Motion in a Straight Line

Question : 2

Total: 27

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 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).

Solution:

(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.
(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.

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