Derive the kinematic equations of motion for constant acceleration.

Explanation

Consider an object moving in a straight line with uniform or constant acceleration `alpha`.
Let u be the velocity of the object at time t = 0, and v be velocity of the body at a later time t.
Velocity - time relation:
(i) he acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
`a=(dv)/(dt)ordv=adt`
Integrating both sides with the condition that as time changes from 0 to 1, the velocity changes from u to v. For the constant acceleration,
`int_(u)^(v)dv=int_(0)^(t)adtimplies[v]_(u)^(v)=a[t]_(0)^(t)`
`v-u=at(or)v=u+at`
Displacement - time relation
(ii) The velocity of the body is given by the first drivative of the displacement with respect to time.
`v=(ds)/(dt)ords=vdt`
and since v=u+at,
We get ds =(u+at)dt
Assume that initially at time 1 = 0, the particle started from the origin. At a later time i, the particle displacement is s. Further assuming that acceleration is time-independent, we have
`int_(0)^(t)ds=int_(0)^(t)udt+int_(0)^(t)atdtors=ut+(1)/(2)at^(2)" "......(2)`
Velocity - displacement relation (iii) The acceleration is given by the first derivative of velocity with respect to time.
`a=(dv)/(dt)=(dv)/(ds)(ds)/(dt)=(dv)/(ds)v` [since ds//dt=v] where s is displacement traversed.
This is rewritten as `a=(1)/(2)(d)/(ds)(v^(2))ords=(1)/(2a)d(v^(2))`
Integrating the above equation, using the fact when the velocity changes from u to v, displacement changes from 0 to s, we get
`int_(0)^(t)ds=(v_(1))/(""^(=)2a)d(v^(2))`
`:.s=(1)/(2a)(v^(2)-u^(2))`
`:.v^(2)=u^(2)+2as" "....(3)`
We can also derive the displacement s in terms of initial velocity u and final velocity v. From equation we can write,
at=v-u
Substitute this in equation, we get
`s=ut+(1)/(2)(v-u)t`
`s=((u+v)t)/(2)" ".......(4)`
The equations are called kinematic equations of motion.