A car starts with a constant acceleration 2 ms^-2 at t=0. Two coins are released from the car at t=3 sec and t=4 sec Each coin takes exactly 1 sec to reach the ground. When coins hit the ground, they stick to it. The separation between the two coins on ground will be

at t=3s, the car has travelled through a distance of ${\mathrm{s}}_3=\frac{1}{2}2{\left(3\right)}^2=9\mathrm{m}$. And moving with a velocity of 6m/s.

So the coin will touch the ground at 9+6(1)=15m from the starting point.

Similarly at t=4s, the car has travelled through a distance of ${\mathrm{s}}_4=\frac{1}{2}2{\left(4\right)}^2=16\mathrm{m}$. And moving with a velocity of 8m/s.

So the second coin will touch the ground at 16+8(1)=24m from the starting point.

So the distance between the coins will be 9m.