pls do this

L = 12 m
v_o = 7 m/s

Friction will act on the cylinder in a backwards direction in its full magnitude.

ma₁=μmg
& mr²Î±/2=μmgr
& 2ma₂=μmg

now at some time t ,
v_cylinder=v₁
angular velocity=ω
v_plank=v₂
and lets assume that slipping process at time T

=> v₂=v₁-ωr
& v_o-μgT-2μgT=μgT/2
& T=2v_o/7μg

=> T=2 sec
We also know that in time T ,
cylinder travels distance=d₁=v_oT- a₁₂T²/2
where a₁₂=accelration of cylinder wrt plank

=>d₁=11m which is less than total length of plank(12m)

After time T,friction stops acting
=>both move with constant velocities
=> v₁=v_o-μgT=5 m/s
&& v₂=μgT/2 =1 m/s

=>Cylinder travels additional 1 m in time t
t= 1/v₁₂=1/4=0.25 sec

=> Total time taken=T+t=2.25 sec

A shorter method might exist ......[6]
but i think this is the best approach[1]