For shown situation draw a graph showing acceleration of A and B on y-axis and time on x-axis `(g=10ms^(-2))`

Explanation

Limiting fricition between `A & B, f_(1)=mum_(A)g=((1)/(2))(2)(10)=10N`
Block B moves due to friction only. So maximum acceleration of
`B, a_("max")=(f_(1))/(m_(B))=(10)/(4)-2.5ms^(-2)`. So both the blokcs move together till the common acceleration becomes `2.5ms^(-2)`, after that acceleration of B will become constant while that of A will go on increasing. Slipping will starts between `A&B` at `2.5 ms^(-2) rArr2.5=(F)/(m_(A)+m_(B))=(3t)/(6)rArrt=5s`
Hence for `le5s,a_(A)=a_(B)=(F)/(m_(A)+m_(B))=(3t)/(6)=(1)/(2)`
and for `tgt5sa_(B)=2.5ms^(-2)a_(A)=(F-f_(1))/(m_(A))=(3t-10)/(2)=(3)/(2)t-5`