A particle moving along a straight line with uniform acceleration a. Let its velocity changes from u to v in an interval t. then the distance covered in t^th second is

Given that a particle is moving in straight line

Uniform acceleration =a

Initial velocity = u

Final velocity = v

Time taken = t

We get v = u + at

The distance travel by it in $t^{th}$ second

$$\begin{gathered} s_t=? \\ We know s_n=u+a\left(n-\frac{1}{2}\right) \\ {s}_t=u+a\left(t-\frac{1}{2}\right) \\ As we know u = v – at \\ {s}_t=v-at+at-\frac{a}{2} \\ \therefore s_t=v-\frac{a}{2} \\ so s_t=v-\frac{a}{2}\left(or\right)u+a\left(t-\frac{1}{2}\right) \end{gathered}$$