JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Kinematics

Area under the $v$–$t$ graph gives displacement when signed and distance when each piece is taken positive. From $0$ to $5$ s the area is $\frac{1}{2}(5)(10)=25$ m; from $5$ to $10$ s it is $(5)(10)=50$ m; from $10$ to $15$ s it is $\frac{1}{2}(10+20)(5)=75$ m; and from $15$ s until the velocity reaches zero at $20$ s it is $\frac{1}{2}(5)(20)=50$ m, all positive, totalling $200$ m. From $20$ to $25$ s the velocity is negative with area $\frac{1}{2}(5)(20)=50$ m. The distance is $200+50=250$ m and the displacement is $200-50=150$ m, so the ratio of distance to displacement is $\frac{250}{150}=\frac{5}{3}$.