +\ at)dt\ =\ u\int_{(n\ -\ 1)}^{n} dt\ +\ a\int_{(n\ -\ 1)}^{n} tdt\] \[\{s\}_{s_{(n\ -\ 1)}}^{s_{n}}\ =\ u\{t\}_{(n\ -\ 1)}^{n}\ +\ a\left\{\frac{t^{2}}{2}\right\}_{(n\ -\ 1)}^{n}\] \[(s_{n}\ -\ s_{n\ -\ 1})\ =\ u\{n\ -\ (n\ -\ 1)\}\ + \frac{1}{2}a\{n^{2}\ -\ (n\ -\ 1)^2\}\] \[s_{n^{th}}\ =\ u\{n\ -\ n\ +\ 1\}\ +\ \frac{1}{2}a \{n^{2}\ -\ n^{2}\ +\ 2n\ -\ 1\}\] \[s_{n^{th}}\ =\ u\ +\ \frac{1}{2}a(2n\ -\ 1)\ \ \ \ \ \ ……(d)\]

Motion by Graphs

General concepts:

We have to study only two types of graph to show the motion:
(a) Straight line graph
(b) Parabola

Straight Line Graph:

An equation of the form of y = mx + c is called straight line equation. Here, = slope of graph with x – axis and, c = intercept on y – axis.
Note: If c = 0, then graph will pass through origin.
The slope m is given by
\[tan\theta\ =\ \frac{y_{2}\ -\ y_{1}}{x_{2}\ -\ x_{1}}\ =\ \frac{\Delta y}{\Delta x}\]
For examples, y = 4x + 3; y = -4x + 5; y = 3x – 1; y = -2x + 3; y = 3x, etc.

Graphs:

(1) $y\ =\ mx\ +\ c;\ \ y\ =\ mx;\ \ y\ =\ mx\ -\ c$

(1) $y\ =\ -mx\ +\ c;\ \ y\ =\ -mx;\ \ y\ =\ -mx\ -\ c$

Note: Straight line graph is applicable in equations like s = vt; V = IR; F = ma; P= mv; ds = vdt; dv = adt; dw = Fdx; v = u + at, etc.

Parabola

An equation of the form y = ax² or x = by² is called parabola.
(a) It is the simplest form of quadratic equation y = ax² + bx + c, from which we obtain parabola by putting b = 0 and c = 0.
(b) This parabola has its vertex at origin (0,0) because when we put x = 0, it gives y = 0.
(c) The graph for y = ax² will be a symmetric parabola about y – axis. The orientation of parabola will be decided by the sign of a. Similarly, The graph for x = by² will be a symmetric parabola about x – axis. The orientation of parabola will be decided by the sign of b.

Graphs

Note:
(i) Here the parabolic equation like y = ax² is valid for positive and negative both value of x, y will always be positive. But, in motion we discuss only about the graph obtained by putting positive value of x only.

(ii) Parabolic graph is applicable in equations like $s\ =\ \frac{1}{2}at^{2};\ \ v^{2}\ =\ u^{2}\ +\ 2as;\ \ w\ =\ \frac{1}{2}Fx^{2}, etc.$

Slope

The slope of the curve at any point is defined by tanθ, where θ is the angle made from x – axis. Here,
\[tan\theta\ =\ \frac{dy}{dx}\]

(a) A straight line graph has a single slope. If line makes angle θ = 0⁰ , tanθ = 0, the graph line is parallel to x – axis. If line makes angle θ < 90⁰ , the slope tanθ is positive. If line makes angle θ > 90⁰ , the slope tanθ is negative.

(i) Zero Slope: θ = 0⁰ , tanθ = 0,

(ii) Positive Slope: θ < 90⁰ , tanθ = +ve

(iii) Negative Slope: θ > 90⁰ , tanθ = -ve

(b) A curve graph has variable of slopes. In a curve with crest upward, slope decreases with increasing value of x and vice versa.

Position – time graph:

In a position – time graph, its slope gives the velocity of the object.
Case I: Zero velocity i.e., the position of the object is fix at all the time, so the object is at rest. In this case,
$Slope\ =\ \frac{dy}{dx}\ =\ tan\theta\ =\ tan0^{0}\ =\ 0$

Therefore, velocity of particle is zero and the position-time graph is a straight line parallel to the time axis.

Case II: Uniform velocity: An object in uniform motion covers equal distances in equal intervals of time. So, the position-time graph for an object in uniform motion along a straight line path is a straight line inclined to the time axis.

Slope of Position-time graph AB
$tan\theta\ =\ \frac{QR}{PR}\ =\ \frac{x_{2}\ -\ x_{1}}{t_{2}\ -\ t_{1}}\ =\ \frac{displacement}{time\ interval}\ =\ velocity$
Hence, the slope of position-time graph gives velocity of the object.

