The position-time $(x-t)$ graphs for two children $A$ and $B$ returning from their school $O$ to their homes $P$ and $Q$ respectively are shown in the figure. Choose the correct entries in the brackets below:
$(a)$ $(A/B)$ lives closer to the school than $(B/A)$
$(b)$ $(A/B)$ starts from the school earlier than $(B/A)$
$(c)$ $(A/B)$ walks faster than $(B/A)$
$(d)$ $A$ and $B$ reach home at the (same/different) time
$(e)$ $(A/B)$ overtakes $(B/A)$ on the road (once/twice)

(A-D) As $OP < OQ$,$A$ lives closer to the school than $B$.
$(b)$ For $x=0$,$t=0$ for $A$,while $t$ has a finite positive value for $B$. Therefore,$A$ starts from the school earlier than $B$.
$(c)$ Since the velocity is equal to the slope of the $x-t$ graph in the case of uniform motion and the slope of the $x-t$ graph for $B$ is greater than that for $A$,hence $B$ walks faster than $A$.
$(d)$ It is clear from the given graph that both $A$ and $B$ reach their respective homes at the same time.
$(e)$ $B$ starts later than $A$ and his/her speed is greater than that of $A$. From the graph,it is clear that $B$ overtakes $A$ only once on the road.