Kinematic Equations for Uniformly Accelerated Motion

Various representations are used to represented motion of any object like pictorial representation, graphs, and verbal representation etc. The branch of mechanics which studied the motion of an object is called kinematics. It describes the concept of motion with graphical representation of its basic term like displacement, velocity, speed, acceleration etc.  And motion equation which gives the relation between these basic terms

Accelerated Motion: The velocity of a vehicle on a road is never uniform. Its velocity Increases and Decreases at random. As the velocity of the vehicle on the road keeps on changing, it is said to have Accelerated Motion.

The time rate of change of velocity of an object is called Acceleration of the object.

Uniformly Accelerated motion: Acceleration is the change in velocity in each unit of time. In case, the change in velocity each unit of time is constant, the object is said to be moving with constant acceleration and such motion is called “Uniformly accelerated motion”.

Non – Uniformly Accelerated motion: If the change in velocity in each unit of time is not constant, the object is said to be moving with variable acceleration and such a motion is called “non – uniformly accelerated motion”.

In General the motion of a vehicle on a road is non – uniformly accelerated as it speeds up and slows down at random.

Kinematic Equations: These Kinematic equations are used to detect the unknown value or variable with the use of known information of variables. These equations describe motion either at constant velocity of at constant acceleration.

First Equation of motion: Consider a particle moving along a straight line with uniform acceleration ‘a’. At t = 0, let the particle be at A and u be its Initial velocity and when t = t, v be its final velocity.First Equation of Motion\(Acceleration=\frac{change\,\,in\,\,velocity}{time}=\frac{v-u}{t}\)

\(a=\frac{v-u}{t}\)

v – u = at

v = u + at

Second equation of motion:

Average Velocity = \(\frac{Total\,distances\,travelled}{Total\,time\,taken}\)      

Average Velocity = \(\frac{S}{t}\)   (1)

Average velocity can also be written as \(\frac{v+u}{2}\)

Average velocity = \(\frac{v+u}{2}\)   … (2)

From equations (1) and (2):

\(\frac{S}{t}=\frac{v+u}{2}\)   … (3)

The first equation of motion is v = u + at

Substituting the value of v in equation (3),

We get

\(\frac{S}{t}=\frac{u+u+at}{2}=\frac{(u+u+at)t}{2}\)

\(=\frac{(2u+at)t}{2}\)

\(=\frac{2ut+a{{t}^{2}}}{2}\)

\(=\frac{2ut}{2}+\frac{a{{t}^{2}}}{2}\)

S = ut + ½ at²

Third equation of motion: The first equation of motion is v = u + at

v – u = at   … (1)

Average velocity = \(\frac{S}{t}\)   … (2)

Average velocity = \(\frac{v+u}{2}\)   … (3)

From equations (2) and (3) we get

\(\frac{v+u}{2}\,=\,\frac{S}{t}\)   … (4)

 Multiplying equations (1) and (4) we get

v² – u² = 2as

(v – u) (v + u) = 2as                              [∵ (a – b) (a + b) = a² – b²]

v² – u² = 2as

Example: A car is moving along a straight highway with speed of 126km/hr is brought to a stop within a distance of 200m then how long did it take for the car to go?

Sol: Here, distance covered, S = 200m

Initial speed of the car, u = 126 km/hr

= 126 x (1000 m) x (60 x 60 sec)¯¹

= 35 m/sec

Final speed of the car, v = 0

Let a be the Retardation of the car

We know that:

v² – u² = 2as

(0)² – (35)² = 2a x 200

\(a\,=\,-\,\frac{{{(35)}^{2}}}{2\times 200}=3.06m/\sec \)

Now,

v = u + at (or) \(0\,=\,35\,+\,\left( -\,\frac{49}{16} \right)t\)

\(t\,=\,\frac{35\times 16}{49}=\frac{80}{7}=11.43\,\sec \).