Case III: Non-uniform velocity (Increasing with time) i.e., acceleration ( a > 0)
In this case, as the time increases, θ also increases.
$\therefore\ \frac{dx}{dt}\ =\ tan\theta$ also increases.
Hence, velocity of particle increases.

Case IV: Non-uniform velocity (Decreasing with time) i.e., retardation ( a < 0)
In this case, as the time decreases, θ also decreases.
$\therefore\ \frac{dx}{dt}\ =\ tan\theta$ also decreases.
Hence, velocity of particle decreases.

Case V: Non-uniform velocity (Object accelerating and then decelerating)
In graph, at point A, object accelerates and at point B, it decelerates.

Note: For non-uniform motion, the slope of the tangent to the curve at a point gives velocity at that instant i.e., instantaneous velocity.

Velocity-time graph

(a) The slope of v – t graph gives the acceleration of the object.
(b) Area under v – t graph gives the displacement traversed by the object.

Case I: Zero acceleration
Velocity is constant i.e., the object travels with uniform velocity. Here, tanθ = 0.
$\therefore\ \frac{dv}{dt}\ =\ 0$ i.e., acceleration is zero.
So, the velocity – time graph is a straight line parallel to the time- axis.

Area under the v – t graph between times t₁ and t₂
= Area of rectangle ABCD
= AD x DC = v(t₂ – t₁)
= Velocity x time
= Displacement
Hence, the area under the v – t graph gives the displacement of the object in the given time interval.

Case II: Uniform acceleration ( θ < 90⁰)
When an object moves with a uniform acceleration, its velocity changes by equal amounts in equal intervals of time. So, the velocity – time graph for a uniformly accelerated motion is a straight line inclined to the time-axis.
(a) In this case, c = 0 i.e., v = at.
$\because$ tanθ = constant
$\therefore\ \frac{dv}{dt}$ = constant
Hence, it shows constant acceleration.

(b) In this case, c = u i.e., v = u + at

Slope of v – t graph AB $=\ tan\theta\ =\ \frac{QR}{PR}\ =\ \frac{v_2\ -\ v_1}{t_2\ -\ t_1}\ =\ \frac{change\ in\ velocity}{time\ interval}\ =\ acceleration$
Hence, the slope of the velocity – time graph gives the acceleration of the object.

Also,
Area under the velocity – time graph AB between time interval (t₂ – t₁)
= Area of trapezium PQST
= $\frac{1}{2}(PT\ +\ QS)\ \times\ TS\ =\ \frac{1}{2}(v_1\ +\ v_2)(t_2\ -\ t_1)$
= Average velocity x time interval
= Distance travelled in time t
Hence, the area under the velocity – time graph gives the distance travelled by the object in the given time interval.

Case III: Uniform retardation ( θ >90⁰ )
$\because $ tanθ = constant and negative.
$\therefore \ \frac{dv}{dt}$ = negative and constant.
Hence, it shows constant retardation.

Acceleration-time graph

Case I: Constant acceleration
In this case, tanθ = 0,
$\therefore $ $\frac{da}{dt}\ =\ 0$
Hence, acceleration is constant.
Area under a-t graph gives the measure of velocity of the object.

Case II: Uniformly increasing acceleration
In this case, θ < 90⁰ i.e., tanθ > 0
$\therefore\ \frac{da}{dt}\ =\ tan\theta$ = constant >0
$\therefore\ \frac{da}{dt}$ = positive constant
Hence, acceleration is uniformly increasing with time.

Case III: Uniformly decreasing acceleration
In this case, θ > 90⁰ i.e., tanθ < 0
$\therefore\ \frac{da}{dt}\ =\ tan\theta$ = constant < 0
$\therefore\ \frac{da}{dt}$ = negative constant
Hence, acceleration is uniformly decreasing with time.

Notes:

(1) In a displacement – time graph
(a) Slope gives velocity.
(b) If position – time graph is a straight line $\implies \ a\ =\ 0$
(c) If it is a parabola opening upward $\implies \ a\ >\ 0$
(d) If it is a parabola opening downward $\implies \ a\ <\ 0$

(e) To convert a displacement-time graph into a distance-time graph, place a mirror at ‘A’ parallel to time axis and take its reflection.

This is because displacement can be negative, but distance can never be negative.

(2) In a velocity – time graph
(a) Slope gives acceleration.
(b) Area under v – t curve gives displacement / distance.
Displacement = +a₁ – a₂
Distance = |+a₁| + |-a₂|
To convert a velocity-time graph into a speed-time graph, place a mirror parallel to time axis and take its reflection.

