A particle with initial velocity $v_0$ moves with constant acceleration in a straight line. Find the distance travelled in $n^{th}$ second.

For given particle, distance travelled in $n^{\text {th }}$ second is, $d=$

(distance travelled in $n$ seconds) - (distance travelled in $(n-1)$ second)

$=\left(v_{0} n+1 / 2 a n^{2}\right)-\left(v_{0}(n-1)+\frac{1}{2} a(n-1)^{2}\right)\left(\text { From equation } d=v_{0} t+\frac{1}{2} a t^{2}\right)$

$=\left(v_{0} n+\frac{1}{2} a n^{2}\right)-\left(v_{0} n-v_{0}+a / 2\left(n^{2}-2 n+1\right)\right)$

$=\left(v_{0} n+\frac{1}{2} a n^{2}-v_{0} n+v_{0}-\frac{1}{2} a n^{2}+a n-\frac{a}{2}\right)$

$=v_{0}+a_{n}-\frac{a}{2}$

$\therefore d=v_{0}+\frac{a}{2}(2 n-1)$