Question

The acceleration (in $m/s^2$) of movable pulley P is and block B is , if acceleration of block $A=1{\text{m/s}}^2$ downwards.

Answer 1

From the figure, we can say:
$SM+ON+RB=constant$
If block B is assumed to move up, pulley P will move down so that the string between them remains taught.
$\therefore v_P+v_P+(-v_B)=0$
$\Rightarrow 2v_P=v_B$
$\Rightarrow 2a_P=a_B$

Also,
$AC+DE+FN=constant$
$v_A+0+(-v_P)=0$
$\Rightarrow a_A=a_P$
$\Rightarrow a_P=1m/s^2$
$\Rightarrow a_B=2m/s^2$