## Kinematics of a particle

Kinematics is the study of particle or a body undergoing a certain motion. A body can travel without acceleration, with uniform acceleration or with non- uniform acceleration. A particle travelling with zero acceleration has constant velocity. Same particle travelling with constant acceleration has uniformly increasing velocity. .

**STRAIGHT LINE MOTION**

A particle travelling in a straight line is considered to be in a one dimensional motion.

CASE - I

Let the acceleration of the particle be equal to zero i.e. it is moving with constant velocity. The distance - time graph for a particle under such motion is shown here.

So, the equations for such motions are:

1) Displacement = Velocity * Time.

2) Initial velocity = Final velocity (Velocity is constant).

CASE - II

Suppose the particle is having an acceleration = a. Its initial speed at time t=0 be u and final velocity at time t be v.

Then the displacement s is related by the following equations:

1) s = ut + 0.5at^{2}

2) v = u + at

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

CASE - III

Now the difficult case is that when the acceleration is non- uniform. Here we have to apply certain instantaneous equations. But don't worry we will be doing some examples and everything will be clear to you.

The equations are:

1) Instantaneous velocity v = ds/dt (use it when s is a function of t)

2) Instantaneous acceleration a= dv/dt (use it when v is a function of t)

3) Instantaneous acceleration a= vdv/dx (use it when v is a function of x)

**2 - DIMENSIONAL MOTION**

Suppose a particle is thrown in a projectile motion. Then, we can visualise it as a body moving in horizontal as well as in vertical direction and the resultant of the two motions is a projectile.

We suppose that the initial velocity of the ball is u and it is thrown at an angle Q with the horizontal. Now resolving the velocity along vertical and horizontal direction we have:

The vertical component of velocity is usinQ and the horizontal is ucosQ.

The acceleration on the particle is g (acc. due to gravity) in downward direction and there is no horizontal acceleration acting on the body (note that there is no air resistance). Now writing the equations of motion for the y- axis and x- axis separately, we have:

Y-axis: X- axis:

1) v_{y}= u_{y}+ (- gt) 1) v_{x}= u_{x}

2) v_{y}^{2 }= u_{y}^{2}+ (-2gs) 2) v_{x}^{2} = u_{x}^{2}

3) s_{y}= u_{y}t - 0.5gt^{2} 3) s_{x}= u_{x}t

With these 6 equations any question related to projectile motion can be solved in one go.

We will now look at some of the examples below to understand the applications of all the above cases.

Example 1:

Consider a body of mass 10 kg acted upon by a force 2 N. The body was initially travelling with a uniform velocity 2 m\s. Find the distance covered by it after 5 sec.

Solution:

In the question we are given initial velocity and time t. For finding the distance covered we should find the acceleration given to the particle.

The acceleration given is:

a = F/m

a = 2/10 m / s2

Now according to the second equation of motion

s = 2 * 5 + 0.5 * (2/10) * 5 * 5

= 10 + 2.5

= 12.5 m

Example 2:

Consider a particle undergoing a motion with speed varying with distance as v = x^{2}. Find the instantaneous acceleration of the particle when distance covered is 5 m.

Solution:

We know that a = vdv/dx.

dv/dx = 2x = 2 * 5 =10 s-1

v (at x = 5) = 5 * 5 = 25 m/s

Hence acceleration = 25 * 10 m/s2

= 250 m/s2

Example 3:

Consider a particle thrown with an initial velocity 10 m/s at an angle 30 with horizontal. Find the time taken by the particle to reach the highest position and also find out the horizontal distance it covers during the motion.

Solution:

Vertical component of velocity = 10 sin 30 = 5 m/s

Horizontal component of velocity = 10 cos 30 = 5 * 3^{1/2}.

Now at the highest point vertical component of velocity is zero at that instant and there is only the horrizontal component.

So v_{y} = u_{y} - gt

0 = 5 - 10t

t = 0.5 s

Time taken by the particle to reach highest point = 0,5 s

Now time taken by the particle to reach the ground = 2 * t = 1 s

Horizontal distance covered = u_{x} * 2t = 5 * 3^{1/2} * 1

= 5 * 1.732

= 8.66 m

So, I have tried my best to help you develop an understanding of Kinematics and I think that this is enough if you want to get a firm grip on the chapter.

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