**Kinematics**

The study of motion of a particle without its cause is called kinematics.

**Terms regarding the motion:****Distance:** The length of actual path followed by the particle between its initial and final position is called distance. It is a scalar quantity. Distance travelled by the particle is always positive.

**Displacement:** The shortest distance of the path between two points in a specified direction is called displacement. It is a vector quantity. Displacement of a particle may be positive, negative or zero.

**Speed:** The rate of covering the distance by the particle is called its speed. It is a scalar quantity. Speed of a particle is equal to or greater than velocity of the particle. Speed of a particle is always positive.

**Velocity:** The time rate of change of displacement is called velocity. It is a vector quantity. Velocity of the particle is equal or less than the speed of the particle. Velocity of a particle may be positive, negative or zero.

**Uniform velocity:** A particle is said to be moving with uniform velocity, if it undergoes equal displacement in equal intervals of time, however small intervals may be.

**Variable velocity:** A particle is said to be moving with variable velocity if either its speed or its direction of motion or both changes with time.

**Average velocity:** The average velocity of a particle is defined as the ratio of its total displacement to the total time interval during which the displacement occurs.

Let us consider a particle moving with variable velocity. Let x_1 and x_2 be the displacement of a particle at time t_1 and t_2 respectively. Then the average velocity of the particle in the time interval t_2-t_1 is given by:

**Instantaneous velocity:** The velocity of a particle at a particular point of its path or at a particular instant of time is called instantaneous velocity and is defined as the limiting value of average velocity of the particle in a small time interval around that instant, when the time interval approaches zero.

Mathematically, v_{ins} = \lim_{\Delta t\rightarrow 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}

**Acceleration:** The time rate of change of velocity is called acceleration. It is denoted by a .

\text{Acceleration}(a) = \frac{\text{Change in velocity}}{\text{Change in time}}

It is a vector quantity which may have positive or negative value. The negative acceleration is sometimes called retardation.

**Uniform acceleration:** A particle is said to be moving with uniform acceleration, if the velocity changes by equal amount in equal intervals of time.

**Variable acceleration:** A particle is said to be moving with variable or non uniform acceleration if either the magnitude of the acceleration or its direction or both changes with time.

**Average acceleration:** When the particle is moving is with variable acceleration, then the average acceleration of the particle for the given motion is defined as the ratio of the total change in velocity of the particle to the total time interval.

Let us consider a particle moving along x-axis. Suppose the velocity of the particle at time t_1 and t_2 becomes v_1 and v_2 respectively. Then the average acceleration of the particle is given by:

**Instantaneous acceleration:** The acceleration of a particle at a particular point of its path or at a particular instant of time is called instantaneous acceleration and is defined as the limiting value of average acceleration of the particle in a small time interval around that instant, when the time interval approaches zero.

Mathematically, a_{ins} = \lim_{\Delta t\rightarrow 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}

**Velocity-time graph:**

A plot showing the variation of velocity as a function of time is called velocity-time (v-t) graph.

i) When a body moves with uniform velocity, then the velocity-time graph will be straight line parallel to the time axis.

ii) When a body moves with positive uniform acceleration, then the velocity-time graph will be a straight line having positive slope.

iii) When a body moves with negative uniform acceleration, then the velocity-time graph will be a straight line having negative slope.

**Equation of motion with uniform acceleration:**

1) v = u + at

Consider a body moving with initial velocity u and acceleration a . v be its final velocity after time t .

Then the acceleration of the body is a = \frac{v-u}{t}

or, at = v - u

Or, v = u + at

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

The displacement s for accelerated motion is

s = \text{area under v-t graph}

= \frac{u+v}{2} \times t

= \frac{2u + at}{2} \times t

3) v^2 = u^2 + 2as

We have, v = u + at

Squaring this equation on both sides, we get

v^2 = u^2 + 2uat + a^2t^2

Or, v^2 = u^2 + 2a\left(ut + \frac{1}{2}at^2\right)

Or, v^2 = u^2 + 2as

**Equations of motion derived from velocity-time graph:**

Let us consider a body moving with uniform acceleration a on a straight line has initial velocity u at time t=0 and v be the velocity of the body after time t . The velocity-time graph for such motion is shown in the figure.

In this figure,

\text{OA} = \text{CD} = u

\text{BD} = v

i) v = u + at

Slope of AB = (Change in velocity)/(Change in time) = \frac{v-u}{t} = a

ii) s = ut + \frac{1}{2}at^2

We have, distance covered (s) = \text{Area of trapezium OABD}

Or, s = \text{Area of triangle ABC} + \text{Area of parallelogram OACD}

= \frac{1}{2} \times \text{BC} \times \text{AC} + \text{OA} \times \text{OD}

= \frac{1}{2} \times at \times t + u \times t

Therefore, s = ut + \frac{1}{2}at^2

iii) v^2 = u^2 + 2as

We have,

distance covered (s) = \text{Area of trapezium OABD}

= \frac{1}{2} \times (\text{BD} + \text{OA}) \times \text{OD}

= \frac{1}{2} \times (\text{BD} + \text{OA}) \times \frac{(\text{BD} - \text{OA})}{a} (Acceleration = Slope of AB)

= \frac{1}{2a} \times (\text{BD}^2 - \text{OA}^2)

Or, s = \frac{1}{2a} \times (v^2 - u^2)

Or, v^2 = u^2 + 2as

**Distance travelled in n^{th} second**

Let s_n and s_{n-1} are distance covered by an object in n and n-1 seconds respectively.

