Two particles $A$ and $B$ start moving simultaneously along the line joining them in the same direction with acceleration of $1 \mathrm{m}/{\mathrm{s}}^2$ and $2 \mathrm{m}/{\mathrm{s}}^2$and speeds $3 \mathrm{m}/\mathrm{s}$ and $1 \mathrm{m}/\mathrm{s}$ respectively. Initially, $A$ is $10 \mathrm{m}$ behind $B$. What is the minimum distance between them?

At minimum distance, their velocities are same, 

$\therefore v_A=v_B$ 

 $u_A+a_{A}t=u_B+a_{3}t$  

$(3+t)=(1+2 t)$ 

$t=2 \mathrm{s}$
Minimum distance, $d_{\text{min }}=$ Initial distance extra displacement of $A$ upto this instant due to its greater speed
$$\begin{gathered} =10-\left(s_A-s_B\right) \\ =10+s_B-s_A \\ =10+\left(u_{B}t+\frac{1}{2}{a}_B{t}^2\right)-\left(u_{A}t+\frac{1}{2}{a}_A{t}^2\right) \\ =10+\left[\left(1\right)\left(2\right)+\frac{1}{2}\left(2\right)(2{)}^2\right]-\left[\left(3\right)\left(2\right)+\frac{1}{2}\left(1\right)(2{)}^2\right] \\ =8 \mathrm{m} \end{gathered}$$