To find out the final equation for constant acceleration, the following formulations are:

S = S_{0}

t =0

v =v_{0}

At point (1.2), the equation stands at dv = a dt

On integrating this equation, the calculation is,

An alternative to this calculation is v – v_{0 }= at

v =v_{0} +at

Again on integrating on the seconds instance, the calculation stands to be,

The next step in this is,

v2 =v_{02} +2a (S –S_{0})

From the equation (1.1),

dS =v dt

dS = (v_{0}+at) dt

so integration equation as per the above is,

S –S0 = v0 t +1/2 at2

S =S0 + v0 t +1/2 at2

P.S.

It is only in case of constant acceleration when equations (1.5), (1.6) and (1.7) are required to be applied.

**Links of Previous Main Topic:-**

- Introduction about distributed forces
- Area moments of inertia in rectangular and polar coordinates
- Mass moment of inertia introduction
- Work done by force
- Kinematics of particles
- Plane motion

**Links of Next Mechanical Engineering Topics:-**

- Analysis for non uniform variable acceleration
- Graphical representation x y plane graph
- Distance travelled in the second in constant acceleration
- Acceleration due to gravity
- Position vector velocity and acceleration
- Plane kinematics of rigid bodies introduction
- Combined motion of translation and rotation
- Rectilinear motion in kinetics of particles