Question

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

Answer 1

At t= n sec, distance traveled by object is

$s_n=u\left(n\right)+\frac{1}{2}a{\left(n\right)}^2$

In (n-1) sec, distance traveled by particle is

$s_{(n-1)}=u(n-1)+\frac{1}{2}a{(n-1)}^2$

Therefore distance covered in $n^{th}$ sec, is

$s_n-s_{(n-1)}=un+\frac{1}{2}a{n}^2-\left[u\right(n-1)+\frac{1}{2}a{(n-1)}^2]=un+\frac{1}{2}a{n}^2-[un-u+\frac{1}{2}a(n^2-2n+1\left)\right]$

$=u+an-\frac{1}{2}a$

$\therefore s_{nth}=u+\frac{1}{2}a[2n-1]$