Kinematics
Kinematics is a branch of mechanics where we study behavior of moving objects. By “behavior”, we mean how the object or a particle moves with time in space under some law which governs the motion of the particle in question. The exact cause of the motion or the cause of the law which governs the motion is subject matter of dynamics. For example, we may say that a particle is moving at a constant speed of 5 m/s or under time varying acceleration which at time instant ‘t’ is given by a(t)=kt where k is constant. But what caused it’s acceleration to be equal to k.t at all time instants is not the subject matter of Kinematics.
In this chapter, we study various types of motion of a particle. By particle, we mean an infinitesimally small object or a point mass. This is an idealistic assumption as in practice it is not possible to find an object which is a truly point mass. Nevertheless, by studying kinematics of point object we will be able to study kinematics of bigger objects by visualizing bigger objects as a system of particles joined together by means of forces.
State of Rest and Motion
A particle (or a body) is said to be in a state of rest if it’s “location” remains fixed at all time instants. Likewise, a particle is in a state of motion if it’s “location” constantly changes with time.
One would note that the above definition is loose unless and until we make the concept of location (or rather change in location) clear. First of all, with respect to what is the location changing. Change of location is something which is reported by an observer, otherwise we would never know about any change in location with time. I may be relaxing on my chair and say that my location is not changing and therefore am in a state of rest. But imagine an observer outside earth who may be looking at me with the help of a telescope and say that I am in state of motion because for him earth is moving. My colleague sitting with me as an observer would always say that I am in state of rest.
It is impossible to define the state of Absolute Rest – we have seen that rest and motion are with respect to an observer. This gives birth to the concept of Frame of Reference relative to which we would be able to measure the location or change in location. The simplest way to define a Frame of Reference is to look at it as coordinate axes x,y,z in space with the observer located at origin (0,0,0). The position of a particle can now be described as a vector in space as the following figure shows:
Having fixed the observer at the origin, what do we need to describe motion? The motion can be fully described if we specify the position vector 
as a function of time ‘t’. Not to confuse matters more, the notion of time is simply what we understand by it in everyday life. It is a continuously increasing variable(which is never negative) and is used to quantify the duration of an event. It is a fundamental quantity just like position or distance.
Though the position vector the position vector is all what we need to completely describe the kind of motion the particle is exhibiting, there are other quantities which can be used to describe motion and require lesser amount of information as explained below:
Displacement: Displacement from point A to point B in space is simply defined to be vector 
. So if particle is at point A at time t=t1 and reached point B at time t=t2, then the effective distance and direction it has moved is captured by the line joining point A and B, and thus .
Velocity: Velocity is defined to be the rate of change of position vector which is 
. So if we write 
, then
where x,y,z coordinates specify the position of particle are of course function of time.
Note that velocity is a vector and so has both a magnitude as well as direction. It’s magnitude is called speed, and this quantity called speed is what we come across in every day life while driving or traveling or walking. Speed is a scalar and has no direction. On the other hand, velocity is a vector and therefore has magnitude as well as direction. So we may say that speed of a particle is 10 km/hr, but we can not say that it’s velocity is 10km/hr – we also need to specify the direction as well, such as 10 km/hr in the N-E direction.
Overall, if the speed of a particle at any instant is not 0, we may infer that the particle is in state of motion.
Acceleration: The rate of change of velocity is called acceleration. Because velocity is a vector, acceleration is also a vector. Note that velocity is a vector and has both speed and direction. Hence it can change in following ways:
(i) Change in magnitude, i.e. speed, without any change in direction. For example, when a car starts at t=0 seconds and acquires speed of 10 km/hr at t=3 seconds on the road without taking any turn(assuming the steering is not touched in the ideal case), then there is a change in speed without change in direction.
(ii) Change in direction without any change in speed. Example of this is a uniform circular motion which we shall study shortly.
(iii) Change in speed as well as direction. Example of this includes a car slowing down in speed
and then taking a turn. Or a better example is a particle undergoing non-uniform circular
motion.
Therefore, we can mathematically define acceleration as the quantity:
where .
Thus, the magnitude of acceleration not only tells that a body is in state of motion, but also gives information about the rate of change of motion.
Acceleration is an important quantity which finds use in Newton’s second law of motion as we shall see.
Let us now see few examples illustrating motion.
Example(Motion under Uniform Acceleration in 1D): A particle moves along a straight line under constant acceleration with magnitude ‘a’. The velocity of particle at t=0 is equal to ‘u’. Calculate the total distance covered by the particle at time t=T, and also it’s speed at t=T.
Let us first try to represent the above scene in a picture form. It’s an example of motion along 1-D and so, all the vectors involved like velocity and acceleration are one dimensional, and hence can be represented by means of single unit vector î.
So, what we are given is the fact that acceleration is 
where ‘a’ is constant and therefore independent of time. Also, it’s velocity at any point is given by 
at time t=t.
Hence, we have
Hence, the speed of the particle at time t=t is given by u + at. This expression is linear in time t and was expected because the rate of change of velocity is constant, and so velocity must be proportional to time t.
To get the total distance the particle traversed, we have
where we used the fact that x(t=0) = 0.
These equations of motion we have come across in class 9. Here we have formally derived all the quantities using vectors and tools of calculus. This way, the chances of making error gets reduced.
