Consider an object moving in a straight line with uniform or constant acceleration ‘a’. 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) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
a = \(\frac{dv}{dt}\) or dv = a dt
Integrating both sides with the condition that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,
Displacement – time relation:
(II) The velocity of the body is given by the first derivative of the displacement with respect to time.
v = \(\frac{ds}{dt}\) or ds = vdt
and since v = u + at,
We get ds = (u + at)dt
Assume that initially at time t = 0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time-independent, we have
Velocity – displacement relation
(III) The acceleration is given by the first derivative of velocity with respect to time.
a = \(\frac{dv}{dt}\) = \(\frac{dv}{ds}\)\(\frac{ds}{dt}\) = \(\frac{dv}{ds}\) v [since ds/dt = v] where s is displacement traversed.
This is rewritten as a = \(\frac{1}{2}\)\(\frac{dv^2}{s}\)or ds = \(\frac{1}{2a}\) d(v2)
Integrating the above equation, using the fact when the velocity changes from u2 to v2 displacement changes from u2 to v2 , we get
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 + \(\frac{1}{2}\) (v - u)t
s =\(\frac{(u +v)t}{2}\)
