A body starts from rest, what is the ratio of the distance travelled by the body during the 4^th and 3^rd $$s$$ ?

Solution :
Distance travelled by the body in $${n^{th}}$$ second is given by
$$\eqalign{ & {s_n} = u + \frac{a}{2}\left( {2n - 1} \right) \cr & {\text{Here,}}\,u = 0 \cr & \therefore {\text{For}}\,{4^{th}}\,s,\,{s_4} = \frac{a}{2}\left( {2 \times 4 - 1} \right) \cr & {\text{and}}\,{\text{For}}\,{3^{rd}}\,s,\,{s_3} = \frac{a}{2}\left( {2 \times 3 - 1} \right) \cr & {\text{Hence,}}\,\frac{{{s_4}}}{{{s_3}}} = \frac{{\left( {2 \times 4 - 1} \right)}}{{\left( {2 \times 3 - 1} \right)}} = \frac{7}{5} \cr} $$