Thank you for reporting, we will resolve it shortly
Q.
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 $6\, s$ is :
$OS =4+\frac{1}{3}=\frac{13}{3}$
$SD =2-\frac{1}{3}=\frac{5}{3}$
Area of OABS is $A_1$
Area of $SCD$ is $A _{2}$
Distance $=\left| A _{1}\right|+\left| A _{2}\right|$
$A _{1}=\frac{1}{2}\left[\frac{13}{3}+1\right] 4=\frac{32}{3}$
$A _{2}=\frac{1}{2} \times \frac{5}{3} \times 2=\frac{5}{3}$
Distance $=\left| A _{1}\right|+\left| A _{2}\right|$
$=\frac{32}{3}+\frac{5}{3}=\frac{37}{3}$
