The distance-time graph of two trains is given below. The trains start simultaneously in the same direction.
$(i)$ How much ahead of $A$ is $B$ when the motion starts?
$(ii)$ What is the speed of $B$?
$(iii)$ When and where will $A$ catch $B$?
$(iv)$ What is the difference between the speeds of $A$ and $B$?
$(v)$ Is the speed of either train uniform or non-uniform? Justify your answer.

(N/A) $(i)$ At $t = 0$,train $A$ is at $0 \ km$ and train $B$ is at $100 \ km$. Thus,$B$ is $100 \ km$ ahead of $A$.
$(ii)$ Speed of $B = \frac{\text{Distance}}{\text{Time}} = \frac{150 \ km - 100 \ km}{2 \ h - 0 \ h} = \frac{50 \ km}{2 \ h} = 25 \ km/h$.
$(iii)$ $A$ catches $B$ at the point of intersection $Q$,which corresponds to $2 \ h$ and $150 \ km$. So,$A$ catches $B$ after $2 \ h$ at a distance of $150 \ km$.
$(iv)$ Speed of $A = \frac{150 \ km - 0 \ km}{2 \ h - 0 \ h} = 75 \ km/h$. Speed of $B = 25 \ km/h$. The difference is $75 \ km/h - 25 \ km/h = 50 \ km/h$.
$(v)$ The speed of both trains is uniform because their distance-time graphs are straight lines,indicating a constant rate of change of distance with respect to time.