Q 11-02-083JEE MainJEE Main 2023 (11 Apr, Shift 1)Medium
From the $v$–$t$ graph shown, the ratio of distance to displacement in $25$ s of motion is
Answer: (C) $\dfrac53$
Areas: $0$–$5$ s: $25$; $5$–$10$ s: $50$; $10$–$15$ s: $75$; $15$–$20$ s: $50$; $20$–$25$ s: $-50$ (m).
Distance $=250\ \text{m}$, displacement $=150\ \text{m}$; ratio $=\dfrac{250}{150}=\dfrac53$.
Solution by Sreeraj P, M.Sc Physics