- Home
- Standard 11
- Physics
The position-time $(x-t)$ graphs for two children $A$ and $B$ returning from their school $O$ to their homes $\mathrm{P}$ and $\mathrm{Q}$ 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 $(B/A)$
$(d)$ $\mathrm{A}$ and $\mathrm{B}$ reach home at the (same/different) time
$(e)$ $(\mathrm{A} / \mathrm{B})$ overtakes $(\mathrm{B} / \mathrm{A})$ on the road (once/twice)
Solution
(a) As $O P < OQ$, A lives closer to the school than $B$
(b) For $x=0, t=0$ for $A ;$ while $t$ has some finite value for $B .$ Therefore, $A$ starts from the school earlier than $B$.
(c) since the velocity is equal to slope of $x -t$ graph in case of uniform motion and slope of $x-t$ graph for $B$ is greater that 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) $B$ moves 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.
