Question:
The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point $S$ is at $4.333$ seconds. The total distance covered by the body in $6 \mathrm{~s}$ is :
Correct Option: , 4
Solution:
$\mathrm{OS}=4+\frac{1}{3}=\frac{13}{3}$
$\mathrm{SD}=2-\frac{1}{3}=\frac{5}{3}$
Area of $\mathrm{OABS}$ is $\mathrm{A}_{1}$
Area of $S C D$ is $A_{2}$
Distance $=\left|\mathrm{A}_{1}\right|+\left|\mathrm{A}_{2}\right|$
$\mathrm{A}_{1}=\frac{1}{2}\left[\frac{13}{3}+1\right] 4=\frac{32}{3}$
Distance $=\left|\mathrm{A}_{1}\right|+\left|\mathrm{A}_{2}\right|$
$=\frac{32}{3}+\frac{5}{3}$
$=\frac{37}{3}$