From now onwards, we fix our attention to problems involving motion in 1D or 2D, that is motion along a straight line or in a plane with respect to the frame of reference. Example 1 already derived equations of uniform rectilinear motion in 1D. Let us solve few more complicated examples using the same tools we used in example 1. Note that we used rigorous vector notation in example 1. But remember that vectors in 1D can also be represented by means of real numbers(recall chapter on vectors). The origin means the real number 0, forward direction means direction of x=1, backward direction means x=-1. Hence, we can represent the vector quantities like displacement, velocity, and acceleration by means of real numbers and drop the unit vector î. So velocity of -5 m/s means that the particle has speed of 5m/s in the backward direction. And acceleration of -2 m/s2 means that the particle’s speed is decreasing at a rate of 2 m/s. Example : A particle starts from rest at t=0 and moves along the straight line according to the law v(t)=u.ekx(t) where k and u>0 are constants, and v(t) is the velocity of particle at t and x(t) is it’s displacement at time t. Find the total distance traveled, speed, and acceleration at time t. Also comment on these quantities as a function of time for different possible values of k. Since, the motion is along a straight line, all the vectors can be represented in terms of real numbers. Thus, the velocity of particle at time t is nothing but v(t) and it’s position vector at time t is x(t), which may be positive or negative. Thus, where C is some constant to be determined by boundary conditions. When t=0, we have x=0 and so substituting t=0 and x=0 we get C=-1/k. Thus, And similarly, Do you find anything surprising in this kind of motion, especially when k>0 ? Yes, when t?1/k.u, x(t)??, v(t)??, a(t)?? when k>0 which means that particle leaves the space at t=1/ku. The following graphs illustrate this phenomenon. 
When k<0, then distance, speed, and acceleration can be rewritten as: In this case, the particle goes all along till infinity with time. As t??, it’s velocity and acceleration both go to 0. The graph below illustrates this: Note that we have plotted all the three curves, distance, velocity, acceleration curves in the same graph. This in general is not proper because the three quantities have different dimensions and so we can not use the same y-axis in the three cases. 
Example(Linear Simple Harmonic Motion): A particle moves along straight line according to law
a(t)=-k2x(t) where a(t) is acceleration and x(t) is it’s displacement at time t, and k is a constant. Assume that particle was at x=0 at t=0 and it’s velocity at t=0 was u. Find the smallest period of time ‘T’ after which it comes back to it’s position which it had at t=0.
Let us first first try to find an expression for x(t). So, we are given
which is a second order linear Ordinary Differential Equation whose general solution is given by
Also, we are given that x=0 at t=0 and so A=0. It’s velocity at any time can be found by differentiating x(t) and so
v(t)=-k (A.sin(k.t) – B.cos(k.t)) and because v(0) = u => B = k/u. Thus, 
which is a periodic function of t. It is equal to 0(at t=T) when sin kT = 0 or T=nπ/k, where n=0,1,2,…Hence, the smallest time after which the particle returns to x=0 is T= π/k.
This type of periodic motion is called Simple Harmonic Motion(SHM). The particle starts from x=0, goes till x=k/u, comes back to x=0 after time T= π/k, then goes in the negative direction and reaches x=-k/u, and again comes back to x=0 thus completing one cycle. The total time period of one complete oscillation is therefore 2T = 2π/k.
In the previous section, we saw that the vector quantities like displacement, velocity, acceleration as well as scalar quantities like distance, speed, magnitude of acceleration provide information about the motion of a particle. However in real life it may be too stringent as a requirement to have information about the velocity or acceleration at each time instant. It may not be possible to describe the velocity and acceleration vectors at each instant of time while the particle is/was in motion. So at times, average of these quantities over time may be sufficient to describe the motion. The average velocity(also called mean velocity) between t=T1 and t=T2 is defined to be the net displacement of the particle from t=T1 to t=T2 divided by T2 – T1. Similarly, average speed between t=T1 and t=T2 is equal to the total distance traveled by the particle from t=T1 to t=T2 divided by time duration T2 – T1.
Thus, we have
Hence, if we know that the average speed of the particle from t=0 to t=10s was 10m/s then we know that the particle was in a state of motion, infact we know that it covered a distance of 100m in 10 seconds. But we do not know what was it’s speed at t=1s or t=2s.
Example: Let us revisit example 1 and calculate the average velocity of the particle in motion in 1D with uniform acceleration ‘a’. The initial velocity of particle is given by u and the particle is at x=0 at t=0.
Average velocity from t=0 to t=T is the total displacement from t=0 to t=T divided by T. We know that
Example: Let us calculate average velocity and average speed of the particle under motion in example 2 from t=0 to t=T for the case k < 0.
We note that in this case v>0 at any point and time, and so speed which is |v| = v. So the average velocity and average speed in this case are equal. We calculated the distance traveled by the particle from t=0 to t=T, which is
And so, the average velocity (or average speed) is given by
Example: A particle traversed half the distance with speed v0. The remaining part of the distance was covered with speed v1 for half the time and with speed v2 for the other half of the time. Find the average speed of the particle over the whole time of motion.
The figure below illustrates equations of motion of particle.
If ‘d’ is the total distance covered by the particle and T is the duration it elapsed while in motion, then as per assumptions, the particle moves a distance of d/2 with constant speed v0 and so takes time d/2v0 to cover half the distance. If d1 and d2 are the distances it covers with speeds v1 and v2 respectively, then we have
Solving the above two equations, we get
and so, the total time of motion is given by
And thus the average speed is given by,
.So far, we have studied examples of rectilinear motion, i.e. motion along a straight line in 1D. Let us now see examples of motion in 2D, particularly angular motion. The simplest kind of angular motion in 2D we can think of is uniform circular motion.
IIT JEE, AIEEE, BITSAT 2011 Full Length Tests