Proportion: Train Speed and Compartments

Reduction in speed $\Delta v = k \sqrt{N}$. When $N = 9$, $\Delta v = 42 - 24 = 18 \text{ kmph}$. So $18 = k \sqrt{9} = 3k \implies k = 6$. To stop the engine completely ($v = 0$), $\Delta v = 42 \implies 6 \sqrt{N} = 42 \implies \sqrt{N} = 7 \implies N = 49$. Since at $N=49$ the train stops, the maximum number of compartments it can carry (move with) is $49 - 1 = 48$.