Kinematics is the domain of Mechanics that deals with simple translational motion. In this domain, we simply discuss the relationships among displacement, velocity and acceleration; without considering the cause of the motion.
We know that velocity is the rate of change of displacement:
v = dx/dt
Hence, assuming the velocity stays constant(uniform motion), the net displacement(x) of an object would be the product of its velocity and the time taken to cover that distance. Hence, if a car is moving on a straight road with constant velocity of 100 km/hr, it would have covered 50 kilometers in half an hour. Since it is moving in a straight line, its net displacement is 50 km from its starting point. Therefore, in general we can say that
x’ = x0 + v × t
where x’ is the final position, x0 is the initial position and v is the constant velocity. Both x’ and v are vectors. Note that we could have arrived at this formula by integrating the equation (v dt = dx) and setting the constant term to x0.
Graphically, we can describe it as
It is all very straightforward and intuitive. However, in practice you need to start the car first, accelerate all the way from 0 km/hr to 100 km/hr and then only you can use the constant velocity formula above. In order to study this kind of motion, you need to take acceleration into account.
Acceleration is the rate of change of velocity:
a = dv/dt
It is similar in form to the displacement-velocity relation above. Let me take the liberty to simply integrate it rather than go into the full discussion from the point of view of physics, in order to save time (You can learn more about the physics behind this here):
∫a dt = ∫dv
If we assume constant acceleration, a comes out of the integration in the left hand side:
a∫dt = ∫dv
⇒ at = v – v0
v = v0 + at
It is easy to see that the constant of integration should be the initial velocity, v0.
Let’s denote this graphically:
In order to calculate displacement, we replace v by dx/dt:
⇒ dx/dt = at + v0
⇒ dx = (at + v0) dt
∫dx = ∫(at + v0) dt
x = (1/2)at2 + v0t
This equation enables us to plot the displacement-time plot directly, even with varying velocity:
In the uniformly accelerated motion with acceleration and displacement in the same direction, you get the curve like (3): concave up curve because displacement is constantly increasing function of time.
Let’s see one last example to understand the graphical representation of these concepts. Let’s say a particle starts from rest, has a uniform acceleration of 3 m/s2 for the next 8 seconds. Then suddenly the acceleration drops to zero(instantaneously) and stays zero for the next 6 seconds, and finally it decelerates with 2 m/s2. We can represent this motion graphically as follows:
It can be seen that the concave up region in plot 4 corresponds to the positive slope line segment in plot 5(from t = 0 to t= 8 seconds), the constant slope region in plot 4 corresponds to the straight horizontal line segment in plot 5(t= 8 to t = 14 seconds) and finally, the concave down region(when the displacement is increasing but more and more slowly) in plot 4 corresponds to the negative slope line segment in plot 5, until finally the velocity reaches zero and particle comes to rest.
Another important relationship can be derived between acceleration and displacement which does not involve time:
we know that
v = v0+at
⇒ t = (v-v0)/a
putting this in equation 2
x = v0(v-v0)/a +(1/2) a((v-v0)/a)2
v2 = v02 +2ax
This is the third equation of Kinematics.
In conclusion, motion of an object can be characterized by its position, velocity and acceleration. If you know the three variables, you can easily describe(and predict) particle trajectory using the three equations of Kinematics.