The distance travelled by a particle starting from rest and moving with an acceleration $$\frac{4}{3}m{s^{ - 2}},$$  in the third second is:

Solution :
Distance travelled in the $$n$$^th second is given by
$$\eqalign{ & {t_n} = u + \frac{a}{2}\left( {2n - 1} \right) \cr & {\text{put}}\,u = 0,a = \frac{4}{3}m{s^{ - 2}},n = 3 \cr & \therefore d = 0 + \frac{4}{{3 \times 2}}\left( {2 \times 3 - 1} \right) = \frac{4}{6} \times 5 = \frac{{10}}{3}m \cr} $$