The distance travelled by a particle starting from rest and moving...

Distance travelled by the particle in $n$th second is $S_{n t h}=u+\frac{1}{2} a(2 n-1)$ where $u$ is initial speed and $a$ is acceleration of the particle. Here, $n=3, u=0, a=\frac{4}{3} \mathrm{~m} / \mathrm{s}^2$$\begin{aligned} \therefore S_{3 \text{ rd }} & =0+\frac{1}{2} \times \frac{4}{3} \times(2 \times 3-1) \\ & =\frac{4}{6} \times 5 \\ & =\frac{10}{3} \mathrm{~m} \end{aligned}$$$Alternative : Distance travelled in the 3rd second = distance travelled in$3 \mathrm{~s}$- distance travelled in$2 \mathrm{~s}$As,$u=0$,$S_{(3 \mathrm{rd} s)}=\frac{1}{2} a \cdot 3^2-\frac{1}{2} a \cdot 2^2=\frac{1}{2} \cdot a \cdot 5$Given$a=\frac{4}{3} \mathrm{~ms}^{-2} \therefore S_{(3 \mathrm{rd} s)}=\frac{1}{2} \times \frac{4}{3} \times 5=\frac{10}{3} \mathrm{~m}$