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The position-time ( $x-t)$ graphs for two children $\mathbf{A}$ and $\mathbf{B}$ returning from their school $\mathrm{O}$ 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).

(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:
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Verified Answer
(a) Because $\mathrm{OP} < \mathrm{OQ}$, hence A lives closer to the school than $B$.
(b) When $x=0, t=0$ for $A$, while $t \neq 0$ for $\mathrm{B}$. Therefore A starts from the school earlier than B.
(c) B walks faster, since the slope of B is more than the slope of $A$.
(d) A and B reach home at the same time.
(e) B overtakes A once during the journey.
(b) When $x=0, t=0$ for $A$, while $t \neq 0$ for $\mathrm{B}$. Therefore A starts from the school earlier than B.
(c) B walks faster, since the slope of B is more than the slope of $A$.
(d) A and B reach home at the same time.
(e) B overtakes A once during the journey.
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