Kinematics is that what tells about the motion. It is a branch of classical mechanics that tells about the description of motion without dealing with the force that causes them. It shows the motion of objects using words, diagrams, numbers, graphs and equations. Kinematics equations tells about the use of equations to describe and represent the motion of objects.

Kinematics equations tells about the use of equations to describe and represent the motion of objects. There are many things existing in nature that tells about the object's motion like displacement, velocity, acceleration and time.
There are basically four kinematics equations
Here V_{o} is initial velocity, V is final velocity, a is acceleration, x_{o} is initial displacement and x is final displacement and t is the time taken.
The 2D kinematics equations gives the kinematics equations in 2 dimensions x and y. There are basically four kinematics equations
V = V_{o} + at
X  X_{o} = V_{o} t + $\frac{1}{2}$ at^{2}
V^{2} = V_{o}^{2} + 2a(X  X_{o})
X  X_{o} = $\frac{1}{2}$ (V_{o} + V) t
X  X_{o} = V_{o} t + $\frac{1}{2}$ at^{2}
V^{2} = V_{o}^{2} + 2a(X  X_{o})
X  X_{o} = $\frac{1}{2}$ (V_{o} + V) t
Here V_{o} is initial velocity, V is final velocity, a is acceleration, x_{o} is initial displacement and x is final displacement and t is the time taken.
In xdirection the kinematics equations are
V_{x} = V_{ox} + a_{x} t
X = X_{o} + V_{xo} t + $\frac{1}{2}$ a_{x} t^{2}
V_{x}^{2} = V_{xo}^{2} + 2a_{x}(X  X_{o})
In ydirection the kinematics equations areX = X_{o} + V_{xo} t + $\frac{1}{2}$ a_{x} t^{2}
V_{x}^{2} = V_{xo}^{2} + 2a_{x}(X  X_{o})
V_{y} = V_{yo} + a_{y} t
y = x_{o} + V_{yo} t + $\frac{1}{2}$ a_{y} t^{2}
V_{y}^{2} = V_{yo}^{2} + 2a_{y}(y  y_{o})
y = x_{o} + V_{yo} t + $\frac{1}{2}$ a_{y} t^{2}
V_{y}^{2} = V_{yo}^{2} + 2a_{y}(y  y_{o})
The word inverse kinematics itself states that its nothing but reverse of kinematics. You already know that kinematics is that what calculates the position of space of the end of the linked structure if the angles are given. Inverse kinematics tells about the angles the joints should have to achieve the end point.
Inverse Kinematics is a branch that tells about the the mathematical way to calculate the different angles of a joint object to achieve a certain configuration. It has wide applications in robotics to building the robots.
Lets compare kinetics versus kinematics:
Kinetics 
Kinematics  
1  It tells about the study of motion and the forces acting on it for the motion.  It focuses only on motion but not the forces that causes it. 
2  It has wide applications in designing automobiles 
It has the applications in studying the movement of celestial bodies and robotics. 
Solved Examples
Question 1: A balloon is raising vertically upwards with a velocity of 6 ms^{1}. A bag of marbles is released from it reaches the ground in 10s. Whats the height of the balloon when bag is released?
Solution:
Let h be the height of the balloon when the sand bag was released.
Initial velocity u = + 6 ms^{1},
Acceleration a = g =  10 ms^{2},
Distance covered s =  h,
Time taken t = 10 s
Distance traveled s = ut + $\frac{1}{2}$ at^{2}
 h = (6)10 + $\frac{1}{2}$ $\times$ (10) $\times$ (10 s)^{2}
= 60 + 0.5 $\times$ (1000)
= 60  500 =  440 m
$\therefore$ Height h = 440 m.
Solution:
Let h be the height of the balloon when the sand bag was released.
Initial velocity u = + 6 ms^{1},
Acceleration a = g =  10 ms^{2},
Distance covered s =  h,
Time taken t = 10 s
Distance traveled s = ut + $\frac{1}{2}$ at^{2}
 h = (6)10 + $\frac{1}{2}$ $\times$ (10) $\times$ (10 s)^{2}
= 60 + 0.5 $\times$ (1000)
= 60  500 =  440 m
$\therefore$ Height h = 440 m.
Question 2: A train moves in a straight track with a velocity of 40 kmph is brought to rest after covering a distance of 100 m. Calculate its retardation.
Solution:
Given: Initial velocity of train u = 40 kmph = 11.11 m/s,
Final velocity v = 0,
Distance traveled s = 100 m
Consider the equation v^{2 }= u^{2} + 2as
0 = (11.11)^{2} + 2 $\times$ a $\times$ 100 m
Acceleration a = $\frac{ 123.43}{200}$ =  0.6175 ms^{2}
$\therefore$ Retardation = 0.6175 m/s^{2}.
Solution:
Given: Initial velocity of train u = 40 kmph = 11.11 m/s,
Final velocity v = 0,
Distance traveled s = 100 m
Consider the equation v^{2 }= u^{2} + 2as
0 = (11.11)^{2} + 2 $\times$ a $\times$ 100 m
Acceleration a = $\frac{ 123.43}{200}$ =  0.6175 ms^{2}
$\therefore$ Retardation = 0.6175 m/s^{2}.