Figure shows a small block $A$ of mass m kept at the left end of a plank $B$ of mass $M=2m$ and length $l$. The system can slide on a horizontal road. The system is started towards right with the initial velocity $v$. The friction coefficients between the road and the plank are $1/2$ and that between the plank and the block is $1/4$. The time elapsed before the block separate from the plank is $t=x\sqrt{\frac{l}{3g}}$. Find the value of $x$.

As given block and plank both moves towards right with velocity v.
Given, friction coefficient between the block and plank, ${\mu }_1=\frac{1}{4}$
Friction coefficient between the road and plank, ${\mu }_2=\frac{1}{2}$
For block $A,N_1=mg$
So, friction force, $f_1={\mu }_1{N}_1=\frac{1}{4}mg$
Acceleration of block $A,a_A=\frac{f_1}{m}{a}_A=\frac{g}{4}$
Friction force on the plank due to motion of block, $f_1=\frac{1}{4}mg$
Friction force on the plank due to road, $f_2={\mu }_2{N}_2{f}_2={\mu }_2(N_1+Mg)=\frac{1}{2}\left(3mg\right)$
Acceleration of plank $B,$
$a_B=\frac{f_2-f_1}{2m}{a}_B=\frac{\frac{1}{2}\left(3m\right)g-\frac{1}{4}mg}{2m}=\frac{5g}{8}$
As the block and the plank moves with same velocity, $u_{rel}=0$
So, relative acceleration of the blocks,
$a_{rel}=\frac{5g}{8}-\frac{g}{4}=\frac{3g}{8}$
Distance travelled by block before it separate from plank, $=l$
$l=\frac{1}{2}\times \frac{3}{8}\times g{t}^2$
So, time elapsed before the block separate from the plank,
$t=4\sqrt{\frac{1}{3g}}x=4$