We consider all three blocks as a single system with total mass $M = m_A + m_B + m_C$. The net external force driving the system is the gravitational force on the hanging blocks, $F_{net} = (m_B + m_C)g$.

Using Newton's second law, $F_{net} = Ma$, we find the acceleration $a$ of the system:

$$ (m_B + m_C)g = (m_A + m_B + m_C)a $$ $$ a = \frac{(m_B + m_C)g}{m_A + m_B + m_C} $$

[Q1] To find the tension $T_{BC}$, we apply Newton's second law to block C alone. Taking the downward direction as positive:

$$ \Sigma F_C = m_C g - T_{BC} = m_C a $$ $$ T_{BC} = m_C(g - a) $$

Substituting the expression for $a$:

$$ T_{BC} = m_C \left( g - \frac{(m_B + m_C)g}{m_A + m_B + m_C} \right) = m_C g \left( \frac{(m_A + m_B + m_C) - (m_B + m_C)}{m_A + m_B + m_C} \right) $$ $$ T_{BC} = \frac{m_A m_C g}{m_A + m_B + m_C} $$

[Q2] Since the system starts from rest ($v_0 = 0$) and has constant acceleration $a$, we use the kinematic equation for displacement:

$$ d = v_0 t + \frac{1}{2} a t^2 = \frac{1}{2} a t^2 $$

Substituting the expression for $a$:

$$ d = \frac{1}{2} \left( \frac{(m_B + m_C)g}{m_A + m_B + m_C} \right) t^2 $$