Therefore, s_n = un + \frac{1}{2}an^2

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

Therefore, total distance travelled in n^{th} second is given by

s_n - s_{n-1} = un + \frac{1}{2}an^2 - \left\{u(n-1) + \frac{1}{2}a(n-1)^2\right\}

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

= u + \frac{1}{2}a(2n - 1)

Therefore, s = u + \frac{1}{2}a(2n-1)

**Equation of motion under gravity:**

For downward motion For upward motion

v = u + gt v = u - gt

s = ut + \frac{1}{2}gt^2 s = ut - \frac{1}{2}gt^2

v^2 = u^2 + 2gh v^2 = u^2 - 2gh

**Projectile:**

Any object thrown into space such that it moves under the effect of gravity alone is called a projectile and its motion is called projectile motion.

It is two dimensional motion. In projectile motion, the velocity of the projectile along horizontal direction remains constant throughout its motion but only vertical velocity changes due to effect of gravity. For projectile motion, air resistance is neglected and acceleration due to gravity is downward and constant.

**Projectile fired horizontally:**

Let a projectile be projected horizontally from height h above the ground with initial velocity u . The projectile is under the action of gravity. So, the horizontal velocity remains constant, whereas its vertical velocity goes on increasing (Initial vertical velocity is zero). The path of the projectile is parabolic.

**Equation of trajectory or path:**

It is the relation between horizontal and vertical distances travelled by the projectile in time t .

Let x and y be the horizontal and vertical distances travelled by the projectile in time t .

Therefore, x = u_xt + \frac{1}{2}a_xt^2

= ut (as a_x = 0 for horizontal motion)

And y = u_yt + \frac{1}{2}a_yt^2

= 0 + \frac{1}{2}(-g)t^2 (as initial vertical velocity u_y = 0 , and a_y = -g for downward motion)

From equations (i) and (ii), we get

y = -\frac{g}{2u^2} \times x^2Or, y = ax^2 where, a = -\frac{g}{2u^2} = constant

This is an equation of parabola. Hence the path of the horizontal projectile is parabolic.

**Time of flight(T):**

Time taken by the projectile to reach the ground is called time of flight. It is denoted by T .

When the projectile reaches the ground, y=h and t=T then

h = \frac{1}{2}gT^2

Or, T = \sqrt{\frac{2h}{g}}

**Horizontal range(R):**

The horizontal distance travelled by the projectile during its time of flight is called horizontal range. It is denoted by R .

That is, horizontal range (R) = \text{Horizontal velocity} \times \text{Time of flight}

Or, R = u \times T = u \times \sqrt{\frac{2h}{g}}

This implies that R \propto \sqrt{h} and R \propto u

**Velocity of projectile at any instant:**

Let v_x and v_y be the horizontal and vertical components of velocity after time t .

v_x = u + a_xt

Or, v_x = u (as a_x = 0 for horizontal motion)

And v_y = u_y + a_yt

Or, v_y = 0 + (-g)t

Or, v_y = -gt

Magnitude of resultant velocity v after time t is given by

v = \sqrt{v_x^2 + v_y^2}

Or, v = \sqrt{u^2 + (-gt)^2}

Let \beta be the angle made by resultant velocity v with horizontal then

\tan \beta = \frac{v_y}{v_x} = \frac{-gt}{u}**Projectile fired at an angle with the horizontal:**

Let a projectile be fired from O from the ground with initial velocity u at an angle \theta with horizontal. The projectile goes up with decreasing velocity, reaches a highest point and falls back to the level of projection.

Let u_x and u_y be the components of initial velocity u along X and Y axis respectively.

Therefore, u_x = u \cos \theta

and u_y = u \sin \theta

**Equation of trajectory:**

Let x and y be the horizontal and vertical distances travelled by the projectile in time t .

Then x = u_xt + \frac{1}{2}a_xt^2

And y = u_yt + \frac{1}{2}a_yt^2

= u \sin \theta \times t - \frac{1}{2}gt^2Now, from equations (i) and (ii), we get

y = \tan \theta \times x - \frac{gx^2}{2u^2\cos^2\theta}

Or, y = ax - bx^2

Where, a = \tan \theta and b = \frac{g}{2u^2\cos^2\theta} are constants for a motion.

This is an equation of parabola. Hence the path of the projectile is parabolic.

**Time of flight(T):**

It is the total time for which the projectile remains in air. It is denoted by T .

When the projectile returns to the ground, height reached, y=0 and time taken t=T .

Then, y = u\sin\theta\times T - \frac{1}{2}gT^2 becomes

0 = u\sin\theta\times T - \frac{1}{2}gT^2

Or, u\sin\theta\times T = \frac{1}{2}gT^2

Or, T = \frac{2u\sin\theta}{g}

**Horizontal range(R):**

The horizontal distance travelled by the projectile during its time of flight is called horizontal range. It is denoted by R .

That is, horizontal range (R) = \text{Horizontal velocity} \times \text{Time of flight}

Or, R = u_x \times T = (u\cos\theta) \times \frac{2u\sin\theta}{g}

**Maximum horizontal range(R_{max}):**

We have, R = \frac{u^2\sin2\theta}{g}

For a given initial velocity u , the horizontal range will be maximum when \sin2\theta is maximum.

Or, R_{max} = \frac{u^2}{g} (as \sin90^\circ = 1 is maximum)

Thus, range of a projectile for velocity of projection u is maximum when angle of projection is 45^\circ .

**Maximum height(h_{max}):**

It is the greatest height to which the projectile rises above the ground. It is denoted by h_{max} .

At maximum height the vertical velocity of the projectile becomes zero.

From the equation of motion,

v^2 = u^2 - 2gh , we have

(0)^2 = (u\sin\theta)^2 - 2gh_{max}

Or, h_{max} = \frac{u^2\sin^2\theta}{2g}

**Time of ascent:** The time taken by the projectile to reach the maximum height from the ground is called time of ascent.

**Time of descent:** The time taken by the projectile to reach the ground from the maximum height is called time of descent.