blocks

Let Force x be applied on A and 2x
y + 3 x = 15

Acceleration of the Lower block = 15 - 3x/10
Acceleration of the upper block = Acceleration of the Lower block
B:
2x/2 = 15 - 3x /10
10x = 15 - 3x
13x = 15
x = 15/13
This is not the possible max = 1m/s
Therefore Force of friction on B is 2N
Now
y + x + 2 = 15
y + x = 13
y = 13-x
Acceleration = 13-x/10
x= 13 - x/10
11x = 13
x = 13/11 N on A due to friction
Acceleration of A is 1.18m/s²
Acceleration of B is 1m/s²

S_a=1.18 * t²/2
S_b=1* t²/2
S_a - S_b = 3
0.18*t²/2 = 3
t² = 6/0.18
t² = 100/3
t = 10/√3 seconds

Am i rite?

I am posting after a long time

hello

i am getting 2√3 s .if correct tell me then i will post my solution.