**Fifth equation**

Let the initial velocity of an object moving with uniform acceleration ‘a’ be v_{0}. We need to find the distance traveled by the object at a particular second. Let the distance traveled at t-th second be s_{t}. It is equal to the difference between the distances traveled by the object in time t second and (t – 1) second.

s_{t} = distance traveled in t sec — distance traveled in (t – 1) sec.

= **(v _{0}t + ½ at^{2}) – {v_{0 }(t-1) + ½ a (t-1)^{2}}** (from forth equation)

= v_{0}t + ½ at^{2} – {v_{0}t – v_{0} + ½ a (t^{2} – 2t + 1)}

= v_{0}t + ½ at^{2} – {v_{0}t – v_{0} + ½ a t^{2} – at + ½ a)}

= v_{0}t + ½ at^{2} – v_{0}t + v_{0} – ½ a t^{2} + at – ½ a

= v_{0} + at – ½ a = v_{0} + ½ a (2t – 1)

so, **s _{t} = v_{0} + ½ a (2t – 1)** … … … (1)

If the object start from rest, then v_{0} = 0;

then, s_{t} = **½ a (2t – 1)**

When the object moves with uniform retardation, equation (1) becomes;

**s _{t} = v_{0} – ½ a (2t – 1)**