Let the mass of the system be m
Initial velocity of bus u=V
Final velocity of bus v=0
Acceleration of the bus a=−Fm
Using v2−u2=2aS
∴ 0−V2=2×(−Fm)×x ⟹ V2=2Fxm ...........(1)
Now the mass of the system is increased by 25 %
∴ New mass of the system M=1.25m
Acceleration of bus a′=−F1.25m
Using v2−u2=2a′x′
∴ 0−V2=2×(−F1.25m)×x′ ⟹ V2=2Fx′1.25m .....(2)
From (1) and (2) we get 2Fx′1.25m =2Fxm
⟹ x′=1.25x