(a) It is clear from the graph that $OQ > OP$, soA lives closer to the school than B. (b) The position-time $(x-t)$ graph of A starts fromthe origin, so $x = 0, t = 0$ for A while the $(x-t)$ graphof B starts from C which shows that B starts laterthan A after a time interval $OC.$ So $A$ starts fromschool $(O)$ earlier than B. (c) The speed is represented by the slope or steepness of the $(x-t)$ graph.More steeper the graph, more will be the speed, so faster will be the childhaving steeper graph. As the $(x-t)$ graph of B is steeper than the $(x-t)$ graphof A, so we conclude that B walks faster than A.(d) Corresponding to P and Q, the value of t from $$(x-t) graphs for A andB is same i.e. OE. So both A and B reach home at the same time. < br>(e) As the$(x-t) $ graphs for A and B intersect each other at one point i.e.D and B starts from the school later, so B overtakes A on the road onlyonce.