JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Kinematics

The distance covered equals the area enclosed between the velocity$-$time graph and the time axis. The motion splits into two parts. From $t=0$ to the point $S$ at $t=\frac{13}{3}$ s the velocity rises and the bounded region is a trapezium whose area is $\frac{1}{2}\left(\frac{13}{3}+1\right)\times 4=\frac{1}{2}\times \frac{16}{3}\times 4=\frac{32}{3}$ m. Beyond $S$, from $t=\frac{13}{3}$ s to $t=6$ s, the velocity falls linearly to zero over an interval of $\frac{5}{3}$ s, giving a triangle of area $\frac{1}{2}\times \frac{5}{3}\times 2=\frac{5}{3}$ m. Adding the two parts, the total distance travelled in $6$ s is $\frac{32}{3}+\frac{5}{3}=\frac{37}{3}$ m.