Since the system (man + trolley) is initially at rest and there is no external force in the horizontal direction, the total linear momentum of the system is conserved .

1. Define Variables: Mass of man ($m$) = $80\text{ kg}$ Mass of trolley ($M$) = $320\text{ kg}$ Velocity of man relative to trolley ($v_{rel}$) = $1\text{ m/s}$ Let $v_T$ be the velocity of the trolley relative to the ground.

• Let $v_m$ be the velocity of the man relative to the ground.

2. Relative Velocity Relation: $v_{rel} = v_m - v_T \implies v_m = v_T + v_{rel} = v_T + 1$.

3. Conservation of Momentum: $P_{initial} = 0$ $P_{final} = m v_m + M v_T = 0$ Substitute $v_m$: $80(v_T + 1) + 320 v_T = 0$ $80 v_T + 80 + 320 v_T = 0$ $400 v_T = -80 \implies v_T = -0.2\text{ m/s}$.

4. Calculate Man's Velocity w.r.t. Ground: $v_m = -0.2 + 1 = 0.8\text{ m/s}$.

5. Calculate Displacement: Displacement = $v_m \times t = 0.8\text{ m/s} \times 4\text{ s} = 3.2\text{ m}$.