(c) To get acceleration from a velocity-time graph at point Q, draw a tangent and normal at Q then sub-normal AB will give acceleration. The value of subnormal will give acceleration at point Q.

(3) In acceleration – time graph
(a) Area under a – t graph gives change in velocity
(b) Area under a – x graph gives $\frac{v^{2}\ -\ u^{2}}{2}$.

Impossible Graphs

(a) Distance travelled by a particle cannot decrease with time. Also it never negative.

(b) Speed also can never be negative.

(d) A quantity cannot change without spending time.

MOTION UNDER GRAVITY

Free Fall

It is defined as the motion of an object where gravity is the only force acting upon it. The acceleration of a freely falling body is called as acceleration due to gravity.
(a) It is denoted by ‘g’.
(b) Its value near the earth surface is 9.8 m/s² or 980 cm/s².
(c) Its direction is being towards the centre of the earth.

For a freely falling body, the equations of motion are:
(i) $v\ =\ u\ +\ gt$
(ii) $s\ =\ ut\ +\ \frac{1}{2}gt^{2}$
(iii) $v^{2}\ =\ u^{2}\ +\ 2as$
(iv) $s_{n^{th}}\ =\ u\ +\ \frac{1}{2}g(2n\ -\ 1)$
……………………………. (1)

Body Falling freely under gravity

In this case, u = 0, a = + g, s = h, then equation (1) become
(i) $v\ =\ gt\ \ \ \ \ \ or\ \ \ \ \ t\ =\ \frac{v}{g}\ \ \ \ \ \ \ \ \ $ (ii) $h\ =\ \frac{1}{2}gt^{2}\ \ \ \ \ or\ \ \ \ \ t\ =\ \sqrt{\frac{2h}{g}}$
(iii) $v\ =\ \sqrt{2gh}\ \ \ \ \ or\ \ \ \ \ h\ =\ \frac{v^{2}}{2g}\ \ \ \ \ $ (iv) $h_{n^{th}}\ =\ \frac{1}{2}g(2n\ -\ 1)$

Time taken to reach the ground, if height from which it falls, is known.
$t\ =\ \sqrt{\frac{2h}{g}}$
Time taken to reach the ground, if velocity with which it strikes the ground, is known
$ t\ =\ \frac{v}{g}$
Velocity with which it strikes the ground, if time of fall is known.
$v\ =\ gt$
Velocity with which it strikes the ground, if height of fall is known.
$v\ =\ \sqrt{2gh}$
Height from which it falls, if time of fall is known.
$h\ =\ \frac{1}{2}gt^{2}$
Height from which it falls, if velocity at the ground just before the reach is known.
$h\ =\ \frac{v^{2}}{2g}$
The distance fallen in the nth second, if n is the time is fall, is given by
$h_{n^{th}}\ =\ \frac{1}{2}g(2n\ -\ 1)$

When a body is thrown vertically upward with an initial velocity u

In this case, v = 0, a = -g, s = h, then equation (1) become
(i) $u\ =\ gt\ \ \ \ \ \ or\ \ \ \ \ t\ =\ \frac{u}{g}\ \ \ \ \ \ \ \ \ $ (ii) $h\ =\ ut\ -\ \frac{1}{2}gt^{2}$
(iii) $u\ =\ \sqrt{2gh}\ \ \ \ \ or\ \ \ \ \ h\ =\ \frac{u^{2}}{2g}\ \ \ \ \ $ (iv) $h_{n^{th}}\ =\ u\ -\ \frac{1}{2}g(2n\ -\ 1)$

1. Time of ascent = Time of descent, $t\ =\ \frac{u}{g}$
2. Total time of flight = $T\ =\ 2\frac{u}{g}$
3. Maximum height attained = $h\ =\ \frac{u^{2}}{2g}$
4. The velocity of fall at the same point = $u$
5. Body projected upwards with a velocity u, from the top of a tower of height H,
(a) Maximum height reached from the ground = $H\ +\ \frac{u^{2}}{2g}$
(b) Total time taken to reach the foot of the tower = $\frac{u}{g}\ \pm\ \frac{1}{g}\sqrt{u^{2}\ +\ 2gH}$
(c) Velocity with which it reaches the ground, $v\ =\ \sqrt{u^{2}\ +\ 2gH}$

CONCEPTS:

1. Calculating distance from $s\ =\ ut\ +\ \frac{1}{2}at^{2}$
Case I: When acceleration and velocity are along the same direction. In such case,
$$Displacement\ =\ Distance$$
Case II: When acceleration and velocity are opposite to each other. In this case, if t₀ is the time at which the velocity becomes zero, then first equation of motion, we get
$\ \ \ \ \ \ \ \ \ \ \ 0\ =\ u\ -\ at_{0}\ \Rightarrow\ t_{0}\ =\ \frac{u}{a}$